Question

A certain buffer solution equals concentration of $X^{-}$ and $HX$. $K_b$ for $X^{-}$ is ${10}^{-10}$. The $pH$ of buffer is:

Answer 1

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

$K_a=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-10}}={10}^{-4}$

$p{K}_a=-log{K}_a=-log{10}^{-4}=4$

$\left[X^{-}\right]=\left[HX\right]$

$pH=pKa+log\frac{\left[X^{-}\right]}{\left[HX\right]}=4+log1=4+0=4$