Answer
$pH = 7.23$
Work Step by Step
1. Calculate the molar mass $(NaOH)$:
22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol
2. Calculate the number of moles $(NaOH)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.425}{ 40}$
$n(moles) = 0.0106$
3. Find the concentration in mol/L $(NaOH)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.0106}{ 2} $
$C(mol/L) = 5.31\times 10^{- 3}$
4. Drawing the ICE table, we get these concentrations at the equilibrium:
$H_2P{O_4}^-(aq) + H_2O(l) \lt -- \gt HP{O_4}^{2-}(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[H_2P{O_4}^-] = 0.132 M - x$
$[HP{O_4}^{2-}] = 0.132M + x$
$[H_3O^+] = 0 + x$
5. Calculate 'x' using the $K_a$ expression.
$ 6.3\times 10^{- 8} = \frac{[HP{O_4}^{2-}][H_3O^+]}{[H_2P{O_4}^-]}$
$ 6.3\times 10^{- 8} = \frac{( 0.132 + x )* x}{ 0.132 - x}$
Considering 'x' has a very small value.
$ 6.3\times 10^{- 8} = \frac{ 0.132 * x}{ 0.132}$
$ 6.3\times 10^{- 8} = 1x$
$\frac{ 6.3\times 10^{- 8}}{ 1} = x$
$x = 6.3\times 10^{- 8}$
Percent dissociation: $\frac{ 6.3\times 10^{- 8}}{ 0.132} \times 100\% = 4.77\times 10^{- 5}\%$
x = $[H_3O^+]$
$[H_2P{O_4}^-] = 0.132 M - x = 0.132 M - 6.3 \times 10^{-8}M \approx 0.132M$
$[HP{O_4}^{2-}] = 0.132M + x = 0.132 M + 6.3 \times 10^{-8}M \approx 0.132M$
$[H_3O^+] = 0 + x = 6.3 \times 10^{-8}M$
6. Since we are adding a strong base, this reaction will occur:
$H_2P{O_4}^-(aq) + OH^-(aq) -- \gt HP{O_4}^{2-}(aq) + H_2O(l)$
And these are the concentrations after this reaction:
Remember: Reactants at equilibrium = Initial Concentration - y
And Products = Initial Concentration + y
Since $NaOH$ is a strong base, y = $[NaOH] = 5.31 \times 10^{-3}M$
$[H_2P{O_4}^-] = 0.132 M - 5.31 \times 10^{-3} = 0.127M$
$[HP{O_4}^{2-}] = 0.132M + 5.31 \times 10^{-3} = 0.137M$
7. Now, calculate the hydronium ion concentration after the addition of the $NaOH$:
$[H_3O^+] = Ka * (\frac{[H_2P{O_4}^-]}{[HP{O_4}^{2-}]})$
$[H_3O^+] = 6.3 \times 10^{-8} * \frac{0.127}{0.137}$
$[H_3O^+] = 6.3 \times 10^{-8} * 0.927$
$[H_3O^+] = 5.84 \times 10^{-8}$
8. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 5.84 \times 10^{- 8})$
$pH = 7.23$
