Question

How many gram of $H_{2}S{O}_4$ should be removed from $1L$ of its aqueous solution of $pH=2$ to increase its $pH$ upto $3$ .

• A. 49 g
• B. 9.8 g
• C. 0.49 g
• D. 0.441 g

Answer 1

Answer: (D)

I case
$pH=2$
$\therefore \left[H^{+}\right]=10^{-2}M$
$\therefore \left[H_{2}S{O}_4\right]=\frac{10^{-2}}{2}M$
II case
$pH=3$
$\therefore \left[H^{+}\right]=10^{-3}M$
$\therefore \left[H_{2}S{O}_4\right]=\frac{10^{-3}}{2}M$
$\therefore n_{H_{2}S{O}_4}=\frac{10^{-2}}{2}\times 1mol$
$\therefore n_{H_{2}S{O}_4}=\frac{10^{-3}}{2}\times 1mol$
Now ATQ,
$\frac{10^{-2}}{2}-n_{H_{2}S{O}_4}=\frac{10^{-3}}{2}$
$\therefore n_{H_{2}S{O}_4}=\frac{1}{2}\times 10^{-2}-\frac{1}{2}\times 10^{-3}$
$=\frac{1}{2}\times 0.009mol$
$\therefore w_{H_{2}S{O}_4}=\frac{1}{2}\times 0.009\times 98g$
$=0.009\times 49g$
$=0.441g$