Question

Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-13 at 25°C)

Answer 1

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

Answer 2

The pH of the solution is 12.38 and the pOH of the solution is  1.62.

Using the Henderson Hasselbalch equation;

pH = pKa + log [A-]/[HA]

Where;

pKa = - log Ka = -log[ 4.2x10-13] = 12.38

[A-] = [PO43–] = 0.285 M

[HA] = [HPO42-] = 0.285 M

Substituting values;

pH =  12.38 + [ 0.285 M]/[ 0.285 M]

pH = 12.38

But;

pH + pOH = 14

pOH = 14 - 12.38

pOH = 1.62

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