Question

Calculate the pH of a solution containing 0.01M HF and 0.002M NaF if the Ka=7.2x10^-4. Calculate the pH if 0.002M of KOH is added to the buffer.

Answer 1

Molarity of $\text{HF}=0.01M$

Molarity of $\text{NAF}=0.002M$

$Ka=7.2\times 10^{-4}$

$P_{k_a}=-log(7.2\times 10^{-4})=4-0.85=3.15$

Then The $P_H$ of the solution formed is given by the following formula:-

$P_H=P_{k_a}+log\dfrac{[\text{salt}]}{[\text{Acid]}}$

$=3.15+log\dfrac{[0.002]}{[0.01]}$

$=3.15+log(0.2)$

=3.15+0.301

=3.451

Now When KOH is added to the buffer KOH dissassociates to form

$KOH\rightarrow K^++OH^-$

$P_{OH}=-log([OH]^{-1})=-log(0.002)=2.7$

Also, $P_H+P_{OH}=14,P_H=14-2.7=11.3$

Hence the $P_{H}$ is 11.3