How to calculate ph value of the solution? Two salt buffer.

Question

solution contains:
c(Na3PO4) = 0,10 mol/L and c(Na2HPO4)=0,05 mol/L

Answer in my book is ph=12.96

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Answer 1 · 2017-01-31T15:41:29

You can do it like this:

Phosphoric acid has 3 ionisations. Here we are dealing with the 3rd:

$sf(HPO_4^(2-)rightleftharpoonsPO_4^(3-)+H^+)$

For which:

$sf(K_(a3)=([PO_4^(2-)][H^+])/([HPO_4^(2-)])=2.2xx10^(-13)color(white)(x)"mol/l")$

You should be provided with this value in the question.

To find the pH we need the concentration of the $sf(H^+)$ ions.

Rearranging gives:

$sf([H^+]=K_(a3)xx[[HPO_4^(2-)]]/[[PO_4^(2-)]])$

$:.$ $sf([H^+]=2.2xx10^(-13)xx0.05/(0.1)color(white)(x)"mol/l")$

$sf([H^+]=1.1xx10^(-13)color(white)(x)"mol/l")$

$sf(pH=-log[H^+]=-log(1.1xx10^(-13)))$

$sf(pH=12.95)$