Question

If 8.50 mg of the salt, KH2PO4 (M.Wt. of 136) is dissolved in 500.0 ml of water with a final of pH 6.65, what will be the final concentrations of K2HPO4? The pKa for H2PO₄^- is 6.86.

Answer 1

Find the initial phosphate concentration as below.

Mass of KH₂PO₄ = 850 mg = (850 mg)*(1 g/1000 mg) = 0.85 g.

Molar mass of KH₂PO₄ = 136 g/mol.

Number of moles of KH₂PO₄ = (0.85 g)/(136 g/mol) = 0.00625 mole.

Volume of the solution = 500.0 mL = (500.0 mL)*(1 L/1000 mL) = 0.5 L.

Initial molar concentration of phosphate = (0.00625 mole)/(0.5 L) = 0.5 mol/L ≈ 0.5 M.

Note that when a buffer is formed with KH₂PO₄ and K₂HPO₄, the total phosphate concentration will remain the same. The buffer is represented as

H₂PO₄^- (aq) <=====> H⁺ (aq) + HPO₄^2- (aq)

Use the Henderson-Hasslebach equation to find out the ratio of the molar concentrations of H₂PO₄^- and HPO₄^2-.

pH = pK_a + log [HPO₄^2-]/[H₂PO₄^-]

====> 6.65 = 6.86 + log [HPO₄^2-]/[H₂PO₄^-]

====> log [HPO₄^2-]/[H₂PO₄^-] = -0.21

====> [HPO₄^2-]/[H₂PO₄^-] = antilog (-0.21) = 0.6166

====> [HPO₄^2-] = 0.6166*[H₂PO₄^-] …..(1)

Since the total phosphate concentration remains the same, we must have

[HPO₄^2-] + [H₂PO₄^-] = 0.5 M

===> 0.6166*[H₂PO₄^-] + [H₂PO₄^-] = 0.5 M

===> 1.6166*[H₂PO₄^-] = 0.5 M

===> [H₂PO₄^-] = (0.5 m)/(1.6166) = 0.309 M ≈ 0.31 M

The final concentration of H₂PO₄^- is 0.31 M (ans).