Question

$0.1M$ of $HA$ is titrated with $0.1MNaOH$, calculate the pH at end point. Given, $K_a(HA)=5\times 10^{-6}\alpha <<1$.

Answer 1

At the end-point, $[A^{-}]=0.05$
$K_b(A^{-})=K_w/K_a=2\times 10^{-9}$
$[O{H}^{-}]=\sqrt{K_{b}C}=\sqrt{2\times 10^{-9}\times 0.05}=10^{-5}$
$pOH=5$ and $pH=9$