Question

Ionisation constant of ${CH}_{3}COOH$ is $1.7\times {10}^{-5}$ and concentration of $H^{+}$ ions is $3.4\times {10}^{-4}$.Then find out initial concentration of ${CH}_{3}COOH$ molecule.

• A. $3.4\times {10}^{-4}$
• B. $3.4\times {10}^{-3}$
• C. $6.8\times {10}^{-4}$
• D. $6.8\times {10}^{-3}$

Answer 1

Answer: (D)

$6.8\times {10}^{-3}$

The ionization reaction for $C{H}_{3}COOH$follows:

$C{H}_{3}COOH\to C{H}_{3}CO{O}^{-}+H^{+}$

At eqm. $x$ $3.4\times {10}^{-4}$ $3.4\times {10}^{-4}$

The ionization constant for the above reaction is written as:

$k=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

Putting values in above equation, we get:

$1.7\times {10}^{-5}=\frac{(3.4\times {10}^{-4})(3.4\times {10}^{-4})}{x}$

$x=6.8\times {10}^{-3}$

Hence, option D is correct.