Question

3) Calculate the pH of a solution by mixing 50.0 mL of 0.200 M Na2HPO4 with 20.0 mL 0.0400 M NaH2PO4. (Hint: Buffer solution. Use Henderson-Hasselbalch equation) H3PO4 5 H+ + H2PO4 pkı = 2.1 H2PO4 5 H+ + HPO42- pK2 = 7.2 HPO42- 5 H+ PO43- pK3 = 12.3

Answer 1

Q3. Concentration Na₂HPO₄ = 0.200 M

volume Na₂HPO₄ solution = 50.0 mL

moles Na₂HPO₄ = (concentration Na₂HPO₄) * (volume Na₂HPO₄ solution)

moles Na₂HPO₄ = (0.200 M) * (50.0 mL)

moles Na₂HPO₄ = 10.0 mmol

moles HPO₄^2- = moles Na₂HPO₄

moles HPO₄^2- = 10.0 mmol

Total volume = (50.0 mL) + (20.0 mL)

Total volume = 70.0 mL

[HPO₄^2-] = (moles HPO₄^2-) / (total volume)

[HPO₄^2-] = (10.0 mmol) / (70.0 mL)

[HPO₄^2-] = 0.143 M

Similarly, [H₂PO₄^-] = (0.0400 M) * (20.0 mL / 70.0 mL)

[H₂PO₄^-] = 0.0114 M

According to Henderson- Hasselbalch equation,

pH = pK + log([conjugate base] / [weak acid])

pH = pK₂ + log([HPO₄^2-] / [H₂PO₄^-])

pH = 7.2 + log(0.143 M / 0.0114 M)

pH = 7.2 + log(12.54)

pH = 7.2 + 1.1

pH = 8.3