Question

$100$ mL of $0.20M$ weak acid, $HA$ is completely neutralized by $0.20M$ $NaOH$. $K_b$ for $A^{-}$ is ${10}^{-5}$, the $pH$ at the equivalence point is:

• A. $4$
• B. $11$
• C. $9$
• D. $4.4$

Answer 1

The correct option is B $11$

Volume $V$ of $NaOH$ required for neutralization $=\frac{100ml\times 0.20M}{0.20M}=100$ mL

The concentration of $A^{-}$ ions in the solution $=\left[A^{-}\right]=\frac{100\times 0.20}{100+100}=0.10M$

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

$0.1$ $0$ $0$

$0.1-x$ $x$ $x$

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

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

$\frac{x\times x}{0.10-x}={10}^{-5}$

$0.10-x\simeq 0.10$

$\frac{x^2}{0.10}={10}^{-5}$

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

$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{{10}^{-3}}={10}^{-11}$

$\text{pH=-log}\left[H^{+}\right]=\text{-log}{10}^{-11}=11$

Hence, the correct option is $\text{B}$