Solutions to Selected End of Chapter 2 Problems
Use your Pencil to Cancel out Units to Check the Equations
1. Simple prediction based on polarity of the molecule: ethane H3C-CH3 vs H3C-CH2OH
The alcohol group of ethanol is highly polar – it lovingly H-bonds with water, whereas the
methyls of ethane are completely not polar and don’t want to have anything to do with water. The
concentration of ethane in water is so low it is difficult to measure. But, as you can get multi-molar
concentrations of ethanol in water. Just imagine what life would be like without beer and wine! No,
don’t.
2 and 3. These are calculator practice converting [H+] to pH and pH to [H+],
2a. Practice with [H+] and pH…and your calculator. [H+] = - log pH
So, if [H+] = 1.75 x 10-5 mole/L, then pH = - log (1.75 x 10-5 M) = - (-4.76) = 4.76
Answers for b = 9.19; c = 4.0, d = 4.82
3a. pH = 3.82, thus pH = -log[H+], [H+] = 10-pH on the calculator it is the 10x button, which is
usually Shift-10x. [H+] = 10-3.82 = 1.51 x 10-4 M
Answers for b = 3.02 x 10-7 M, c = 7.76 x 10-12 M
Note: a. the data has 3 significant figures, so the answer should. b. pH has no units, it is a
logarithm which do not have units, but concentration does have units of moles/liter.
See how easy this is…fun to be back practicing General Chemistry.
4. Stomach acidity. Data: hospital lab has 10 ml of stomach juice (hours after a meal, so no food is
there as a buffer) and it was titrated to neutrality with 7.2 mL of 0.1 M NaOH. What is the pH of
stomach juice (HCl)? First convert 7.2 mL to Liters = 0.0072L
The equivalents of base that did the titration = (0.0072 L)(0.1 mole/L) = 0.00072 moles
This is the amount of protons in 10mL (= 0.010L) = 0.00072 moles/0.01L = 0.072 moles H+ /L
So, the pH is then = - log (0.072) = -(-1.1) = 1.1
5. Even easier Gen Ghem! a. HCL  H+ + Cl-.
c. NaOH  Na+ + OH-.
b. pH= – log 5 x 10-4 M = 3.3.
d. pH + pOH = 14. pOH = -log[OH-] = -log 7 x 10-5 M = 4.15
pH = 14 – pOH = 14 – 4.15 = 9.85
8. Strongest acid here is HCl, the others are weak acids and do not complete dissociate, so they hold
on to their hydrogens, where as HCl completely dissociates. Now between formic acid, pKa = 3.75
and acetic acid, pKa 4.76….the one with the lowest pKa is the stronger weak acid. (Note they had a
typo for the real pKa of acetic acid, it is not 4.86)
11. Oh goodie, this is super easy. It is the conjugate base that releases the proton to the seductive
nature of water. The conjugate bases are: a. RCOO-; b. R-NH2; c. H2PO4-; and d. HCO3-.
12. Henderson-Hasselbalch territory. Know this: there is always a H-H question on Exam 1 and the
Final, this means it is important. Hey, they got the right pKa for acetic acid (pKa = 4.76) in this
question!
pH = pKa + log
[A−]
[HA]
lets just do parts a and b.
a. where ratio of A- to HA is 2:1. pH = 4.76 + log 2 = 4.76 + 0.30 = 5.06.
b. where ratio of A- to HA is 1:3. pH = 4.76 + log 0.33 = 4.76 + (-0.48) = 4.28
Now do these, did you get: c. = 5.46.
d. = 4.76 (this is where pH = pKa). e. 3.76
13. This is all about getting certain molecules dissolved. For some it takes adding base, for others
adding acid, the idea is to get the molecule to have an ionic charge. Beginning with the classic
benzoic acid (naturally present in cranberry juice) is an antimicrobial and the reason why some drink
cranberry juice for urinary bladder infections. But to get it to work it must be solubilized (small
intestine) and taken up and then excreted (kidneys) to the urinary bladder. The whole trick is getting a
charge on each of the molecules a, b and c by either getting the pH below or above the pKa.
That is, for all the weak acids here, carboxylic acid and amino groups, the charge is either 0 and +1,
or 0 and -1. The chemistry of the weak acid makes it pos or neg when charged, whether it is charged
or not depends on the pH.
14. Poison ivy and poison oak cause a rash (local inflammation) based interaction with our natural
immune response. The figure shows the main catechol with the dissociable H in bold. The treatment
that works the best is d. Because soap helps get the undissociated catechol, but the presence of
sodium bicarbonate will cause much of the catechol to dissociate and solubilize so it will be washed
off.
