Question

what is the ph of a 0.500 m solution of nacn (kₐ of hcn is 4.9×10⁻¹⁰)?

Explanation:

Step1: Find \(K_b\)

We know that \(K_w = K_a\times K_b\), where \(K_w=1.0\times 10^{-14}\). Given \(K_a = 4.9\times 10^{-10}\) for \(HCN\), then \(K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{4.9\times 10^{-10}}\approx2.04\times 10^{-5}\)

Step2: Set up the hydrolysis equation

\(CN^-+H_2O
ightleftharpoons HCN + OH^-\)
Let \(x\) be the concentration of \(OH^-\) and \(HCN\) at equilibrium. The initial concentration of \(CN^-\) is \(c = 0.500M\), and the equilibrium concentration of \(CN^-\) is \(c - x\approx c\) (since \(K_b\) is small, \(x\) is negligible compared to \(c\)).

Using the formula \(K_b=\frac{[HCN][OH^-]}{[CN^-]}\), substituting the values we get \(K_b=\frac{x\cdot x}{c}\)

Step3: Solve for \(x\) (\([OH^-]\))

\(x = [OH^-]=\sqrt{K_b\times c}\)
Substituting \(K_b = 2.04\times 10^{-5}\) and \(c = 0.500M\)
\(x=\sqrt{2.04\times 10^{-5}\times0.500}=\sqrt{1.02\times 10^{-5}}\approx3.19\times 10^{-3}M\)

Step4: Calculate \(pOH\)

\(pOH=-\log[OH^-]=-\log(3.19\times 10^{-3})\approx 2.49\)

Step5: Calculate \(pH\)

Since \(pH + pOH=14\), then \(pH=14 - pOH\)
\(pH=14 - 2.49 = 11.51\)

Answer:

\(11.51\)