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.)
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-] + [HPO42-] = 0.30 M ...(1)
According to Henderson - Hasselbalch equation,
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa + log([HPO42-] / [H2PO4-])
8.0 = 7.21 + log([HPO42-] / [H2PO4-])
log([HPO42-] / [H2PO4-]) = 8.0 - 7.21
log([HPO42-] / [H2PO4-]) = 0.79
[HPO42-] / [H2PO4-] = 100.79
[HPO42-] / [H2PO4-] = 6.2
[HPO42-] = 6.2 * [H2PO4-] ...(2)
Solving equations (1) and (2), we get
[H2PO4-] = 0.0417 M
[HPO42-] = 0.2583 M
moles H2PO4- = (concentration H2PO4-) * (volume buffer)
moles H2PO4- = (0.0417 M) * (52 mL)
moles H2PO4- = 2.17 mmol
Similarly, moles HPO42- = 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 HPO42- = (initial moles HPO42-) - (moles HCl added)
new moles HPO42- = (13.43 mmol) - (0.385 mmol)
new moles HPO42- = 13.045 mmol
According to Henderson - Hasselbalch equation,
pH = pKa + log([HPO42-] / [H2PO4-])
pH = 7.21 + log(13.045 mmol / 2.555 mmol)
pH = 7.92
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