Two solutions A and B, each of 100 ; L was made by dissolving 4 ;… - Vidyadip

MCQ

Two solutions $A$ and $B$, each of $100\; L$ 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 $\mathrm{B}$ is

A
$7.3$
B
$8.7$
C
$9.6$
✓
$10.6$

✓

Answer

Correct option: D.

$10.6$

d
$4 \mathrm{gm}$ of $\mathrm{NaOH}$ in $100 \mathrm{L}$ sol. $\Rightarrow 10^{-3} \mathrm{M}$ sol.

$9.8 \mathrm{gm}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $100 \mathrm{L}$ sol. $\Rightarrow 10^{-3} \mathrm{M}$ sol.

Mixture : $40 L$ of $10^{-3} \mathrm{M} \mathrm{NaOH}$ and $10 \mathrm{L}$ of

$10^{-3} \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ sol.

Final Conc. of $OH^-=\frac{10^{-3}(40 \times 1-10 \times 1 \times 2)}{40+10}=6 \times 10^{-4} \mathrm{M}$

$\mathrm{pOH}=-\log \left(6 \times 10^{-4}\right)$

$=4-\log 6=4-0.60=3.40$

$\mathrm{pH}=14-3.40=10.60$

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