Hydrocyanic acid is a weak acid with dissociation constant Ka = 4.93×10–10 — Analytical Chemistry Chemistry Question
Hydrocyanic acid
Hydrocyanic acid is a weak acid with dissociation constant Ka = 4.93×10–10
Find the pH of a 1.00 M solution of HCN.
Model Answer
HCN ⇌ H+ + CN-
c - x x x
x^2 + Ka x - Ka c = 0
⇒ x = (–Ka + √(Ka^2 + 4Kac)) / 2
[H+] = c x = 2.22×10−5 ⇒ pH = 4.65. Acceptable to ignore [OH–]
10 dm3 of pure water is accidentally contaminated by NaCN. The pH is found to be 7.40. Deduce the concentrations of each of the species, Na+, H+, OH–, CN–, HCN, and hence calculate the mass of NaCN added.
Model Answer
(1) [H+ ][CN− ] = Ka[HCN]
(2) [H+ ][OH− ] = Kw
(3) [H+ ] + [Na+ ] = [CN− ] + [OH− ]
(4) [Na+ ] = [CN− ] + [HCN]
(5) [H+ ] = 3.98×10−8
From (2): [OH− ] = 2.51×10−7
From (1): [HCN] = [H+][CN−] / Ka = 80.8 [CN−]
From (3): [Na+] = [CN−] + 2.11×10−7
From (4) [HCN] = 2.11×10−7
Hence c(CN−) = 2.62×10−9 mol dm-3, c(Na+) = 2.14×10−7 mol dm-3
Hence 10 dm3 contains 2.14×10–6 mol NaCN, i. e. = 0.105 mg NaCN