1.
(a) What is the pH of a mixture of 5 ml of 0.1M sodium acetate and 4 ml of 0.1M acetic acid ? (pK a of acetic acid is 4.76). pH = pKa + log[base]/[acid] calculate concentration of base: total volume is 9 ml therefore, 5 ml of 0.1M base = 0.5/9 M calculate concentration of acid: total volume is 9 ml therefore, 4 ml of 0.1M acid = 0.4/9 M
Thus, pH = 4.76 +0.5/0.4 (volume is constant and canceled out)
= 4.86
(b) What is the pH change upon the addition of 1 ml of (i) 0.1M HCl, and (ii)
0.1M NaOH to the mixture? i) new volume = 10 ml. amount acid added = 0.01M (final) amount base = [5/10 x 0.1] – [0.01] = 0.04M amount acid = [4/10 x 0.1] + [0.01] = 0.05M pH = 4.76-0.097
= 4.66
ii) reason as above pH =4.76 + 0.3
= 5.06
(c ) What is the pH if 1 ml of (i) 0.1M HCl, and (ii) 0.1M NaOH is added to 9 ml of water? i) total volume = 10ml pH = -log[H+]
= -log[0.01]
= 2 ii) total volume = 10 ml pH + pOH = 14 pH = 12.
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2.
Consider two solutions of lactic acid and sodium lactate (pK a
, 3.86, total lactate,
0.1M).
Lactic Acid
Sodium lactate
Calculate the pHs of buffer A and B.
A
0.05M
0.05M
B
0.015M
0.085M
Add NaOH to each solution in an amount equivalent to 0.005M. What is the pH change in each of the solutions? What conclusions can you draw from these calculations?
Solution A = 0.05M acid and 0.05M base. pH = 3.86 + log[0.05]/[0.05] = 3.86
Solution B = 0.015 M acid and 0.085M base. pH = 3.86 + log[0.085]/[0.015] = 4.61
When 0.005M (final) amount of NaOH added, solution A will lose 0.005M of acid and gain equivalent amount of base.
Solution A: pH = 3.86 + log[0.055]/[0.045] = 3.947
Solution B: pH = 3.86 + log[0.09]/[0.01] = 4.81
Thus, pH change is lower in A than in B.
[change in solution A ~ 0.087 pH and change in solution B ~ 0.2 pH unit)
Solution B is at the basic range and adding more base (past the pKa of 3.86) will continue to shift upwards. B is better used as a buffer for acid than base.
3.
Drug X has a pK a of 6.5 and is absorbed into the blood through the cells lining the stomach (pH 1.5) and the upper small intestine (pH 6.0). Calculate the amount of
X absorbed in each of the sites. Which is the major absorption route?
HX

H+ + X- pH = pKa + log[base]/[acid]
1.5 = 6.5 + log [X-]/[HX]
-5 = log [X-]/[HX] thus very little amount of dissociated form (base). Mostly acid form (less dissociated)
Intestine: [X-]/ [HX] = 0.32.
% HX = 1/(1 + 0.32) x 100 = 75.7%
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4.
The amino acid glycine is often used as the main ingredient of buffers in biochemical experiments.
(a) What are the buffering groups of glycine? State their pK a s.
(b) In what range can glycine be used as an effective buffer due to its amino and carboxyl groups?
(c) In a 0.1M solution of glycine at pH 9.0, what fraction of glycine has its amino group in the –NH
3
+
form?
(d) In order to have 99% of the glycine in its –NH
3
+
form, what must the pH of the solution be? a) pKa of glycine – found in the book. – COOH pKa = 2.3; NH2 group pKa=9.6 b) +/- pH c) 9 = 9.6 + log[base]/[acid]
-0.6 = log [NH2]/[NH3+]
0.25 = [NH2]/[NH3+] or, 0.25/[1 + 0.25] x 100 = 20% of total amino group is in the form of NH2 and 80% in the protonated form (NH3+). d) pH = 9.6 + log [0.01]/[0.99] = 7.6
5.
One of the most useful buffers is tris(hydroxymethyl)aminoethane (R
3
-C-NH
2
, where R = CH
2
OH) (TRIS) with a pK a
of 8.2.
(a) Write the equilibrium expression for the ionization of Tris starting from the acidic form
(b) What is the effective buffering range for TRIS ?
(c) What is the ratio of the basic to acidic form of TRIS at pH 7.4
(d) What percent of TRIS is in the basic form at pH 7.4?
(e) At pH 7.4, is TRIS a better buffer against acid or against base? a) –NH3+ 
NH2 b) pH buffering range = 8.2 +/- 1 c) -0.8 = log [-NH2]/[-NH3+]
0.158 = [-NH2]/[-NH3+] d) 0.158/(1 + 0.158) x 100 = 13.6%. e) against base as more – NH3+ is available.
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