Two solutions, A and B, each of


Two solutions, $A$ and $B$, each of $100 \mathrm{~L}$ was made by dissolving $4 \mathrm{~g}$ of $\mathrm{NaOH}$ and $9.8 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in water, respectively. The $\mathrm{pH}$ of the resultant solution obtained from mixing $40 \mathrm{~L}$ of solution $A$ and $10 \mathrm{~L}$ of solution. $B$ is _________________


(10.60) $\mathrm{M}_{\mathrm{H}_{2} \mathrm{SO}_{4}}=\frac{9.8}{98 \times 100}=10^{-3} \mathrm{M}$

$\mathrm{M}_{\mathrm{NaOH}}=\frac{4}{40 \times 100}=10^{-3} \mathrm{M}$

After neutralisation $\left[\mathrm{OH}^{-}\right]$can be calculated as

$\left[\mathrm{OH}^{-}\right]=\frac{\left(40 \times 10^{-3}\right)-\left(2 \times 10^{-3} \times 10\right)}{50}$

$=\frac{20}{50} \times 10^{-3}$

$\left[\mathrm{OH}^{-}\right]=\frac{2}{5} \times 10^{-3}$




Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now