Question

Calculate the change in $pH$ when $4$ gm of solid $NaOH$ & $10$ mmol of $H_{2}S{O}_4$ are added to a solution of volume $5$ litre, which was prepared by mixing $20$ mmol of $HCl,40$ mmol of $H_{2}S{O}_4$ and $2$ gm of $NaOH$ and sufficient water. Ignore the volume change.

$(log2=0.3,log3=0.48)$

• A. $9.78$
• B. $5.26$
• C. $4.86$
• D. $8.84$

Answer 1

The correct option is A $9.78$

Initial solution contains $20$ $mmol$ of $HCl$, $40$ $mmol$ of $H_{2}S{O}_4$ and $2$ $gm$ i.e., $50$ $mmol$ of $NaOH$

Total $mmol$ of $H^{+}=20+2\times 40=100$ $mmol$

Total $mmol$ of $O{H}^{-}=50$ $mmol$

$\therefore mmol$ of $H^{+}$ left $=100-50=50$ $mmol$

Volume $=5$ $L$

$\therefore \left[H^{+}\right]=\frac{50\times {10}^{-3}}{5}={10}^{-2}$

$\therefore pH$ initial $=2$

$4$ $gm$ i.e., $100$ $mmol$ of $NaOH$ and $10$ $mmol$ of $H_{2}S{O}_4$ are added.

$\therefore$ Total $mmol$ of $O{H}^{-}=100$ $mmol$

Total $mmol$ of $H^{+}=mmol$ of $H^{+}$ left $+mmol$ of $H^{+}$ added $=50+2\times 10=70$ $mmol$

$\therefore mmol$ of $O{H}^{-}$ left $=100-70=30$ $mmol$

$\left[O{H}^{-}\right]=\frac{30}{5}\times {10}^{-3}=6\times {10}^{-3}$ $M$

$\therefore pOH=-\log (6\times {10}^{-3})=2.22$

$\therefore pH=14-pOH=11.78$

$\therefore$ Final $pH=11.78$

$\therefore$ Change in $pH=11.78-2=9.78$