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- Phenylketonuria is an inherited disorder in which phenylacetic acid, C 6 H 5 CH 2 COOH, (simplified here to HPAc) accumulates in the blood. A study of the acid shows that the pH of a 0.12 M HPAc solution is 2.60. What is the p K a of phenylacetic acid?
As HPAc is a weak acid, the equilibrium for its dissociation can be studied using an ‘ICE’ table:
HPAc PAc –^ H+ initial 0. 12 0 0 change - x + x + x final 0. 12 – x x x
By definition, pH = - log 10 [H+(aq)] so [H+(aq)] = 10-2.60^ M. From the reaction table, x = [H+(aq)]eq so:
[HPAc]eq = 0.12 – x = (0.12 – 10 - 2.60) M = 0.12 M (to 2 s.f.) [H+(aq)]eq = x = 10-2.60^ M [PAc-(aq)]eq = x = 10-2.60^ M
The equilibrium constant K a is given by:
K a =
𝐇𝐏𝐚𝐜!^ [𝐇!] [𝐇𝐏𝐀𝐜]
(𝟏𝟎!𝟐.𝟔𝟎)(𝟏𝟎!𝟐.𝟔𝟎) (𝟎.𝟏𝟐)
= 5.26 × 10 -^5
By definition, p K a = - log 10 K a so: p K a = - log 10 (5.26 × 10 -^5 ) = 4.
Answer: 4.
- A sample of hydrofluoric acid (0.10 M, 25.0 mL) is titrated with 0.10 M NaOH. The p K a of hydrofluoric acid, HF, is 3.17. Calculate the pH at the following four points.
Marks 7
before any NaOH is added
At this point, the solution contains only a weak acid. As HF is a weak acid, [H+] must be calculated by considering the equilibrium:
HF F–^ H+
initial 0. 10 0 0 change - x + x + x final 0. 10 – x x x
The equilibrium constant K a is given by:
K a =
𝐅!^ [𝐇!] [𝐇𝐅]
𝒙𝟐 (𝟎.𝟏𝟎!𝒙)
As p K a = 3.17, K a = 10−^ 3.17. K a is very small so 0.1 0 – x ~ 0. 1 0 and hence:
x^2 = 0. 10 × 10 – 3.1^7 or x = 0.00 822 M = [H+^ ]
Hence, the pH is given by: pH = −log 10 [H+^ ] = −log 10 [0.00 822 ] = 2.
pH = 2.
when half of the HF has been neutralised
At this point, half of the original HF has been converted to its conjugate base F-. The pH can be calculated using the Henderson-Hasselbalch equation:
pH = p K a + log
[𝐛𝐚𝐬𝐞] [𝐚𝐜𝐢𝐝]
= 3.17 + log
[𝐅!] [𝐇𝐅] = 3.17 + + log^ 𝟏^ = 3.
pH = 3.
at the equivalence point
At this point, all of the original HF has been converted to F-. The number of moles of HF originally present is:
number of moles of HF = concentration × volume = (0.10 mol L-^1 ) × (0.025 L) = 0.0025 mol
This is equal to the amount of F-^ present at equivalence. As 25.0 mL of NaOH has been added at this point, the total volume is now (25.0 + 25.0) mL = 50.0 mL. The concentration of F-^ is therefore:
ANSWER CONTINUES ON THE NEXT PAGE
[OH-] = number of moles / volume = (0.00125 mol) / (0.0625 L) = 0.020 mol L-^1
Hence,
pOH - = - log 10 [OH-] = - log 10 (0.020) = 1.
Lastly, pH = 14.00 – pOH:
pH = 14.00 – 1.70 = 12.
pH = 12. THIS QUESTION CONTINUES ON THE NEXT PAGE.
Sketch the titration curve. Marks 2
Putting the 4 points from 2014-N-7 together gives the titration curve:
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
pH
3
7
0 0 12.5 25.0 NaOH (mL)
- What amount of NaOH (in mol) needs to be added to 250 mL of 0.10 M acetic acid to give a solution with a pH of 5.00? The p K a of acetic acid is 4.76.
Marks 3
If the concentration of OH-^ which is added is x M then this will react with acetic acid to produce its conjugate base, acetate, so that:
[acid] = (0.1 0 – x ) M and [base] = x M
The Henderson-Hasselbalch equation can be used to work out the ratio of these needed for a pH of 5.00:
pH = p K a + log
[𝐛𝐚𝐬𝐞] [𝐚𝐜𝐢𝐝]
5.00 = 4.76 + log
𝒙 𝟎.𝟏𝟎!𝒙
Hence:
𝒙 𝟎.𝟏𝟎!𝒙
= 10 0.24^ so x = 0.
