Question

$40mL$ of $0.12M$ $H_3{PO}_4+40mL$ of $0.18M$ $NaOH$ (write the value to the nearest integer)

Answer 1

In mixing of $H_{3}P{O}_4$ and $NaOH$, there will be there equivalence points owing to tribasic nature of $H_{3}P{O}_4$ millimoles of $O{H}^{-}=40\times 0.18=7.2$

Millimoles of $H^{+}=3\times 40\times 0.12=4.8\times 3=14.4$

$1^{st}$ equivalence point will be reached when $4.8$ $mmol$ of acid will react with $4.8$ $mmol$ of base.

Hence, $7.2-4.8=2.4$ $mmol$ of base is left.

$2^{nd}$ equivalence point will have $4.8$ $mmol$ of base reacting with $4.8$ $mmol$ of acid.

But $mmol$ of base left $=2.4$ $mmol$

$\therefore mmol$ of salt formed $=2.4$ mmol= mmol$ of acid left.

$pH={p{K}_a}_2+\log \frac{\left[salt\right]}{\left[acid\right]}$

$\because \left[salt\right]=\left[acid\right]$ [Volume same for both]

$pH={p{K}_a}_2=7.2$

$\therefore$ Nearest integer $=7$