Question

What is the pH of a solution that is 0.24 M K2HPO4 and 0.074 M K3PO4? Use the Acid-Base Table.

Answer 1

K₂HPO₄ ----------> 2K⁺ + HPO₄^2-.

K₃PO₄ ----------> 3K⁺ + PO₄^3-.

This forms a buffer in which we have the ionic equilibrium between HPO₄^2- and PO₄^3- species as,

HPO₄^2- < ------- > H⁺ + PO₄^3- ………(with pKa of HPO₄^2- = 12.66)

HPO4^2- is acid and PO4^3-conjugate base (salt).

As both K₂HPO₄ and K₃PO₄ are strong electrolytes we can have,

[HPO₄^2-] = [K₂HPO₄] = 0.24 M.

[PO₄^3-] = [K₃PO₄] = 0.074 M

Henderson equation,

pH = pKa + log{[salt]/[acid]}

pH = pKa + log{[PO₄^3-] / [HPO₄^2-]}

pH = 12.66 + log(0.074 / 0.24)

pH = 12.66 + log(0.3083)

pH = 12.66 + (-0.51)

pH = 12.15.

pH of a solution of 0.24 M K₂HPO₄ and 0.074 M K₃PO₄ solution is 12.15.

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