If you add 2 mL of 1M HCl to 600 mL of 25 mM phosphate buffer...

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If you add 2 mL of 1M HCl to 600 mL of 25 mM phosphate buffer...

If you add 2 mL of 1M HCl to 600 mL of 25 mM phosphate buffer (pH 8.0), what will be the resulting pH? (pKa of phosphate = 7.21)

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Answer 1

Answer #1

Total concentration of phosphate = 25 mM

[H₂PO₄^-] + [HPO₄^2-] = 25 mM ...(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^-] = 6.166 ...(2)

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

[H₂PO₄^-] = 3.49 mM

[HPO₄^2-] = 21.51 mM

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

moles H₂PO₄^- = (3.49 mM) * (0.600 L)

moles H₂PO₄^- = 2.1 mmol

moles HPO₄^2- = (concentration HPO₄^2-) * (volume of buffer)

moles HPO₄^2- = (21.51 mM) * (0.600 L)

moles HPO₄^2- = 12.9 mmol

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

moles HCl added = (1 M) * (2 mL)

moles HCl added = 2 mmol

HCl will convert HPO₄^2- to H₂PO₄^-

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

new moles H₂PO₄^- = (2.1 mmol) + (2 mmol)

new moles H₂PO₄^- = 4.1 mmol

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

new moles HPO₄^2- = (12.9 mmol) - (2 mmol)

new moles HPO₄^2- = 10.9 mmol

Again using Henderson-Hasselbalch equation,

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

pH = pK_a + log(new moles HPO₄^2- / new moles H₂PO₄^-)

pH = 7.21 + log(10.9 mmol / 4.1 mmol)

pH = 7.21 + log(2.65)

pH = 7.21 + 0.42

pH = 7.63

Resulting pH is 7.63

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Answer 2

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