Question

The value of dissociation constant for a certain weak acid is $1.0\times {10}^{-4}$. The equilibrium constant for its reaction with a strong base is:

• A. $1.0\times {10}^{-4}$
• B. $1.0\times {10}^{-10}$
• C. $1.0\times {10}^{10}$
• D. $1.0\times {10}^{-14}$

Answer 1

The correct option is C $1.0\times {10}^{10}$

The expression for the dissociation of weak acid $\left(HA\right)$ is given below:
$HA\rightleftharpoons H^{+}+A^{-}$

The expression for its acid dissociation constant is given below:
$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$ $.....\left(i\right)$

The expression for the equilibrium reaction of weak acid $\left(HA\right)$ with strong base $\left(B^{+}\right)$ is as given below:
$HA+B^{+}+O{H}^{-}\rightleftharpoons B^{+}+A^{-}+H_{2}O$

The expression for the equilibrium constant is as follows:
$K_{eq}=\frac{\left[H_{2}O\right]\left[A^{-}\right]}{\left[HA\right]\left[O{H}^{-}\right]}$ $....\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, it can be concluded that
$\frac{K_{eq}}{K_a}=\frac{\left[H_{2}O\right]}{\left[H^{+}\right]\left[O{H}^{-}\right]}=\frac{1}{K_w}$

Hence, $K_{eq}=\frac{K_a}{K_w}=\frac{{10}^{-4}}{{10}^{-14}}=1\times {10}^{10}$

Therefore, the value of equilibrium constant for a weak acid in its reaction with a strong base is $1.0\times {10}^{10}$.