Question

Find the $pH$ of a $500mL$ solution containing 2.5 moles of $KCN$ and 2.5 moles of $HCN$ at ${25}^{o}C$.
Given:The $p{K}_b$ of cyanide ion $\left(C{N}^{-}\right)$ is 4.7

• A. 9.3
• B. 7.3
• C. 10.3
• D. 8.3

Answer 1

The correct option is A 9.3

Milli-moles of salt $KCN=2.5\times 500=1250$
Milli-moles of $HCN=2.5\times 500=1250$
Therefore in the buffer solution [Anion of Salt] = [Acid]
So, this quantity is the ratio of concentration of the anion of the salt and the acid present in the mixture will be equal to 1.
$\frac{\left[\text{Anion of Salt}\right]}{\left[\text{Acid}\right]}=1$
$pH=p{K}_a+{\log }_{10}\left(\frac{\left[\text{Anion of Salt}\right]}{\left[\text{Acid}\right]}\right)$
or, $pH=p{K}_a$
$p{K}_a+p{K}_b=14$ at ${25}^{o}C$
$p{K}_{a}ofHCN=14-p{K}_b=14-4.7=9.3pH=p{K}_a=9.3$