Question

what is the ph of a 0.0570 m solution of hydrocyanic acid, hcn (kₐ = 4.9 × 10⁻¹⁰)?

Explanation:

Step1: Write the dissociation equation and \(K_a\) expression

Hydrocyanic acid (\(HCN\)) dissociates as \(HCN(aq)
ightleftharpoons H^{+}(aq)+CN^{-}(aq)\). The \(K_a\) expression is \(K_a=\frac{[H^{+}][CN^{-}]}{[HCN]}\). Let \(x = [H^{+}]=[CN^{-}]\), and \([HCN]=0.0570 - x\). Since \(K_a = 4.9\times10^{-10}\) is very small, \(0.0570 - x\approx0.0570\).

Step2: Substitute into \(K_a\) expression and solve for \(x\)

Substitute into \(K_a=\frac{x\times x}{0.0570}\), so \(4.9\times 10^{-10}=\frac{x^{2}}{0.0570}\). Then \(x^{2}=4.9\times 10^{-10}\times0.0570\), \(x^{2}=2.793\times 10^{-11}\), \(x=\sqrt{2.793\times 10^{-11}}\approx5.28\times 10^{-6}\)

Step3: Calculate \(pH\)

Since \(pH =-\log[H^{+}]\), and \([H^{+}]=x = 5.28\times 10^{-6}\), then \(pH=-\log(5.28\times 10^{-6})\)
\(pH = 6-\log(5.28)\approx6 - 0.723=5.28\)

Answer:

\(5.28\)