To achieve [OH-(aq)] = 0. 0634 mol L-^1 in 25 0 mL, the number of moles of NaOH that must be added is:
number of moles = concentration × volume = 0. 0634 mol L-^1 × 0. 25 0 L = 0.016 mol
Answer: 0.016 mol
- Explain the following terms or concepts.
Marks 1 Isoelectric point
The pH at which a protein or amino acid has no net charge, i.e. it contains equal amounts of positive and negative charges.
- Buffer 1 is a solution containing 0.08 M NH 4 Cl and 0.12 M NH 3. Buffer 2 is a solution containing 0.15 M NH 4 Cl and 0.05 M NH 3. The acid dissociation constant of the ammonium ion is 5.50 × 10 –^10. What are the pH values of each of the buffer solutions?
Marks 4
By definition, p K a = - log 10 K a so:
p K a = - log 10 (5.50 × 10 –^10 ) = 9.
Using the Henderson-Hasselbalch equation,
pH = p K a + log
[𝐛𝐚𝐬𝐞] [𝐚𝐜𝐢𝐝]
= 9.290 + log
[𝐍𝐇𝟑] [𝐍𝐇𝟑𝐂𝐥]
For buffer 1:
pH = 9.26 + log 𝟎.𝟏𝟐 𝟎.𝟎𝟖 = 9.
For buffer 2:
pH = 9.26 + log 𝟎.𝟎𝟓 𝟎.𝟏𝟓 = 8.
Buffer 1 pH = 9.44 Buffer 2 pH = 8.
Which buffer is better able to maintain a steady pH on the addition of small amounts of both a strong acid and strong base? Explain.
Buffer 1 is better able to maintain a steady pH because its pH is closer to the p K a of NH 4 +. This is because it has relatively high concentrations of both NH 4 + and NH 3 which can react with any added OH–^ or H+^ respectively.
- Explain the role played by the lungs and the kidneys in maintaining blood pH at a constant value of 7.4.
Marks 4
The most important buffer system in the blood is the hydrogencarbonate / carbonic acid system:
HCO 3 – (aq) + H +^ (aq) H 2 CO 3 (aq)
If the amount of H +^ exceeds the capacity of the buffering system ( e.g. during vigorous exercise), the lungs can help by removing CO 2 (g). CO 2 is linked to the buffer system via
H 2 CO 3 (aq) H 2 O + CO 2 (g) Thus removal of CO 2 (g) will shift the HCO 3 –^ /H 2 CO 3 equilibrium to the right, reducing H+^. If the blood becomes too basic, the kidneys can help by excreting HCO 3 –^. This will shift the buffer equilibrium to the left, producing more H+^.
As F-^ is a weak base. [OH - ] must be calculated using a reaction table.
F-^ H 2 O HF OH -
initial 0.050 large 0 0 change - y negligible + y + y final 0.050 – y large y y
The equilibrium constant K b is given by:
K b =
ሾ۶۴ሿሾ۶۽ ష^ ሿ
ሾ۴ ష^ ሿ
For an acid and its conjugate base:
p K a + p K b = 14.
p K b = 14.00 – 3.17 = 10.
As p K b = 10.83, K b = 10 - 10.83^. K b is very small so 0.050 – y ~ 0.050 and hence:
y^2 = 0.050 × 10 –10.83^ or y = 8.59 × 10-7^ M = [OH - ]
Hence, the pOH is given by:
pOH = -log 10 [OH - ] = log 10 [8.59 × 10 -7^ ] = 6.
Finally, pH + pOH = 14.00 so
pH = 14.00 – 6.07 = 7.
(iv) At this point, there is excess strong base present. Addition of 1.5 times the equivalence volume corresponds to addition of (1.5 × 25.0) mL = 37.5 mL. This volume of 0.10 M NaOH contains
number of moles of NaOH = concentration × volume = (0.10 mol L -1^ ) × (0.0375 L) = 0.00375 mol
From (iii), there was original 0.0025 mol of HF present so the excess of OH-^ is:
excess moles of OH -^ = (0.00375 – 0.0025) mol = 0.00125 mol
This is present in a total volume of (25.0 + 37.5) mL = 62.5 mL, so its concentration is:
[OH - ] = number of moles / volume = (0.00125 mol) / (0.0625 L) = 0.020 mol L -
ANSWER CONTINUES ON THE NEXT PAGE
Hence,
pOH -= -log 10 [OH - ] = -log 10 (0.020) = 1.
