SciHawks

AP Chem- took the most challenging test of the year: a combo buffer/titration/salt equilibria exam. The number of steps required to get the answer and the level of rational thought involved on this test is unparalleled compared to any other test. Only the Bonding/Intermolecular Forces test involves as much knowledge of chemistry.



Bio 6/7- we illustrated how to draw out a pedigree chart; pedigree charts are useful for tracking and/or inferring the genotypes that exist in members of a family tree. The charts are also used to predict the probability that a given offspring will inherit a given trait from her/his parents based on the parents' and grandparents' phenotypes. Be careful, some of these charts show the genotype of the family members but others merely show the phenotypes from which the genotypes must be inferred. It may take some trial and error to deduce/infer the genotypes of the family tree members!
We noted that it is easier/more logical to start from the part of the chart that has the RECESSIVE allele expressed in the phenotype, i.e. begin anywhere in the chart that shows the homozygous recessive genotype.
We continued the human trait pedigree lab and finished up any questions on the prior two labs.

Bio 8- we illustrated how to draw out a pedigree chart; pedigree charts are useful for tracking and/or inferring the genotypes that exist in members of a family tree. The charts are also used to predict the probability that a given offspring will inherit a given trait from her/his parents based on the parents' and grandparents' phenotypes. Be careful, some of these charts show the genotype of the family members but others merely show the phenotypes from which the genotypes must be inferred. It may take some trial and error to deduce/infer the genotypes of the family tree members!
We noted that it is easier/more logical to start from the part of the chart that has the RECESSIVE allele expressed in the phenotype, i.e. begin anywhere in the chart that shows the homozygous recessive genotype.

AP Chem- THE test is tomorrow; the one that I have been crowing about all year. Please do as many practice problems as possible so that you do not run out of time or make any of the myriad possible errors (as pointed out in class) on tomorrow's test.

Discussed the 3rd Law of Thermodynamics, which allows us to calculate the ACTUAL absolute entropies of substances because a perfect crystal of a substance at zero Kelvin actually has zero absolute entropy (not an arbitrarily chosen reference value).
We then mimicked J. Willard Gibbs, the grandaddy of Thermo, in the derivation of Gibbs Free Energy from the 2nd Law of Thermo; we distilled that formula into one involving only measurable changes in a system's enthalpy and entropy. We discussed the FOUR permutations of delta H and delta S, showing the conditions that would net a negative delta G (spontaneous reaction or process) or positive delta G (not spontaneous reaction or process).
We discussed the qualitative and quantitative meanings of delta G as well as the definition of STANDARD Gibbs free energy conditions.
We then used that definition to apply the equivalent of Hess's Law to solve for the standard CHANGE in Gibbs Free Energy for a given reaction (see notes for the answer to today's final question).

Bio 6- discussed the notation and deductive reasoning involved in determining GENOTYPES from a series of generations on a PEDIGREE chart.
We saw how to spot a "sex-linked" trait: GIVEN normal PHENOTYPE parents, we notice that ONLY the male offspring have the RECESSIVE allele caused PHENOTYPE show up. This is because the Y chromosome that causes an offspring to be male does not contain the complementary allele found on the X chromosome. So, if the mother passes on a recessive allele on that X chromosome to her son, the son MUST display the phenotype caused by that recessive allele.

Bio 7/8- discussed the notation and deductive reasoning involved in determining GENOTYPES from a series of generations on a PEDIGREE chart.
We saw how to spot a "sex-linked" trait: GIVEN normal PHENOTYPE parents, we notice that ONLY the male offspring have the RECESSIVE allele caused PHENOTYPE show up. This is because the Y chromosome that causes an offspring to be male does not contain the complementary allele found on the X chromosome. So, if the mother passes on a recessive allele on that X chromosome to her son, the son MUST display the phenotype caused by that recessive allele.
We then finished up the dominant/recessive human pedigree lab.

AP Chem- explained four factors that can cause a change in entropy: change of phase, change in the number of mole of substance(s), change in temperature of substance, and change of the substance to a more or less complex substance.
We then discussed the 2nd Law of Thermodynamics and used it to calculate the change in entropy of the universe for the Haber process (making ammonia).
We then began a Ksp and Titration review sheet to which I will post answers this weekend (and I should have those podcast videos done also; update- the buffer podcast video is now available).

Bio 6/7- we did several problems showing expected blood types as a result of crosses between various pairs of blood genotypes.
We then discussed sex-linked traits, which are any traits determined by genes that are located on the X-chromosome of the sex chromosomes. Females have two X chromosomes whereas males have one X and one Y chromosome. Sex-linked phenotype ratios MAY differ from other chromosome phenotypes due to the fact that the Y chromosome does not carry many genes because it is smaller/consists of less DNA.
We saw that the genetic sex of offspring is determined by the male parent; half of all male gametes contain an X chromosome and the other half of all male gametes contain a Y chromosome. When X chromosome sperm cells fertilize an egg, a female is produced but when a Y chromosome sperm cell fertilizes an egg, a male is produced.
We then saw that the evidence for a sex-linked trait is that, if there is a cross between a "normal" male and a heterozygous-genotype with "normal"-phenotype female, the trait phenotype is EXPRESSED/SEEN exclusively in males.
We then finished up our last two labs. The write-ups for these labs are due on Tuesday.

Bio 8-we did several problems showing expected blood types as a result of crosses between various pairs of blood genotypes.
We then discussed sex-linked traits, which are any traits determined by genes that are located on the X-chromosome of the sex chromosomes. Females have two X chromosomes whereas males have one X and one Y chromosome. Sex-linked phenotype ratios MAY differ from other chromosome phenotypes due to the fact that the Y chromosome does not carry many genes because it is smaller/consists of less DNA.
We saw that the genetic sex of offspring is determined by the male parent; half of all male gametes contain an X chromosome and the other half of all male gametes contain a Y chromosome. When X chromosome sperm cells fertilize an egg, a female is produced but when a Y chromosome sperm cell fertilizes an egg, a male is produced.
We then saw that the evidence for a sex-linked trait is that, if there is a cross between a "normal" male and a heterozygous-genotype with "normal"-phenotype female, the trait phenotype is EXPRESSED/SEEN exclusively in males.

