Biochemical reactions take place in buffered aqueous environments. For example, the pH of the blood — Analytical Chemistry Chemistry Question
Solubility equilibrium in a buffer solution
Biochemical reactions take place in buffered aqueous environments. For example, the pH of the blood is maintained around 7.4 by the buffering action of carbonate, phosphate, and proteins. Many chemical reactions in the laboratory are also carried out in buffer solutions. In this problem, let’s consider the solubility equilibrium in a buffer solution.
H2S gas with a volume of 440 cm3 at STP can be dissolved in 100 cm3 of water at 25 °C. Calculate the concentration of H2S (mol dm–3) in water saturated with H2S. Assume that there is no volume change in water upon dissolution of H2S.
Model Answer
440 cm3 H2S in 100 cm3 of water = 4.4 dm3 H2S in 1 dm3 of water = 0.20 mol dm–3
Assume that equilibrium is established after a FeCl2 solution with a concentration of 1.0×10–8 mol dm–3 is saturated with H2S by continuously bubbling of H2S into the solution.
Ksp(FeS) = [Fe2+][S2–] = 8.0×10–19 at 25 °C (1)
For acid dissociation of H2S,
K1 = [H+][HS–] / [H2S] = 9.5×10–8 (2)
K2 = [H+][S2–] / [HS–] = 1.3×10–14 (3)
For autoprotolysis of water:
Kw = [H+][OH–] = 1×10–14 (4)
In the solution, the positive charge is balanced by the negative charge:
[H+] + 2 [Fe2+] = [Cl–] + [OH–] + [HS–] + 2 [S2–] (5)
Cross out terms that are negligibly small in the charge balance equation (5), in order to determine [H+] and [Fe2+]. Would you increase or decrease the pH of the solution to precipitate more FeS? How does the increase of pH by 1 affect the concentration of Fe2+ ion?
Model Answer
For approximation, the concentration of all anions with the exception of [Cl–], which is 0.02, can be omitted in eq. (5). Thus
[H+] + 2 [Fe2+] = [Cl–] = 0.020 (6)
Combine (2) and (3): [H+]2 [S2–] / [H2S] = 1.24×10–21
Since [H2S] = 0.2, one gets [H+]2 [S2–] = 2.48×10–22 (7)
Combine (1) and (7): [H+]2 (8.0×10–19 ) / [Fe2+] = 2.48×10–22
[H+]2 = 3.1×10–4 [Fe2+] (8)
Combine (6) and (8) and solve for [H+]:
[H+] = 0.0176 pH = 1.75
[Fe2+] = (0.020 – 0.0176) / 2 = 0.0012 (12 % remains in solution)
Check: [HS–] = (9.5×10–8) [H2S] / [H+] = 1.1 × 10–6 << [Cl–] = 0.02
[S2–] = (1.3×10–14) [HS–] / [H+] = 2.5×10–18
[OH–] = 5.7×10–13
Eq. (8) shows that 10–fold decrease in [H+] increases [Fe2+] 180–fold.
How would you adjust the final pH of the solution saturated with H2S to reduce the concentration of Fe2+ from 0.010 mol dm–3 to 1.0×10–8 mol dm–3?
Model Answer
From [H+]2 = 3.1×10–4 [Fe2+],
[H+] = [(3.1×10–4) (1×10–8)]1/2 = 1.76×10–6 pH = 5.75
You want to use acetic acid (HAc) / sodium acetate (NaAc) buffer to achieve concentration of Fe2+ 1.0×10–8 mol dm–3 as described above. Suppose that you are making the buffer by mixing acetic acid and sodium acetate in water in a volumetric flask. Enough acetic acid was added to make the initial concentration 0.10 mol dm–3. Considering that the precipitation reaction produces H+ (Fe2++ H2S→ FeS(s) + 2 H+), how would you adjust the initial concentration of sodium acetate to obtain concentration of Fe2+ equal to 1.0×10–8 mol dm–3 after equilibrium is established?
The dissociation constant for acetic acid at 25 °C is 1.8×10–5.
Model Answer
Original n(HAc) = 0.10 mol dm–3 × 100 cm3 = 10 mmol
Henderson–Hasselbach equation for the HAc / Ac– buffer at pH = 5.75:
pH = 5.75 = pK + log [Ac–] / [HAc] = 4.74 + log [Ac–] / [HAc]
initial n(Fe2+) = 0.01 mol dm–3 × 100 cm3 = 1 mmol
n(H+) produced upon precipitation of 1 mmol Fe2+ = 2 mmol Ac– consumed by H+ produced = 2 mmol
log [Ac–] / [HAc] = 5.75 – 4.74 = 1.01
Let x = original n(Ac–)
(x – 2) / (10 + 2) = 10^1.01 = 10.2,
x = 124 mmol [Ac–] = 124 mmol / 100 cm3 = 1.24 mol dm–3
What is the pH of the buffer before H2S is introduced and FeS is precipitated?
Model Answer
pH = 4.74 + log (1.24 / 0.10) = 5.83