Question

The $p{{K}_{a}}$ of HCN is 9.3. The pH of a solution prepared by mixing 2.5 moles of HCN and 0.25 moles of KCN in water and making of the total volume to 500 ml is:
A. 9.3
B. 7.3
C. 10.3
D. 8.3

Answer 1

There is a formula to calculate the pH of the solution by using $p{{K}_{a}}$ and it is as follows.
$pH=p{{K}_{a}}+\log \dfrac{[salt]}{[acid]}$
pH = pH of the solution
$p{{K}_{a}}$ = acid ionization constant
[salt] = concentration of the salt
[acid] = concentration of the acid

Complete step by step solution:
- In the question it is given that 2.5 moles of HCN and 0.25 moles of KCN in water and made the solution to 500 ml by using water.
- We have to calculate the pH of the resulting solution.
- It is given that Acid ionization constant $p{{K}_{a}}$ = 9.3
- Concentration of the salt (KCN) = 0.25 moles
- Concentration of the acid (HCN) = 2.5moles
- Substitute all the known values in the below formula to get the pH of the resulting solution.
$pH=p{{K}_{a}}+\log \dfrac{[salt]}{[acid]}$
pH = pH of the solution
$p{{K}_{a}}$ = acid ionization constant = 9.3
[salt] = concentration of the salt = 0.25 moles
[acid] = concentration of the acid = 2.5 moles

& pH=p{{K}_{a}}+\log \dfrac{[salt]}{[acid]} \\\ & pH=9.3+\log \dfrac{0.25}{2.5} \\\ & pH=8.3 \\\ \end{aligned}$$ \- Therefore the pH of the resulting mixture is 8.3. **So, the correct option is D.** **Note:** $HCN$ (hydrocyanic acid) is a weak acid and $KCN$ (potassium cyanide) is salt. By mixing these two we will get an acidic buffer. $KCN$ is going to be formed by the reaction of $HCN$ with $KOH$ and it is as follows. $$HCN+KOH\to KCN+{{H}_{2}}O$$