Answer
(a) $pH = 7.03$
Work Step by Step
$H_2{PO_4}^-(aq) + H_2O(l) \lt -- \gt HP{O_4}^{2-}(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[H_2{PO_4}^-] = 0.15 M - x$
$[HP{O_4}^{2-}] = 0.10M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 6.2\times 10^{- 8} = \frac{[HP{O_4}^{2-}][H_3O^+]}{[H_2{PO_4}^-]}$
$ 6.2\times 10^{- 8} = \frac{( 0.1 + x )* x}{ 0.15 - x}$
Considering 'x' has a very small value.
$ 6.2\times 10^{- 8} = \frac{ 0.1 * x}{ 0.15}$
$ 6.2\times 10^{- 8} = 0.67x$
$\frac{ 6.2\times 10^{- 8}}{ 0.67} = x$
$x = 9.3 \times 10^{- 8}$
Percent dissociation: $\frac{ 9\times 10^{- 8}}{ 0.15} \times 100\% = 6.2\times 10^{- 5}\%$
x = $[H_3O^+]$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 9.3 \times 10^{- 8})$
$pH = 7.03$