Note: I'm making up Thursday morning's extra help with a Friday morning (03/27) extra help session from 8:15 to 9 AM in Room 308. See you there.
AP Chem- began our Thermo unit by reviewing the main Laws of the Universe: the 0th and 1st Laws of Thermodynamics i.e. thermal energy always flows (net) from higher temperature matter to lower temperature matter and that energy can neither be created nor destroyed, respectively.
We then discussed the mathematical and physical meaning of entropy and its units of measurement, Joules per K; molar entropy is in Joules per mole-K.
We reasoned entropy changes for various processes such as melting, freezing, sublimation, etc.

The Ksp/Salt Solubility/Buffer-Titration exam will be given next Tuesday.

Bio 6- we discussed the two possible crosses between an organism that shows the DOMINANT phenotype (either the TT genotype or the Tt genotype will cause the dominant phenotype of tallness in pea plants) and an organism that shows the RECESSIVE phenotype (which MUST be the tt genotype, only).By analyzing the percentage of offspring with the dominant or the recessive phenotype, we could infer/conclude the GENOTYPES of the parents even though we cannot physically see the genes because they are at the molecular size level.We then saw that not all genes come in just the dominant and recessive forms.There are two distinct types of INTERMEDIATE DOMINANCE:1. CODOMINANCE: occurs when BOTH alleles are expressed fully and SEPARATELY (NO BLENDING!) phenotypically. The notation used is a capital letter for the trait with a SUPERSCRIPT of the actual allele type; for example, a roan horse has both red AND white hairs, so the alleles are written CR and CW.2. INCOMPLETE DOMINANCE: occurs when BOTH alleles BLEND together to form a SINGLE intermediate trait. The notation is just a capital letter of each allele type; for example, a RED carnation (RR) crossed with a WHITE carnation (WW) will form PINK carnation (RW) offspring because both traits are expressed and blended together (NOT SEPARATE AS IN CODOMINANCE).

Bio 7/8-we discussed the two possible crosses between an organism that shows the DOMINANT phenotype (either the TT genotype or the Tt genotype will cause the dominant phenotype of tallness in pea plants) and an organism that shows the RECESSIVE phenotype (which MUST be the tt genotype, only).By analyzing the percentage of offspring with the dominant or the recessive phenotype, we could infer/conclude the GENOTYPES of the parents even though we cannot physically see the genes because they are at the molecular size level.We then saw that not all genes come in just the dominant and recessive forms.There are two distinct types of INTERMEDIATE DOMINANCE:1. CODOMINANCE: occurs when BOTH alleles are expressed fully and SEPARATELY (NO BLENDING!) phenotypically. The notation used is a capital letter for the trait with a SUPERSCRIPT of the actual allele type; for example, a roan horse has both red AND white hairs, so the alleles are written CR and CW.2. INCOMPLETE DOMINANCE: occurs when BOTH alleles BLEND together to form a SINGLE intermediate trait. The notation is just a capital letter of each allele type; for example, a RED carnation (RR) crossed with a WHITE carnation (WW) will form PINK carnation (RW) offspring because both traits are expressed and blended together (NOT SEPARATE AS IN CODOMINANCE).
We then began a lab on human pedigree charts based on dominant/recessive alleles for certain traits.

AP Chem- did another selective precipitation problem; this time, we added a cation, Ag+, in order to selectively precipitate either the chloride or the phosphate ion from solution.
We then did the typical calculations:
-which anion precipitates first
-what is the concentration of the precipitating ion JUST as the other ion is about to precipitate
-was there "complete" precipitation of the less soluble salt's anion at this point

We then discussed how and why certain salts (and hydroxide bases) are made more soluble in acidic solutions. For ANY salt that has anions that are strong enough to react with H+, i.e. ALL SALTS EXCEPT for chlorides, bromides, iodides, nitrates, and perchlorates, the loss of the anions will cause a Le Chatelier shift that results in more dissolving of the salt.

We introduced Thermodynamics, which we will complete next week.

Bio 6/7- Mendel's main observations are summarized in the Law of Dominance, the Law of Segregation and Recombination, and the Law of Independent Assortment.
We saw that the genes for certain traits can come in two or more forms/alleles. From Mendel's experiments, he saw that typically one allele DOMINATES (shows up in offsprings' phenotype) over the other (recessive) allele. He also showed, based on the percentage of offspring with a specific phenotype, that the alleles (within a given parent) segregate/separate as the sex cells are made; we now know that this occurs during anaphase II of meiosis as the chromatids, each carrying one allele per gene) are pulled apart and go to separate sex cells.
Alleles are then "recombined" when the diploid number of chromosomes is restored via fertilization of an ovum by a sperm to form a zygote.
We discussed the difference between GENOTYPE and PHENOTYPE. Genotype refers LITERALLY to the specific alleles that an individual has for a given trait.
For genes that have only two alleles, one dominant and one recessive, there are THREE possible GENOTYPES:
HOMOZYGOUS DOMINANT, e.g. TT
HETEROZYGOUS, e.g. Tt or tT (hetero means different; this type has two DIFFERENT alleles)
HOMOZYGOUS RECESSIVE, e.g. tt

A PHenotype is what you PHysically SEE expressed by the organism! In order to have the recessive alleles appear physically (expressed physically), an organism MUST have the homozygous RECESSIVE GENOTYPE, otherwise, BY DEFINITION, the dominant allele will be expressed in the phenotype INSTEAD of the recessive allele.
Therefore, a Tt or a TT GENOTYPE will be expressed as the "TALL" PHENOTYPE (you can SEE the tall characteristics even though you can't physically see the genes that cause the phenotype). However, a "tt" genotype will form/be expressed as the "SHORT" PHENOTYPE.
We discussed and showed, using Punnett squares, the cause of the genotype and phenotype ratios that are expected in the F1 and F2 generations for alleles that exhibit a dominant/recessive relationship. We also introduced the two types of intermediate dominance: codominance (both alleles separately are expressed) and incomplete dominance (an average blend of the two parent phenotypes is seen).
We then finished up questions from the past two labs. These writeups are due on Monday so that I can enter your lab grades.

