Question

Using a 0.30 M phosphate buffer with a pH of 8.0, you add 0.70 mL of 0.55 M HCl to 52 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)
  

Using a 0.30 M phosphate buffer with a pH of 8.0, you add 0.70 mL of 0.55 M NaOH to 52 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Answer 1

Q1. 0.55 M HCl

new pH = 7.92

Q2. 0.55 M NaOH

new pH = 8.10

Explanation

Q1. Total phosphate concentration = 0.30 M

[H₂PO₄^-] + [HPO₄^2-] = 0.30 M ...(1)

According to Henderson - Hasselbalch equation,

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

pH = pK_a + log([HPO₄^2-] / [H₂PO₄^-])

8.0 = 7.21 + log([HPO₄^2-] / [H₂PO₄^-])

log([HPO₄^2-] / [H₂PO₄^-]) = 8.0 - 7.21

log([HPO₄^2-] / [H₂PO₄^-]) = 0.79

[HPO₄^2-] / [H₂PO₄^-] = 10^0.79

[HPO₄^2-] / [H₂PO₄^-] = 6.2

[HPO₄^2-] = 6.2 * [H₂PO₄^-]   ...(2)

Solving equations (1) and (2), we get

[H₂PO₄^-] = 0.0417 M

[HPO₄^2-] = 0.2583 M

moles H₂PO₄^- = (concentration H₂PO₄^-) * (volume buffer)

moles H₂PO₄^- = (0.0417 M) * (52 mL)

moles H₂PO₄^- = 2.17 mmol

Similarly, moles HPO₄^2- = 13.43 mmol

moles HCl added = (concentration HCl) * (volume HCl)

moles HCl added = (0.55 M) * (0.70 mL)

moles HCl added = 0.385 mmol

new moles H₂PO₄^- = (initial moles H₂PO₄^-) + (moles HCl added)

new moles H₂PO₄^- = (2.17 mmol) + (0.385 mmol)

new moles H₂PO₄^- = 2.555 mmol

new moles HPO₄^2- = (initial moles HPO₄^2-) - (moles HCl added)

new moles HPO₄^2- = (13.43 mmol) - (0.385 mmol)

new moles HPO₄^2- = 13.045 mmol

According to Henderson - Hasselbalch equation,

pH = pK_a + log([HPO₄^2-] / [H₂PO₄^-])

pH = 7.21 + log(13.045 mmol / 2.555 mmol)

pH = 7.92