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? $\left(\right.pK_{b}=4.7447\left.\right)$

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

Answer 1

Answer: (A)

$\underset{4 \, m m o l}{H N O_{3}}+\underset{6 \, m m o l}{N H_{3}} \rightarrow \underset{4 \, m m o l}{N H_{4} N O_{3}}$
[Since $W_{B}\left(\right.NH_{3}\left.\right)$ is left and salt $\left(\right.NH_{4}NO_{3}\left.\right)$ 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 $=pK_{b}+log_{10}\frac{\left[\right. S a l t \left]\right.}{\left[\right. B a s e \left]\right.}$
So, pOH $=4.7447+log4/2=5.0457$
and pH = 14 - 5.0457 = 8.95.