The Following questions pertain to a 2.2M solution of hydrocyanic acid at 25°C. pKa = 9.21 at 25°C. Find the concentrations of a

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The Following questions pertain to a 2.2M solution of hydrocyanic acid at 25°C. pKa = 9.21 at 25°C. Find the concentrations of a

ll species present in the solution at equilibrium. Find the pH of the solution. Identify the strongest base in this system.

Chemistry

1 answer:

horsena [70] · Answer 1 · 2022-03-21T13:55:03+03:00

1) Chemical reaction: HCN + H₂O → CN⁻ + H₃O⁺.
c(HCN) = 2,2 M = 2,2 mol/L.
pKa(HCN) = 9,21.
Ka = 6,16·10⁻¹⁰.
[CN⁻] = [H₃O⁺] = x.
[HCN<span>] = 2,2 M - x.
</span>Ka = [CN⁻] · [H₃O⁺] / [HCN].
6,16·10⁻¹⁰ = x² / 2,2 M -x.
Solve quadratic equation: [CN⁻] = [H₃O⁺] = 0,0000346 M.
[HCN] = 2,2 M - 0,0000346 M = 2,199 M.

2) pH = - log[H₃O⁺].
pH = -log( 0,0000346 M).
pH = 4,46.
Hydrocyanic acid and hydronium ion (H₃O⁺) are acids. Cyanide anion (CN⁻) is the strongest base in the system, cyanide anion accept protons in chemical reaction.
pKb = pKw - pKa.
pKb = 14 - 9,21 = 4,79.