Question

$pH$ of $0.5M$ aqueous solution of $HF$ $(K_a=2\times {10}^{-3})$ is:

• A. $1.5$
• B. $4$
• C. $6$
• D. $10$

Answer 1

The correct option is A $1.5$

$K_a=2\times {10}^{-3}$

We know that,

$K_a=\frac{\left[H^{+}\right]\left[F^{-}\right]}{\left[HF\right]}$

$\Rightarrow 2\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{0.5-x}$

$\Rightarrow 2\times {10}^{-3}=\frac{x^2}{0.5-x}$

$\Rightarrow x^2=(0.5-x)(2\times {10}^{-3})$

$\Rightarrow x^2=0.5\times 2\times {10}^{-3}$ [If $x$ is small]

$\Rightarrow x^2=1\times {10}^{-3}$

$\Rightarrow x=0.031623$

$pH=-log\left[H^{+}\right]$

$=-log\left[0.031623\right]$

$=1.499$

$pH$ of $0.5M$ aqueous solution of $HF$ is $1.499$.