Question

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

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

Answer 1

Answer: (D)

0.441 g

Given that

$p{H}_1=2\Rightarrow -\log \left[H^{+}\right]=2$

$\log \left[H^{+}\right]=-2$

$\left[H^{+}\right]={10}^{-2}M$

$H_{2}S{O}_4⟶2H^{+}+S{O}_4^{2-}$

$[H_{2}S{O}_4{]}_1=\frac{\left[H^{+}\right]}{2}=\frac{{10}^{-2}}{2}M$

$=0.005M=M_1$

$p{H}_2=3\Rightarrow [H^{+}{]}_2={10}^{-3}M$

$[H_{2}S{O}_4{]}_2=\frac{[H^{+}{]}_2}{2}=\frac{{10}^{-3}}{2}M$

$=0.0005M=M_2$

$\therefore$ Moles of $H_{2}S{O}_4$ removed, $\Delta n$ $=M_{1}V-M_{2}V$

$=(M_1-M_2)V$

$(0.005-0.0005)\times 1$

$=4.5\times {10}^{-3}moles$

$\therefore$ Moles mass of $H_{2}S{O}_4\left(M\right)=98g/mol$

$\therefore$ Mass of acid removed $=\Delta n\times M$

$=4.5\times {10}^{-3}\times 98g$

$=0.441g$