Question

The concentration of acetic acid $(K_a=1.8\times {10}^{-5})$ required to give $3.5\times {10}^{-4}mol/L{H}^{+}$, is

• A. $6.8mol/L$
• B. $6.8\times {10}^{-3}mol/L$
• C. $1.94\times {10}^{-1}mol/L$
• D. $194mol/L$

Answer 1

The correct option is B $6.8\times {10}^{-3}mol/L$

${CH}_{3}COOH\rightleftharpoons {CH}_3{COO}^{-}+H^{+}$

$K_a=\frac{\left[H^{+}\right]\left[{CH}_3{COO}^{-}\right]}{\left[{CH}_{3}COOH\right]}$

Since 1 mole of ${CH}_{3}COOH$ gives 1 mole of ${CH}_3{COO}^{-}$ and 1 mole of $H^{+}$ ions.

$\left[{CH}_3{COO}^{-}\right]=\left[H^{+}\right]$

$K_a=\frac{{\left[H^{+}\right]}^2}{\left[{CH}_{3}COOH\right]}$

$\because$ $K_a=1.8\times {10}^{-5},H^{+}=3.5\times {10}^{-4}mol/L$

$1.8\times {10}^{-5}=\frac{{[3.5\times {10}^{-4}]}^2}{\left[{CH}_{3}COOH\right]}$

$\left[{CH}_{3}COOH\right]=\frac{12.25\times {10}^{-8}}{1.8\times {10}^{-5}}$

$=6.8\times {10}^{-3}mol/L$