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Q.
How many gram of $H _{2} SO _{4}$ should be removed from $1\, L$ of its aqueous solution of $pH =2$ to increase its $pH$ upto $3$ .
Equilibrium
Solution:
I case
$pH =2$
$\therefore \left[ H ^{+}\right]=10^{-2} M$
$\therefore \left[ H _{2} SO _{4}\right]=\frac{10^{-2}}{2} M$
II case
$pH =3$
$\therefore \left[ H ^{+}\right]=10^{-3} M $
$\therefore \left[ H _{2} SO _{4}\right]=\frac{10^{-3}}{2} M $
$\therefore n _{ H _{2} SO _{4}}=\frac{10^{-2}}{2} \times 1 mol$
$\therefore n _{ H _{2} SO _{4}}=\frac{10^{-3}}{2} \times 1 mol$
Now ATQ,
$\frac{10^{-2}}{2}- n _{ H _{2} SO _{4}}=\frac{10^{-3}}{2}$
$\therefore n _{ H _{2} SO _{4}}=\frac{1}{2} \times 10^{-2}-\frac{1}{2} \times 10^{-3}$
$=\frac{1}{2} \times 0.009\, mol$
$\therefore w _{ H _{2} SO _{4}}=\frac{1}{2} \times 0.009 \times 98\, g$
$=0.009 \times 49\, g$
$=0.441\, g$