Question

A week monobasic (0.1 M) has a pH of 3 at particular temperature $\left({25}^{0}C\right)$. When this acid is neutralised by strong base (NaOH), what is the value of equilibrium constant at equivalent point at ${25}^{0}C$.

• A. ${10}^{-5}$
• B. ${10}^9$
• C. ${10}^{-7}$
• D. ${10}^{-14}$

Answer 1

The correct option is B ${10}^9$

Given,

Base - $NaOH$

pH - $3$

Let weak monobasic - $HA$

$HA+NaOH\to H_{2}O+A^{-}+N{a}^{+}$

We can write $O{H}^{-}$ instead of $NaOH$

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

$K_{eq}=\frac{\left[A^{-}\right]}{\left[HA\right]\left[O{H}^{-}\right]}$

When we dissociate $HA$

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

equilibrium of this is $K_a$

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

pH = $3$

pH - $-\log \left[H^{+}\right]$

$-\log \left[H^{+}\right]=3$

$\left[H^{+}\right]={10}^{-3}$

$\left[H^{+}\right]=\left[A^{-}\right]={10}^{-3}$

$\left[HA\right]=0.1$

$K_a=\frac{{10}^{-3}\times {10}^{-3}}{0.1}={10}^{-5}$

ionic constant $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

$\frac{K_a}{K_w}=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}\times \frac{1}{\left[H^{+}\right]\left[O{H}^{-}\right]}$

= $\frac{\left[A^{-}\right]}{\left[HA\right]\left[O{H}^{-}\right]}$ = $K_{eq}$

$K_{eq}=\frac{K_a}{K_w}$

$K_a={10}^{-5}$

$K_w={10}^{-14}$

$K_{eq}=\frac{{10}^{-5}}{{10}^{-14}}={10}^9$

So, option (B) is the correct answer.