15. This is our first Case Study. The Aspirin overdose. The pKa of the carboxyl is 3.5, so the sodium
bicarbonate will raise the pH of the stomach (~1.5 or so) above the pKa to solubilize it by making
aspirin have a negative charge…and in our Case Study it will help with the stomach pump to remove
the massive amount of aspirin the patient took.
16. The ammonium pKa is 9.25. Having a solution of 0.12 mole/L NH4Cl (which means 0.12 mole/L
NH4+ which is the HA) and adding NaOH (strong base) at 0.03 mole/L The OH- will convert 0.03
moles of NH4+ to NH3 which is the conjugate base (and H2O). A- was not present in the original
solution. So using H-H equation:
pH = 9.25 + log(
0.03
0.09
) = 9.25 + (-0.48) = 8.77
17. What happens when you convert a weak acid, HA, to A- by adding a strong base (or to convert
the conjugate base A- to HA by adding a strong acid)? The weak acid here is not known, but has a
pKa ot 7.4. Even an unknown weak acid behaves the same H-H rules!
Data: 100 mL of a 1.0 M solution of the weak acid at pH 8.0 will get 30 mL of 1 M HCl. So the
amount of strong acid will convert the same amount of A- to HA.
Begin first at pH 8.0
pH = pKa + log (A/HA).
8.0 = 7.4 + log (A-/HA)
log(A- / HA) = 0.6
[A-] / [HA] = 3.98
So, 3.98 [HA] = [A-] and we know that: [A-] + [HA] = 1.0M So we can substitute
into this last equation: 3.98[HA] + HA = 1.0 M
4.98 HA = 1.0 M
So
HA = 1.0 M / 4.98 = 0.20 M at pH 8.0
And at this 8.0 pH, A- = 0.80 M.
Check to see if this make sense? At a pH above the pKa A - must
be more than HA. And, A + HA = total (0.80M + 0.20 M= 1.0M)
Now lets add the 30 ml of 1M HCl. This will convert an equivalent amount of A - to HA. That is it will
convert 30 meq of the 80 meg of A- to HA leaving 50 meq of A- and forming 50 meq of HA. This
makes the (A-/HA) term = 1.0 so that the H-H becomes:
pH = pKa + log (A-/HA)
pH = 7.4 + log (0.5 / 0.5) = 7.4 + log 1 = 7.4 + 0 = 7.4
18. We will be doing this Glycine buffer problem in Class.
20. Histidine a polyprotic (has 3 pKa’s) weak acid: pKa1 carboxylic acid = 1.8, pKa2 the imidazole
group = 6.0, and pKa3 the amino group = 9.2. Data: 100 mL of 0.1M histidine at pH 5.4. This is close
to pKa2 so we need to find out how much of the imidazole group is dissociated (His) and how much is
associated (HisH+).
First the total Histidine (amount) is 0.1 mole/L x 0.1L = 0.01 mole = 10 mmoles
So, at pH 5.4:
5.4 = 6.0 + log(His / HisH+)
-0.6 = log(His / HisH+)
His / HisH+ = 0.25
Consequently 0.25 HisH+ = His
His + HisH+ = total Histidine = 10 mmoles
0.25 HisH+ + HisH+ = 10 mmoles
1.25 HisH+ = 10 mmoles
HisH+ = 10 mmoles / 1.25 = 8 mmoles as HisH+ and His = 2 mmoles
The problem goes on to what happens after adding 40 mL of 0.1 M HCl ? This means adding:
0.04 L x 0.1 mole/L = 0.004 moles of H+ which is 4 mmoles.
Lets conceptualize this: adding HCL will lower the pH, so it will go down to the first pKa. BUT,
first it will convert His to HisH+ that will take 2 mmoles of H+ of the 4 mmoles leaving 2 mmoles
of H+. Now, the HisH+ at pH 5.4 has the carboxyl dissociated (R-COO-) because this is
several pH units above the carboxyl pKa of 1.8. The remaining 2 mmoles of H + will convert the
ionized carboxyl of HisH+ to the associated form of the carboxyl (R-COOH).
Thus the final pH will be: pH = 1.8 + log (8 / 2) = 1.8 + 0.6 = 2.4.
26. The choice is really easy…get the one with a pKa nearest the desired pH. In this case the
answer is acetic acid….buffers are best holding pH’s near the pKa…the flattest portion of the titration
curve.