Bio 8- His main observations are summarized in the Law of Dominance, the Law of Segregation and Recombination, and the Law of Independent Assortment.
We saw that the genes for certain traits can come in two or more forms/alleles. From Mendel's experiments, he saw that typically one allele DOMINATES (shows up in offsprings' phenotype) over the other (recessive) allele. He also showed, based on the percentage of offspring with a specific phenotype, that the alleles (within a given parent) segregate/separate as the sex cells are made; we now know that this occurs during anaphase II of meiosis as the chromatids, each carrying one allele per gene) are pulled apart and go to separate sex cells.
Alleles are then "recombined" when the diploid number of chromosomes is restored via fertilization of an ovum by a sperm to form a zygote.
We discussed the difference between GENOTYPE and PHENOTYPE. Genotype refers LITERALLY to the specific alleles that an individual has for a given trait.
For genes that have only two alleles, one dominant and one recessive, there are THREE possible GENOTYPES:
HOMOZYGOUS DOMINANT, e.g. TT
HETEROZYGOUS, e.g. Tt or tT (hetero means different; this type has two DIFFERENT alleles)
HOMOZYGOUS RECESSIVE, e.g. tt

A PHenotype is what you PHysically SEE expressed by the organism! In order to have the recessive alleles appear physically (expressed physically), an organism MUST have the homozygous RECESSIVE GENOTYPE, otherwise, BY DEFINITION, the dominant allele will be expressed in the phenotype INSTEAD of the recessive allele.
Therefore, a Tt or a TT GENOTYPE will be expressed as the "TALL" PHENOTYPE (you can SEE the tall characteristics even though you can't physically see the genes that cause the phenotype). However, a "tt" genotype will form/be expressed as the "SHORT" PHENOTYPE.
We discussed and showed, using Punnett squares, the cause of the genotype and phenotype ratios that are expected in the F1 and F2 generations for alleles that exhibit a dominant/recessive relationship.

AP Chem- ATTENTION: THERE WILL BE A DESCRIPTIVE CHEM QUIZ DURING WEDNESDAY'S CLASS.
There are a number of Ksp/Salt Equilibria files posted so that you can practice a lot for the next exam.
We continued our salt equilibrium problem by calculating the "percent precipitation", which is just the percent formula. We divide the final concentration of a given ion still in solution (from the ICE table) by the INITIAL concentration of that ion (JUST AFTER the solutions were mixed BUT unreacted!); we multiply that fraction by 100 to get the percentage.

We then did our first (of two) problems on "selective precipitation". These problems are very difficult especially IF you do not draw out the potential reactions that are occurring in solution!
The problem involves two "competing" equilibria; you will be asked to calculate whether the ions of one element can be SELECTIVELY removed/precipitated from solution WHILE the ions of another element MOSTLY/COMPLETELY remain in solution.
You first must find the TWO DIFFERENT concentrations of an ion (assuming two possible precipitates) that can be added to solution that will JUST begin to cause precipitation of each ion. Whichever REQUIRED concentration is LOWER will be for the LESS SOLUBLE PRECIPITATE.

THEN, at the HIGHER concentration of added ion to JUST BEGIN to cause precipitation of the MORE soluble salt, the concentration of the ion that has ALREADY been precipitating is calculated by using its precipitate's Ksp expression.

Finally, we calculate whether one of the ions is "completely precipitated" at the point at which the OTHER ion JUST are ABOUT TO precipitate.

Bio 6- we defined Mendel's Law of Dominance (due to the suppression of the translation of one allele by the other allele on a homologous chromosome) and Law of Segregation (really due to anaphase I and II of meiosis). We saw that gamete formation causes segregation of alleles that can be re-combined during fertilization so that traits that do not show up in a given generation (due to DOMINANT/recessive alleles) WILL show up in the next generation.
We showed the modern understanding of Mendel's laws, which are caused by what occurs to chromosomes/alleles during meiosis/gamete formation. We saw that Punnet squares really just show the TWO possible gametes formed with respect to a given trait and how those gametes combine to form offspring with a particular combination of alleles for a given trait.

Bio 7/8 - we finished discussing Mendel's scientific process that led to his discoveries of patterns of inheritance. We defined Mendel's Law of Dominance (due to the suppression of the translation of one allele by the other allele on a homologous chromosome) and Law of Segregation (really due to anaphase I and II of meiosis). We saw that gamete formation causes segregation of alleles that can be re-combined during fertilization so that traits that do not show up in a given generation (due to DOMINANT/recessive alleles) WILL show up in the next generation. We showed the modern understanding of Mendel's laws, which are caused by what occurs to chromosomes/alleles during meiosis/gamete formation. We saw that Punnet squares really just show the TWO possible gametes formed with respect to a given trait and how those gametes combine to form offspring with a particular combination of alleles for a given trait.

