Question

How much water should be added to 10.00 g of acetic acid to give a hydrogen ion concentration equal to $1.0\times {10}^{-3}M?$

$(Given:{pK}_a=4.74)$

• A. $3l$
• B. $230ml$
• C. $30ml$
• D. $330ml$

Answer 1

The correct option is A $3l$

Let mass of water added is $V$ L.

Molarity= $\frac{10}{60V}=\frac{1}{6V}$

$\Rightarrow \left[H^{+}\right]=(K_{a}M{)}^{1/2}$

$\Rightarrow M=\frac{[H^{+}{]}^2}{K_a}$ $\left[H^{+}\right]={10}^{-3};p{K}_a=-\log K_a=4.74$

$\therefore M=\frac{({10}^{-3}{)}^2}{1.8\times {10}^{-5}}$ $\therefore K_a=1.8\times {10}^{-5}$

$\frac{1}{6V}=\frac{{10}^{-6}}{1.8\times {10}^{-5}}\Rightarrow V=3L$