Question

40 mL sample of 0.1 M solution of nitric acid is added to 20 mL of 0.3 M aqueous ammonia. What is the pH of the resulting solution? $(p{K}_b=4.7447)$

• A. 8.95
• B. 89.5
• C. 0.895
• D. 4.48

Answer 1

Answer: (A)

$\underset{4mmol}{HN{O}_3}+\underset{6mmol}{N{H}_3}\to \underset{4mmol}{N{H}_{4}N{O}_3}$
[Since $W_B(N{H}_3)$ is left and salt $(N{H}_{4}N{O}_3)$ is formed so it is a basic buffer]
Base left = 6 - 4 = 2 mmol
[Acid] = $40\times 0.1$ = 4 mmol
[Base] $=20\times 0.3=6$
pOH $=p{K}_b+lo{g}_{10}\frac{[Salt]}{[Base]}$
So, pOH $=4.7447+log4/2=5.0457$
and pH = 14 - 5.0457 = 8.95.