AP Chem- saw how to quickly compare the solubilities of various salts by comparing their Ksp values IF (!!!) the salts have the same SUBSCRIPTS in their empirical formulas i.e. Ksp expression EXPONENTS. If the salts have different empirical formulas, then a qualitative estimate cannot easily be calculated!
We then calculated the solubility of a salt by itself and then WITH a COMMON ION in solution; the result can be predicted from Le Chatelier's Rule. A common ion will shift equilibrium towards the undissolved precipitate. Thus, when a common ion is in solution, the SOLUBILITY of a sparingly soluble salt is depressed/lowered.
We then played the "will a precipitate form?" game by comparing the given Ksp with the current Qsp of the salt in solution. For precipitates that did form, we calculated the MOST DIFFICULT PROBLEM TYPE FOR THIS UNIT: the number of grams of precipitate formed AND the percent precipitation/ determination of "complete" precipitation.
For this part of the problem, we precipitated/use up ALL OF ONE OF THE IONS (SRFC TABLE!) and then, naturally, some/ a WEE BIT of the precipitate re-dissolves (ICE TABLE!).
The problem is that STOICHIOMETRY must be adhered to in these tables AND in the Ksp expressions!
It is important to DRAW OUT the ions in solution so that you can SEE what is going on; remember the ERRORS that were caused on the last test by not SEEING/DRAWING what was occurring.

Bio 6/7/8- we reviewed the Repro/Meiosis exam, discussing the various test-taking skills that should have been applied; these practices can ONLY HELP YOUR SCORE! This is INDISPUTABLE and is supported by a mountain of evidence and pure logic, NOT CONJECTURE OR OPINION! These skills (especially identifying and USING the key words/data in a question as well as DRAWING things out) should ALWAYS be used and practiced so that they are a natural/ not time-consuming way for you to take tests.
We then began our new unit on Genetics by discussing Mendel's scientific method that led to the discovery of some of the basic laws of heredity/genetics.

AP Chem- went through an example in which a salt's solubility was determined (a saturated solution must be formed); this value was then converted to the salt's molar solubility. A salt's molar solubility is used to find, by stoichiometry, the EQUILIBRIUM concentrations of its dissolved ions; these concentrations can then be plugged into the salt's Ksp expression in order to find its equilibrium constant at a given temperature.
We completed every practical permutation of empirical salt formulas with respect to Ksp and molar solubility, "s".
Next week, we will do the most difficult equilibrium problems in this course. Much practice is required.

Bio 6/7/8- took the Meiosis/Reproduction exam.

AP Chem- took the buffer/titration exam.
We will cover a lot of the sparingly soluble salt unit in one period tomorrow so that you can do some of those problems this weekend. I will be giving a test on this Ksp topic late next week, but not for a full double period.

Bio 6/7/8- I posted SEVEN review sheets on Blackboard to supplement your study for Friday's unit exam. Do them tonight and tomorrow before the test as part of your review.

We reviewed for the Reproduction/Development/Meiosis exam and then continued to watch the video on this topic.

AP Chem-study tips! During study and during the test tomorrow, help yourself out by DRAWING a beaker with any and all substances that are in solution! Start with any and all INITIAL quantities/substances and then SEPARATELY draw an "after" picture to see what is reacting (conjugate acid w/ base added or vice-versa or simple hydrolysis, etc.); THEN, write your ICE or SRFC table and see whether or not you can use simplifying assumptions or the H-H equation.
Truth in advertising:
your test tomorrow will have-
1. a polyprotic acid problem (as seen in the notes) in which all eq. quantities are solved for.
2. a FULL titration of either a weak acid or a weak base and calculations at the typical points of interest.
3. a buffer solution pH calculation before and after addition of a strong acid or base (exactly as done in the notes)
4. Lewis acid and base theory in words and Lewis structures.
5. Titration curves and selection of appropriate indicators.
6. Selecting the correct substances in order to make a buffer of a particular pH.
Be prepared and do well on this challenging exam because the next test is even more challenging for sure.

Finished the acid-base unit by discussing the polyprotic acid titration curve and noting the various buffer regions. We also saw how buffers of a particular pH can be made by choosing suitable combinations of salts of the conjugate amphoteric bases of the polyprotic acid.
Depending on the desired pH, you should choose a salt that has an acidic (even though the ion is amphoteric, just consider it acting as an acid in this case) ion that has a pKa near the desired pH along with a salt that contains the conjugate base of that acidic ion.

We then began the ultimate challenge (as you quickly saw, perhaps?) of our course: sparingly soluble salt equilibria (NOT on tomorrow's exam).
We introduced this topic by painstakingly pointing out the DIFFERENCE between moles of solid DISSOLVED to form a saturated solution and the resulting MOLARITY of the dissolved IONS IN SOLUTION.
It is crucial for you to see/feel this major difference, which is due to the fact that we NEVER calculate or even care about the "molarity" of the solid! Solids, by definition, are not aqueous so the molarity cannot change; thus we ONLY ever consider the grams or moles of solid DISSOLVED to form certain concentrations of the DISSOLVED ions.
We reviewed how to write K expressions, which we will call Ksp for sparingly soluble salts. We also reviewed that the CHARGE DENSITY of the ions in the salt lattice is the major determinant of the salt's solubility in polar solvents.

Bio 6- we discussed the process of embryonic development from fertilization to blastula formation to gastrulation, noting that blastula cells are identical cells formed via MITOSIS. However, a gastrula soon forms containing different types of cells in the three layers: endoderm, mesoderm, and ectoderm. These cell have different structures and form different organ systems of the body because the cells of the blastula begin to experience different chemical and physical environments, which causes certain genes to be turned off or turned on in a given cell. Thus a cell with different genes turned on will translate different proteins and have a different structure and function than another cell that has other genes turned on.
We then began to watch the amazing videography in the video about reproduction and development of the humans.

Bio 7/8- we discussed the process of embryonic development from fertilization to blastula formation to gastrulation, noting that blastula cells are identical cells formed via MITOSIS. However, a gastrula soon forms containing different types of cells in the three layers: endoderm, mesoderm, and ectoderm. These cell have different structures and form different organ systems of the body because the cells of the blastula begin to experience different chemical and physical environments, which causes certain genes to be turned off or turned on in a given cell. Thus a cell with different genes turned on will translate different proteins and have a different structure and function than another cell that has other genes turned on.
We then began to watch the amazing videography in the video about reproduction and development of the humans.

