Answer
$[HCN] \approx 0.025M$
$[CN^-] \approx 0.010M$
$[H_3O^+] = 1.2\times 10^{- 9}M$
$[OH^-] = 8.3 \times 10^{-6} M$
$pH = 8.91$
Percent dissociation: $4.9\times 10^{- 6}\%$
Work Step by Step
1. Drawing the ICE table we get these concentrations at the equilibrium:
$HCN(aq) + H_2O(l) \lt -- \gt CN^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[HCN] = 0.025 M - x$
$[CN^-] = 0.01M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 4.9\times 10^{- 10} = \frac{[CN^-][H_3O^+]}{[HCN]}$
$ 4.9\times 10^{- 10} = \frac{( 0.01 + x )* x}{ 0.025 - x}$
Considering 'x' has a very small value.
$ 4.9\times 10^{- 10} = \frac{ 0.01 * x}{ 0.025}$
$ 4.9\times 10^{- 10} = 0.4x$
$\frac{ 4.9\times 10^{- 10}}{ 0.4} = x$
$x = 1.2\times 10^{- 9}$
Percent dissociation: $\frac{ 1.2\times 10^{- 9}}{ 0.025} \times 100\% = 4.9\times 10^{- 6}\%$
x = $[H_3O^+]$
$[HCN] = 0.025M - 1.2\times 10^{- 9}M \approx 0.025M$
$[CN^-] = 0.010M - 1.2\times 10^{- 9}M \approx 0.010M$
$[H_3O^+] = 1.2\times 10^{- 9}M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.2 \times 10^{- 9})$
$pH = 8.91$
4. Calculate the hydroxide ion concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$1.2 \times 10^{-9} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{1.2 \times 10^{-9}}$
$[OH^-] = 8.3 \times 10^{-6} M$
