Question

The degree of ionization of $0.10M$ lactic acid is $4.0\%$.
$H_{3}C-\stackrel{\stackrel{H}{|}}{\underset{\underset{OH\left(aq\right)}{|}}{C}}-COOH\rightleftharpoons H_{\left(aq\right)}^{+}+H_{3}C-\stackrel{\stackrel{H}{|}}{\underset{\underset{OH\left(aq\right)}{|}}{C}}-{COO}^{-}$

The value of $K_c$ is:

• A. $1.66\times {10}^{-5}$
• B. $1.66\times {10}^{-4}$
• C. $1.66\times {10}^{-3}$
• D. $1.66\times {10}^{-2}$

Answer 1

The correct option is B $1.66\times {10}^{-4}$

$HA\rightleftharpoons H^{+}+A^{-}$

The initial concentrations are as follows:

$\left[HA\right]=0.10$ M

$\left[H^{+}\right]=\left[A^{-}\right]=0$ M

$HA$ is $4.0\%$ dissociated.

$4.0\%$ of $0.10M$ corresponds to $\frac{0.10\times 4.0}{100}=0.0040M$

The equilibrium concentrations are:

$\left[HA\right]=0.10-0.0040=0.096$ M

$\left[H^{+}\right]=\left[A^{-}\right]=0.0040$ M

The equilibrium constant expression is:

$K_c=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$K_c=\frac{0.0040\times 0.0040}{0.096}$

$K_c=1.66\times {10}^{-4}$