AP Chem- two days until buffer/titration test day- use the posted practice worksheets/tests/tutorials! You MUST see similar questions on Thursday's test! No tricks!
Speaking of no tricks, today's descriptive chem quiz was written exactly as described and the substances chosen were relatively easy and common.

We finished the titration of weak base by strong acid problem involving the common points of interest. We then discussed acid-base indicators, noting that these indicators are actually colored weak acids that have conjugate bases of a different color! Thus, we only use a drop or two of indicator so as not to significantly affect the pH of the titration solution.
We saw how to choose a suitable/viable indicator for a given titration by looking for an indicator pKa that is within ONE UNIT of the expected equivalence point pH; this way, the indicator will be in the middle of a color change at the equivalence/end point of the titration.

We then discussed the major features of a polyprotic acid titration curve (titrated by a weak base).
Tomorrow, we begin our next-to-penultimate topic on slightly soluble salts.

Bio 6/7 - NOTE: omit objectives 14-17 in your homework because we did not significantly discuss those answers yet in class. Also, you can PREVIEW the reproduction video from PBS at this link.

We discussed the chromosomal mutation of NONDISJUNCTION that can occur during either anaphase I or anaphase II of meiosis. This meiosis "mistake" usually results in gametes that have one MORE or one LESS chromosome than the normal haploid number of chromosomes.
If these gametes fertilize an egg (or are fertilized) the resulting zygote/embryo is either not viable (won't live) or it will develop abnormally.
We briefly discussed embryonic development and will continue with that tomorrow.
We then did a meiosis lab simulation with gummy worms, showing the results of meiosis I and meiosis II with and without nondisjunction.

Bio 8- We discussed the chromosomal mutation of NONDISJUNCTION that can occur during either anaphase I or anaphase II of meiosis. This meiosis "mistake" usually results in gametes that have one MORE or one LESS chromosome than the normal haploid number of chromosomes.
If these gametes fertilize an egg (or are fertilized) the resulting zygote/embryo is either not viable (won't live) or it will develop abnormally.
We briefly discussed embryonic development and will continue with that tomorrow.

This site has good buffer and titration simulations/tutorials. Just scroll towards the bottom of the page. Also, this site has a virtual titration simulator; just follow steps 1 through 5, in order, and then do the calculation before you type in your answer to step 6: good for extra visual review.

Buffer/Acid Strength/Titration/Salts tutorials:

AP Chem- NOTE: because we did not yet cover acid-base indicator selection in titrations, the hw questions involving such selection/calculation can be omitted.
Finished our quantitative titration of a weak acid by a strong base; the equivalence point pH calculation involves the most steps because we separated the events into the neutralization of the acid by the added titrant base (SRFC TABLE) during which the solution volume changes and then the back hydrolysis of water by the conjugate A- (ICE TABLE).
We saw that, JUST barely past the equivalence point (about 1.oo mL past), there is a drastic increase in pH because there is practically no more HA to react with the added base so the effect is practically the same as adding strong base to WATER.
The limiting/final pH is just due to the given initial (and ultimately final/limiting) concentration of the titrant/hydroxide i.e. just take the negative log of its concentration to get the pOH.

We then did the same routine but with a weak base titration by a strong acid; we will finish that up tomorrow after a brief descriptive chem quiz:
Possible reactions tested-
1. any of the four gas forming reactions
2. double replacement - with precipitation or acid-base neutralization
3. redox- cationic or anionic single replacement
4. Lewis acid-base: metal oxide with water, non-metal oxide with water, metal-ligand complex, boron compound reactions
5. organic reaction types: redox/combustion, substitution, addition, esterification
6. synthesis/redox- metal + non-metal forms a salt

Bio 6- reviewed the mechanism/sources of variety in meiosis I and the finished with meiosis II.
We noted that the sources of genetic variation/recombination occur mainly during PROPHASE I (crossing over of parts of homologous pairs of chromosomes during synapsis) and during METAPHASE I (each homologous pair lines up double file) INDEPENDENTLY of all other homologous pairs and can line up in either order (right-left or left-right).
The other major source of genetic recombination/variety occurs when the egg and sperm unite to form a genetically unique individual (the sperm and egg typically have many alleles that are not the same and they combine to give the individual a unique set of traits).

Bio 7/8- finished watching the animations of meiosis I and meiosis II. We then drew step-by-step to see that the sources of genetic variation/recombination occur mainly during PROPHASE I (crossing over of parts of homologous pairs of chromosomes during synapsis) and during METAPHASE I (each homologous pair lines up double file) INDEPENDENTLY of all other homologous pairs and can line up in either order (right-left or left-right).
The other major source of genetic recombination/variety occurs when the egg and sperm unite to form a genetically unique individual (the sperm and egg typically have many alleles that are not the same and they combine to give the individual a unique set of traits).

We did a first run of our gummy-worm meiosis simulation; I just wanted to get you used to manipulating the worms as chromosomes- we will discuss nondisjunction and then edit and refine the process on Wednesday.

As you can tell from the massive number of worksheets, solved hw problems, and tutorials posted on Blackboard, I am indicating that the next exam (and the subsequent one, also) is the most challenging one in this course. If you found the past few exams difficult, you will find this upcoming exam even more difficult. The problems require more steps and more visualization and analysis with respect to what is actually reacting in the "container".
Knowing that, in addition to the regularly scheduled morning extra help, I'll be staying after school each day this week. On Tuesday and Wednesday, I'll give a practice test to those who can stay for it; if you cannot stay for the test, I'll give you a copy but I will not post answers because I do not want you to think that you are prepared from taking just one practice test- you won't be in both the long term and the short term. Instead, if you cannot go to the practice test, there is more than enough extra help from the files posted on Blackboard; in fact, there is much more posted on Blackboard than could be covered on a given practice test. Furthermore, I will try to put up some "vodcasts" also for the more extensive problems.
If you get through the next two exams in good shape, you are in good shape in this course and you are in good shape for the May 12th AP Chem exam!

