Question

$40mL$ of $0.10M$ $H_3{PO}_4+40mL$ of $0.25M$ $NaOH$

Answer 1

At $1^{st}$ equivalence point, $40\times 0.1=4$ $mmol$ of $H_{3}P{O}_4$ will react with $4$ $mmol$ of $NaOH$.

Same will be the case for $2^{nd}$ equivalence point.

At $3^{rd}$ equivalence point, however, $4$ $mmol$ of $NaOH$ is needed but only $2$ $mmol$ are left.

Hence, $2$ $mmol$ of salt will be formed.

Similarly, $4-2=2$ $mmol$ of acid will be left.

$pH={p{K}_a}_3+\log \frac{\left[salt\right]}{\left[acid\right]}$

$⟹pH={p{K}_a}_3=12\{\therefore [salt]=[acid\left]\right\}$