Lastly, pH = 14.00 – pOH:
pH = 14.00 – 1.70 = 12.
Putting these 4 points together gives the titration curve:
pH
0 12.5 25.0 NaOH (mL)
Calculate the pH of a 0.10 mol L–^1 solution of HF. (The p K a of HF is 3.17.) Marks 6
HF is a weak acid so the equilibrium concentrations need to be calculated using a reaction table:
HF H+(aq) F-(aq) Initial 0.10 0 0 Change - x + x + x Equilibrium 0.10 – x x x
As p K a = - log 10 K a, at equilibrium,
K a =
𝐇+^ 𝐚𝐪 [𝐅−^ 𝐚𝐪 ]
[𝐇𝐅 𝐚𝐪 ]
As K a is so small, x will be tiny and 0.10 – x ~ 0.10 and so
x^2 = 10-3.17^ × 0.10 or x = [H+(aq)] = 0.00822 M
As pH = - log 10 [H+(aq)],
pH = - log 10 (0.00822) = 2.
Answer: 2.
What mass of NaF needs to be added to 100.0 mL of the above solution to make a buffer with a pH of 3.00?
Using the Henderson-Hasselbalch equation for the pH of the buffer:
pH = p K a + log
[𝐛𝐚𝐬𝐞]
[𝐚𝐜𝐢𝐝]
To make a buffer with pH = 3.00 and [acid] = [HF] = 0.10 M:
3.00 = 3.17 + log
[𝐅−]
or log
[𝐅−]
Hence,
[F-] = 0.10 × 10 - 0.17^ = 0.068 M
The number of moles in 100.0 mL is thus 0.0068 mol. As NaF will dissolve to give one F-^ per formula unit, this is also the number of moles of NaF required.
ANSWER CONTINUES ON THE NEXT PAGE
The formula mass of NaF is (22.99 (Na) + 19.00 (F)) g mol-^1 = 41.99 g mol-^1. The mass of NaF required is thus:
mass = number of moles × formula mass = (0.0068 mol) × (41.99 g mol-^1 ) = 0.28 g
Answer: 0.28 g
Explain why HCl is a much stronger acid than HF.
Cl is a much larger atom than F and is less electronegative. The H–Cl bond is therefore much longer and weaker than the H–F bond. The H–Cl bond is therefore easier to break and it is the stronger acid.
HF is actually a weak acid. F is smaller and more electronegative than O, so the H–F bond is stronger than the O–H bond. There is consequently little dissociation of HF when it is dissolved in water.
- Explain why the acidity of hydrogen halides increases with increasing halogen size ( i.e. , K a (HCl) < K a (HBr) < K a (HI)), while the acidity of hypohalous acids decreases with increasing halogen size ( i.e. , K a (HOCl) > K a (HOBr) > K a (HOI)).
3
For the hydrogen halides, the length of the H-X bond increases and hence gets weaker as the halogen gets bigger,. The weaker the bond, the more easily the H+ dissociates.
For the hypohalous acids, as the electronegativity of the halide increases, the more electron density it pulls from the O-H bond towards itself. This results in the O-H bond becoming more polar and increasing the ease with which the H+ will be lost.
The K a of benzoic acid is 6.3 × 10–^5 M at 25 C.
Calculate the pH of a 0.0100 M aqueous solution of sodium benzoate (C 6 H 5 COONa).
Marks 5
As benzoic acid is a weak acid, its conjugate base, C 6 H 5 COO-, is a weak base and so [OH–] must be calculated using the reaction table:
C 6 H 5 COO –^ H 2 O OH–^ C 6 H 5 COOH initial 0.0100 large 0 0 change - x negligible +x +x final 0.0100 – x large x x
The equilibrium constant K b is given by:
K b =
2 6 5 6 5
[OH ][C H COOH] (^) x [C H COO ] 0.0100^ x
For an acid and its conjugate base in aqueous solution, K a × K b = K w = 10-^14. Hence,
K b =
14 5
^
= 1.6 × 10-^10
As K b is very small, 0.0100 – x ~ 0.0100 and hence:
x^2 = 0.0100 (1.6 × 10-^10 ) or x = 1.3 × 10-^6 M = [OH (aq)]
Hence, the pOH is given by:
pOH = log 10 [OH ] = log 10 [1.3 × 10-^6 ] = 5.
Finally, pH + pOH = 14.0 so
pH = 14.0 – 5.9 = 8.
Answer: pH = 8.
Answer: pH = 4. ANSWER CONTINUES ON THE NEXT PAGE