AP Chem- DUE TUESDAY (same day as the quick descriptive chem quiz!), do the following 8 text questions for hw:

Chapter 14, pages 715-716: #100 and #126

Chapter 15, pages 782-787: #24, 39, 56, 60, 70 , 116
Also, on Blackboard, check the answers to today's descriptive chem quiz.

We drew out the various titration curves that result from titrations of acids with bases and vice versa. We emphasized certain "tells" or "dead giveaways" that the acid or base that is titrated is strong or weak (i.e. initial pH from the given concentration or INITIAL sharp change(or lack thereof) in pH.
We also saw that, for weak acids or weak bases, we could SEE the pKa or pKb value (respectively) by looking at the center of the "buffer region"/flat part of the titration curve and locating the pH on the y-axis at that point of inflection.

We then began a LONG full, quantitative titration problem involving a weak acid and a strong base/titrant. The FOUR main points of interest, typically, are:
1. initial pH
2. pH at halfway to the equivalence point = pKa !!!
3. pH AT the equivalence point (MUST DO SRFC AND THEN ICE!!!)
4. pH past the equivalence point (or the limiting final pH).

Bio 6/7- we discussed the evolutionary advantages and disadvantages of sexual reproduction and then we saw a video that detailed the phases of meiosis I and meiosis II (I will post that video on Blackboard for your viewing pleasure/review.)
We then drew out a detailed step-by-step picture showing the source of variety/genetic recombination/"genetic mixing" that occurs during PROPHASE I and METAPHASE I of meiosis.
We will continue this on Monday but, in order to save time, I will give you the template for the rest of the phases.

Bio 8- we discussed the evolutionary advantages and disadvantages of sexual reproduction and then we saw a video that detailed the phases of meiosis I; we will continue with that video on Monday but I will post it on Blackboard for your viewing pleasure.

AP Chem- determined the effect on conjugate concentrations and on pH when a given quantity of strong acid or strong base is mixed with a buffer solution.
We also learned how to choose a suitable pair of conjugates, given their Ka and Kb values, to make a buffer of a particular pH: we had to choose a pair of conjugates that can be combined in a molarity ratio WITHIN the 1:10 or 10:1 buffer range such that the H-H equation yields the desired pH.

Bio 6- discussed the structure and function of the various parts of the female reproductive system. We then described the sequence of events and changes that occur during fertilization and implantation (pregnancy). We discussed the structure and function of the placenta and umbilical cord, both of which stem from the developing embryo embedded in the uterine wall;
thus, the developing fetus can obtain nutrients from the mother and get rid of wastes to the mother without the mixing of the blood cells between the mother and fetus.

Bio 7/8- discussed the structure and function of the various parts of the female reproductive system. We then described the sequence of events and changes that occur during fertilization and implantation (pregnancy). We discussed the structure and function of the placenta and umbilical cord, both of which stem from the developing embryo embedded in the uterine wall;
thus, the developing fetus can obtain nutrients from the mother and get rid of wastes to the mother without the mixing of the blood cells between the mother and fetus.

We did a lab activity in which we sequenced the various changes in the uterus during the menstrual cycle when the egg is fertilized and when the egg is not fertilized.

AP Chem- we derived the Henderson-Hasselbalch equation and saw that it is really just a rearranged version of what we have always done: Ka expression filled with the values from the ICE table.
The "easy" part is that, in the H-H form, we can ignore the insignificant amount of hydrolysis that occurs when the parts of a buffer are mixed AS LONG AS THE CONJUGATE ACID TO BASE RATIO IS WITHIN the 1:10 or 10:1 RANGE!
We did a sample buffer problem and saw that:
1. the pH = pKa when the conjugates are at equal concentrations
2. the pH will go up a bit or down a bit depending on which conjugate is in greater concentration.

We then derived the base version of the H-H equation and made a basic buffer solution and calculated the conjugate concentrations, pH, etc.

We tried a couple of problems in which the final concentrations of the conjugates are not given; that is, we mixed two SEPARATE solutions of the conjugates that, when mixed, DILUTE their initial concentrations. Once we calculate the final concentrations (thanks to the MILLIMOLE), we plug those values into the H-H equation (if within buffer range, otherwise you must use an ICE table!) and solve for the buffer pH.

Lastly, we began to discuss acid-base titrations (there are four combinations of which we will discuss three). We started by looking at a characteristic strong-strong acid-base titration curve (profile).

Bio 6/7- we reviewed, with a comprehensive graph and picture, step-by step, the stages and hormonal regulation of the menstrual cycle.
We then did a manipulation lab in which we sequenced and noted the changes in the various stages of the menstrual cycle.

Bio 8- we reviewed, with a comprehensive graph and picture, step-by step, the stages and hormonal regulation of the menstrual cycle.

AP Chem- we drew out and explained how to make a buffer solution (4 different methods) and saw how a given buffer can "absorb"/react with a given quantity of added strong acid or strong base so that the pH does not change as significantly as it would if the same strong acid or strong base were added to water.
Buffers work by having (within a 10:1 or 1:10 molar ratio) available weak acid molecules and their conjugate bases so that one or the other will react with/neutralize added acid or base (same deal for basic buffers that have a weak base and its conjugate acid ions available).

We then quantitatively showed the pH difference between a solution of ethanoic acid and a buffer solution of ethanoic acid mixed with an equimolar quantity of sodium ethanoate.
We noted that the ICE tables in buffer problems have BOTH acid AND conjugate base quantities in the "I" initial concentration line. We let the concentration of H+ initially be ZERO as a convenience because we know that ONLY the "E" equilibrium line amounts are involved in our calculations with Ka.
We then saw a convenient little equation that is used for buffers: the Henderson-Hasselbalch equation.

