Question

If the dissociation constant of a weak acid is $1.0\times {10}^{-5}$, then the equilibrium constant for the reaction of the acid with a strong base is:

• A. $1.0\times {10}^{-5}$
• B. $1.0\times {10}^{-9}$
• C. $1.0\times {10}^9$
• D. $1.0\times {10}^{14}$

Answer 1

The correct option is C $1.0\times {10}^9$

The equilibrium reaction for the dissociation of the weak acid is as follows:

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

The expression for the dissociation constant $\left(K_a\right)$ is as given below:

$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$ $.....\left(i\right)$

The reaction for the neutralization of weak acid $HA$ with strong base $O{H}^{-}$ is as follows:

$HA+O{H}^{-}\rightleftharpoons A^{-}+H_{2}O$

The expression for the equilibrium constant is as given below:

$K=\frac{\left[A^{-}\right]}{\left[HA\right]\left[O{H}^{-}\right]}$ $......\left(ii\right)$

On dividing equation $\left(i\right)$ by equation $\left(ii\right)$, we get

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

Hence, $\frac{K_a}{K}=\left[H^{+}\right]\left[O{H}^{-}\right]=K_w={10}^{-14}$

Thus, $K=\frac{K_a}{K_w}=\frac{{10}^{-5}}{{10}^{-14}}={10}^9$

Hence, the equilibrium constant for the reaction of the acid with a strong base is $1.0\times {10}^9$.