Question

$pH$ of mixture of $HA$ and $A^{-}$buffer is $5.K_b$ of $A^{-}=10^{-10}$. Hence $[HA]/\left[A^{-}\right]$will be:

• A. 1
• B. 10
• C. 0.1
• D. 100

Answer 1

Answer: (C)

$K_b\left(A^{-}\right)=10^{-10}$
$p{K}_b=10$
$p{K}_a(HA)=14-10=4$
$pH=p{K}_a+\log \frac{\left[A^{-}\right]}{[HA]}$
$5=4+\log \frac{\left[A^{-}\right]}{[HA]}$
$1=\log \frac{\left[A^{-}\right]}{[HA]}$
$\therefore \frac{\left[A^{-}\right]}{[HA]}=10$
$\frac{[HA]}{\left[A^{-}\right]}=\frac{1}{10}=0.1$