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Oxyacids of Phosphorus

In the titrat...

Question

In the titration of a solution of a weak acid $H X$ with $N a O H$ , the $p H$ is $5.8$ after $10.0 m L$ of $N a O H$ solution has been added and $6.402$ after $20.0 m L$ of $N a O H$ has been added. What is the ionization constant of $H X$ ?

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Solution

Let $M_{1}$ and $M_{2}$ be the molarities of acid $H X$ and base $N a O H$
$v =$ volume of acid.
According to Henderson's Equation,
$p H = p K_{a} + l o g \frac{S a l t}{A c i d}$
$5.8 = p K_{a} ° + l o g (\frac{(10 m l) \times M_{2}}{{V M}_{1} - 20 \times M_{2}}) . . . . . . . . . . . . . . . . . . . (1)$
$6.402 = p K_{a} ° + l o g (\frac{20 m l \times M_{2}}{{V M}_{1} - 20 \times M_{2}}) . . . . . . . . . . . . . . . . . . (2)$
Solving $(1)$ and $(2)$
$0.602 = l o g [\frac{20 M_{2}}{{V M}_{1} - 20 M_{2}} . \frac{{V M}_{1} - 10 M_{2}}{10 M_{2}}]$
$\frac{V M_{1}}{M_{2}} = \frac{60}{2} = 30 m l$
Substitute in equation $(1)$
$p K_{a} = 5.8 - l o g (\frac{10}{30 - 10})$
$= 5.8 + 0.30$
$= 6.10$
ionization constant of $H X$ is $6.10$

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