Bio 6- finished discussing the structure and function of the female reproductive system and then began our discussion of the fascinating, multiple hormonal feedback regulated menstrual cycle.
The video is posted on Blackboard.

Bio 7/8- we discussed the structure and function of the female reproductive system and then began our discussion of the fascinating, multiple hormonal feedback regulated menstrual cycle.
The video is posted on Blackboard.
We did some microscopy in lab but were unsuccessful (good thing?) in finding dust mites on the slides.

AP Chem- we discussed the naming and formula writing of halide salts of amine-ium ions (a cation of an amine that results when an H+ is added to the amine). These salts are most commonly seen in/produced in the pharmaceutical industry e.g. oxycodone hydrochloride but the hydro in this name belongs to/is part of the amine-ium cation (as the H+ that bonded to the amine) so the name is somewhat misleading.

We began the relatively lengthy (but loaded with shortcuts) calculations of the various ion concentrations in solutions of polyprotic acids. Except for sulfuric acid, most polyprotic acids have a STEEP decline in subsequent ionization of other H+ ions/protons from the acidic anions that form after the initial H+ ionizes (because, via Coulomb's Law, much energy is required to separate an H+ from a negative ion).

This steep decline in subsequent ionization makes it easier to make simplifying assumptions in calculations involving Ka and the various ion concentrations. Also, in polyprotic acid problems, on paper, we can do the series of ionizations step-by-step and get the same result as if we considered all ionization steps occurring at once!

We did examples with sulfuric acid (first ionization step goes to completion so no Ka is needed or given) and also with oxalic acid.
We reviewed the meaning and examples of the Bronsted term "amphiprotic" or "amphoteric": typically these are H-containing anions of conjugate acids that can either donate or accept H+ depending on the other ions or molecules in solution.
We then started from the opposite type of substance, a conjugate base of a polyprotic acid. We saw that phosphate is a relatively strong base (though not "STRONG", by definition) and will hydrolyze water thus forming hydrogen phosphate ion and hydroxide. HPO4 2- will further hydrolyze water to form H2PO4- and a little more hydroxide; the H2PO4 - will hydrolyze a weeeeee bit of water to form some phosphoric acid and the tiniest insignificant bit of extra hydroxide. Mainly, we saw that, for BOTH polyprotic acids and their conjugates, MOST of the ionization/hydrolysis/influence on the pH comes from the FIRST step.

Bio 6/7- we continued describing and explaining the various structures of the male reproductive system, noting how each structure contributes to increasing the likelihood of internal fertilization of the egg/ovum. We noted that maturation and production of the parts of the male reproductive system is hormonally controlled by the testes and the pituitary gland.
We examined the parts of the female reproductive system: the ovaries, oviducts/Fallopian tubes, uterus, cervix, and vagina/birth canal. We will discuss the hormone-controlled feedback loop that determines when an egg can develop and be fertilized and what happens when an egg is not fertilized.
We then worked on handouts, labeling the parts of the male and female reproductive systems. We finished the questions and graphs from the reaction time lab.

Bio 8- we reviewed the various structures of the male reproductive system, noting how each structure contributes to increasing the likelihood of internal fertilization of the egg/ovum. We noted that maturation and production of the parts of the male reproductive system is hormonally controlled by the testes and the pituitary gland.
We examined the parts of the female reproductive system: the ovaries, oviducts/Fallopian tubes, uterus, cervix, and vagina/birth canal. We will discuss the hormone-controlled feedback loop that determines when an egg can develop and be fertilized and what happens when an egg is not fertilized.
We then worked on handouts, labeling the parts of female reproductive system.

AP Chem- we drew out the mechanism and discussed in detail several examples of Lewis Acid-Base reactions: a non-metal oxide in water, a metal oxide in water, boron compounds with amines, and metal-ligand complex formation.
We then did quantitative calculations (pH, etc.) for acidic salts and basic salts; for these questions for these types of salts, ONLY one of the ions will be involved in hydrolysis of water so that the calculation doesn't involve more than one ICE table/equilibrium calculation.

Bio 6- began our new unit on sexual reproduction/meiosis in humans. We first discussed the adaptive features of the male reproductive system that aid in the production of gametes and their likely transport to/fertilization of female gametes.

Bio 7/8- began our new unit on sexual reproduction/meiosis in humans. We first discussed the adaptive features of the male reproductive system that aid in the production of gametes and their likely transport to/fertilization of female gametes.
We finished up the graphs and questions on our reaction time lab.

AP Chem- took the unit exam on Acids/Bases/A-B Equilibria/Organic- very comprehensive. Good for you for being able to prepare for this important exam.

Bio 6/7- took the Nervous System unit exam.
We then completed the quantitative work on the reaction time lab.

Bio 8- took the Nervous System unit exam.

AP Chem- TEST: INSIDE INFO! This exam covers a lot of material so I want you to be able to focus your study by doing as many of the following types of practice problems as is possible:
(these question types and ONLY these question types will appear on tomorrow's exam)

not in this order necessarily-

1. pH calculation of a strong MONOPROTIC acid of given/known concentration
2.pH calculation of a strong base (either MONOhydroxy OR DIhydroxy!!!-beware trap!!!-see notes) of given/known concentration
(For 3-6, an organizer/ICE Table is suggested and probably helpful; either way, all work must be shown, step by step).
3. given Ka (OR pKa) AND concentration, determine pH AND percent ionization of a weak acid
4. given (equilibrium) pH AND concentration of a weak acid, determine its Ka OR pKa
5. given Kb (OR pKb) AND concentration, determine pH AND percent ionization of a weak base
6. given (equilibrium) pH AND concentration of a weak base, determine its Kb OR pKb
7. EXPLAIN, in terms of (1) H-X bond strength and (2) H-X bond polarity, the relative strengths of a series of acids.
8. explain, in terms of water hydrolysis by an anion, cation, or both ions, how a given salts forms an acidic, basic, or neutral solution. For solutions in which BOTH ions cause hydrolysis, determine, based upon given pKa, Ka, pKb, or Kb of the ions OR their CONJUGATES, whether the solution will be SLIGHTLY acidic or SLIGHTLY basic.
9. be able to write examples (names, formulas, and balancing) of organic substitution, addition, combustion, esterification, and neutralization reactions.
10. be able to name, draw, and write the formula of each type of organic molecule (I won't put more than 6 carbons in the formula): alkanes, alkenes, alkynes, alkyl halides, alcohols, ethers, aldehydes, ketones, organic acids, esters, and amines.
That's enough! There won't necessarily be 10 questions but I formed the test from the above CollegeBoard objectives.
Obviously, this material should not be crammed and the upcoming tests will cover as much material or, if less material is covered, the material will be even more complex (how is that possible? I know).

