Question

$50mL$ of $0.12M$ $H_3{PO}_4+40mL$ of $0.15M$ $NaOH$

Answer 1

The number of mmol of $H_{3}P{O}_4=50mL\times 0.12M=6mmol$
The number of mmol of $NaOH=40mL\times 0.15M=6mmol$
6 mmol of NaOH will neutralise 6 mmol of $H_{3}P{O}_4$ to form 6 mmol of $Na{H}_{2}P{O}_4$.
This correspondce to equivalence point.
$\left[Na{H}_{2}P{O}_4\right]=\frac{6mmol}{(50+40)mL}=0.0667M$
$Na{H}_{2}P{O}_4+H_{2}O\to NaHP{O}_4^{-}+H_3{O}^{+}$

Let x M $Na{H}_{2}P{O}_4$ dissociate to form x M $NaHP{O}_4^{-}$ and x M$H_3{O}^{+}$. $0.0667-x$ M $Na{H}_{2}P{O}_4$ will remain

$K_{a2}=\frac{\left[NaHP{O}_4^{-}\right]\left[H_3{O}^{+}\right]}{\left[Na{H}_{2}P{O}_4\right]}$
$6.2\times {10}^{-8}=\frac{x\times x}{0.0667-x}$
Since $K_{a2}$ is very small, we approximate
$0.0667-x$ to 0.0667.
$6.2\times {10}^{-8}=\frac{x\times x}{0.0667}$
$x=6.4\times {10}^{-5}$
$pH=-lo{g}_{10}\left[H_3{O}^{+}\right]=-lo{g}_{10}x=-lo{g}_{10}6.4\times {10}^{-5}=4.2$