Question

An unknown volume and unknown concentration of a weak acid $HX$ is titrated with $NaOH$ of unknown concentration. After addition of $10.0{\text{cm}}^3$ of the $NaOH$ solution, the $pH$ of the solution is 5.7. After the addition of $20.0{\text{cm}}^3$ of the $NaOH$ solution, the $pH$ is $6.3$. Calculate the $p{K}_a$ of the weak acid, HX. (Given: log2 = 0.3, log = 0.5)

Answer 1

Let the molarity and volume of $HX$ be $M_1$ and $V_1$ respectively, For $NaOH$ assume the molarity to be $M_2$
Now the reaction involving the titration will be
$HX+NaOH\to NaX+H_{2}O$
Since the complete neutralization has not taken place, the solution will act as an acidic buffer:
$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$
When 10 ml of $NaOH$ is used
$5.7=p{K}_a+log\frac{\left[10M_2\right]}{[M_1{V}_1-10M_2]}$
Lets say $V_2$ volume of $NaOH$ is needed for complete neutralization and $M_1{V}_1=M_2{V}_2$, so
$5.7=p{K}_a+log\frac{\left[10M_2\right]}{[M_2{V}_2-10M_2]}$
$\therefore 5.7=p{K}_a+log\frac{\left[10\right]}{[V_2-10]}$ (i)
similarly for 20 ml $NaOH$ used
$6.3=p{K}_a+log\frac{\left[20\right]}{[V_2-20]}$ (ii)
Now from (i) and (ii) we have
$6.3-5.7=log\frac{20(V_2-10)}{10(V_2-20)}$
$2\times 0.3=2log2=log4=log\frac{2(V_2-10)}{1(V_2-20)}$
Hence $V_2=30ml$;
$5.7=p{K}_a+log\frac{\left[10\right]}{[V_2-10]}$
$5.7=p{K}_a+log\frac{\left[10\right]}{\left[20\right]}$
$5.7=p{K}_a-log2$
$\therefore p{K}_a=5.7+0.3=6$