Bio 6- we reviewed and analyzed HOW to write a THOROUGH answer to scientific investigation questions. Study for tomorrow's exam by reviewing your notes, the posted hw objectives, the practice fill-in quiz, AND by practicing your test-taking skills: underlining key words, DRAWING and labeling your answer along with a brief description that refers to your diagram; READING your answer and MAKING SURE that YOUR keywords match/answer the key terms in the question.
Practicing the above, you will do very well tomorrow.

Bio 7/8- we reviewed and analyzed HOW to write a THOROUGH answer to scientific investigation questions.
We finished the "reaction time" lab by completing the "distraction" part; at the neuron level, distractions are just MORE stimuli that cause your receptors to send more electrochemical impulses to your spinal cord and to your brain for processing. One neural pathway may slow down or interfere with another neural pathway leading to a slower response/"reaction time".

Study for tomorrow's exam by reviewing your notes, the posted hw objectives, the practice fill-in quiz, AND by practicing your test-taking skills: underlining key words, DRAWING and labeling your answer along with a brief description that refers to your diagram; READING your answer and MAKING SURE that YOUR keywords match/answer the key terms in the question.
Practicing the above, you will do very well tomorrow.

AP Chem- continued explaining the relative strengths of oxo-acids based on the increasing O-H bond polarity due to an increasing number of electron-attracting/withdrawing O's attached to the central atom in a given series of oxo-acids. We then compared the strengths of oxo-acids that have the SAME number of oxygens attached to the central atom but a DIFFERENT central atom. We say that, the more electronegative the central atom, the more polar the O-H bond due to the electron attracting/withdrawing effect of the central atom and therefore, the more acidic the molecule.
We then reviewed the overall explanation of the strength of any acid or relative strengths of any comparable series of acids. To write a correct explanation:
1. Draw the correct Lewis structures of each acid molecule.
2. Compare the Bond POLARITY of the H-X bond (in an oxo-acid, the H-O bond) based on the electronegativity of X or based on the number of electron withdrawing groups attached to the central atom of the molecule.
3. Compare the Bond STRENGTH of the H-X bond; the smaller the distance between the two nuclei (H and X) sharing the electrons, and the greater the Zeff of X on the shared electrons, the STRONGER the bond and the more energy required for the acid to ionize i.e. the weaker the acid.
That's what you have to do; don't leave out a factor even if you end up writing that it is not significant in causing the difference acidity i.e. always consider and write about each factor.

We discussed the REAL reason that a given salt is acidic, basic, or neutral. Salts that contain ions that are conjugates of STRONG acids and bases will form neutral solutions. Why? Because conjugates of STRONG acids or bases are so weak that they are unable to either donate an H+ to water or accept an H+ from water i.e. they do not react with/hydrolyze water thus the solution formed has the same pH as pure water.
We saw that salts that dissociate into ions that are conjugates of WEAK acids or bases will react with/hydrolyze water to form either an acidic or basic solution.
We will discuss more examples tomorrow. Acid-base test is on Day 1 Thursday; there will be questions on ORGANIC chemistry from your winter break assignment so know how to draw and name examples of each type of organic compound and know the various organic reaction types that were assigned.

Bio 6/7- we did a concept map showing the relationship between various stimuli and the receptor cells that are specialized to detect/sense these stimuli.
We then discussed types of drugs and the reason for their general effects on the nervous system.
We discussed the mechanism for drug tolerance i.e. SATURATION of the nerve/dendrite receptors with the drug so that a much higher dosage is required to have any increased effect.
We discussed the difference between physiological and psychological addiction. We also defined drug withdrawal.

Bio 8- we reviewed the structure and functions of the various parts of the brain. Tomorrow we will talk about the relationship between various drugs and neurotransmitters.

To make up for the snow day, I will have extra help tomorrow morning, Tuesday, in Room 308 from 8:00 AM to 8:45 AM.
Enjoy your extended weekend snow day!

AP Chem- the AP exam waits for no person! Therefore, we still have work today as we bask by a warm hearth.
I posted a few more released AP questions. Do them within the prescribed time limit and then check your work when the answers are posted. I assume/expect that each student is keeping apace of these problems so that s/he may earn "free points" on upcoming exams when these past topic question types appear.

For copious review: I have posted a massive number of acid/base and acid/base equilibrium text questions (from a different AP Chem text). Do those questions and check your answers against the posted full solutions. I also posted an acid "strength" tutorial and explanations. Use this as part of your study for Thursday's unit exam.

The acid-base unit exam is now on Day 1- Thursday, so the after school, timed practice exam will be given on Tuesday. You will not be given a test unless you can stay for the entire timed test- there are infinitely more practice questions on Blackboard than there are questions on the practice test; you should take the practice test after school if you feel the need for timed test-taking skill practice.

Bio- I do not know for sure but my guess is that the Nervous System unit exam will still be on Thursday, March 5.
I have now posted the Nervous System videos (from class) on Blackboard; review them.
Therefore, the hw objectives are still due on Tuesday. However, objectives 19 through 23 will not be graded because we did not get to them on Friday.