what is the h+(aq) concentration in 0.05 m hcn(aq)? (the ka for hcn is 5.0 x 10^-10.) a 2.5 x 10^-11 m b 2.5…

Explanation:

Step1: Write dissociation equation

$HCN(aq)\rightleftharpoons H^{+}(aq)+CN^{-}(aq)$ Let $[H^{+}]=x$. Since $[H^{+}]=[CN^{-}]$ (from dissociation) and initial $[HCN]=0.05M$, at equilibrium $[HCN]=0.05 - x$.

Step2: Write Ka expression

$K_{a}=\frac{[H^{+}][CN^{-}]}{[HCN]}$. Given $K_{a}=5.0\times 10^{-10}$, so $5.0\times 10^{-10}=\frac{x\cdot x}{0.05 - x}$. Since $K_{a}$ is very small, $0.05 - x\approx0.05$. Then $5.0\times 10^{-10}=\frac{x^{2}}{0.05}$.

Step3: Solve for x

$x^{2}=5.0\times 10^{-10}\times0.05 = 2.5\times 10^{-11}$. $x=\sqrt{2.5\times 10^{-11}}=\sqrt{25\times 10^{-12}} = 5.0\times 10^{-6}M$.

Answer:

D. $5.0\times 10^{-6}M$