Chemistry - Ionic Equilibrium Question with Solution | TestHub

Given $K_b$ for $X^{-}=10^{-10}$.

For a conjugate acid-base pair at $298$ K, $K_a\times K_b=K_w=10^{-14}$.

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

The $p{K}_a$ of the weak acid $HX$ is:

$p{K}_a=-\log (K_a)=-\log (10^{-4})=4$

Using the Henderson-Hasselbalch equation for an acidic buffer:

$\text{pH}=p{K}_a+\log \left(\frac{[X^{-}]}{[HX]}\right)$

Since the concentrations of $X^{-}$ and $HX$ are equal, $[X^{-}]=[HX]$.

$\text{pH}=4+\log (1)=4+0=4$

Answer: $4$