Question

Calculate the pH of 500 ml of 1M hydrazoic acid $\left(H{N}_3\right)$. $(K_a=2.5\times {10}^{-5},\text{log 5}=0.7)$

• A. 2.1
• B. 2.2
• C. 2.3
• D. 2.4

Answer 1

The correct option is C 2.3

$\begin{array}{ccccc}H{N}_3& \rightleftharpoons & H^{+}& +& N_3^{-}\\ 1M& & 0& & 0\\ [1-\alpha ]& & \alpha & & \alpha \end{array}$
$\frac{{\alpha }^2}{1-\alpha }=K_a\because \alpha <<1$
${\alpha }^2=2.5\times {10}^{-5}\Rightarrow \alpha =5\times {10}^{-3}$
$\left[H^{+}\right]=5\times {10}^{-3}\therefore pH=-log[H{]}^{+}\Rightarrow pH=3-\text{log 5}$
$\Rightarrow pH=2.3$