Question

Calculate the pH of that buffer soln which is obtained by dissolving 0.2 moles of $N{H}_{4}OH$ & 0.25 mole of $N{H}_{4}Cl$ in water.
$K_b=1.8\times {10}^{–5}$

• A. 4.85
• B. 9.15
• C. 3.5
• D. no option

Answer 1

The correct option is B 9.15

Given: - 0.2 moles ${NH}_4$OH dissolved with

0.25 mole $NH4Cl$ in water.

To calculate:-pH of buffer obtained

$k_b=1.8\times {10}^{-5}$

According to Henderson - Hasel balch equation

$pOH=p{k}_b+\log \left[\frac{\text{salt}}{\text{base}}\right]$

$\therefore p{k}_b=\log \left(k_b\right)=\log (1.8\times {10}^{-5})=4.745$

$\begin{array}{cc}\text{So,}pOH& =4.745+\log \left[\frac{0.25}{0.2}\right]\\ & p_{OH}=4.745+0.0969=4.8419\end{array}$

$\therefore pH=14-pOH=14-4.8419=9.1581$

$50,$ B) 9.15 is correct answer.