Question

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

Answer 1

Volume V of NaOH required for neutralization $=\frac{100ml\times 0.20M}{0.20M}=100ml$
The concentration of $A^{-}$ ions in the solution is $\left[A^{-}\right]=\frac{100\times 0.20}{100+100}=0.10M$
$A^{-}+HOH\rightleftharpoons HA+O{H}^{-}$
0.1 M 0 0
$0.1-x$ x x
$K_b={10}^{-5}$
$\frac{\left[O{H}^{-}\right]\left[HA\right]}{\left[A^{-}\right]}=K_a$
$\frac{x\times x}{0.10-x}={10}^{-5}$
$0.10-x\simeq 0.10$
$\frac{x^2}{0.10}={10}^{-5}$
$\left[O{H}^{-}\right]=x={10}^{-3}$
$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{{10}^{-3}}={10}^{-11}$
$pH=-log\left[H^{+}\right]=-log{10}^{-11}=11$