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When 100.0 mL of a weak acid was titrated with 0.09381 M NaOH, 27.63 mL were...

When 100.0 mL of a weak acid was titrated with 0.09381 M NaOH, 27.63 mL were required to reach the equivalence point. The pH at the equivalence point was 10.99. What was the pH when only 19.47 mL of NaOH had been added?

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Answer #1

millimoles of NaOH added = 0.09381 x 27.63 = 2.592

2.592 millimoles acid must be present.

2.592 = M x 100.0

M = 0.02592 M

[acid] = 0.02592 M

after equivalence point reached

[salt] = 2.592 / 100 + 27.63 = 0.0203 M

pH = 1/2 [pKw + pKa + logC]

10.99 = 1/2 [14 + pKa + log 0.0203]

21.98 = pKa + 12.31

pKa = 9.67

millimoles of NaOH added = 19.47 x 0.09381 = 1.826

2.592 - 1.826 = 0.766 millimoles acid left

1.826 millimoles salt formed

[acid] = 0.766 / 100 + 19.47 = 0.0064 M

[salt] = 1.826 / 119.47 = 0.0153 M

pH = pKa + log [salt] / [acid]

pH = 9.67 + log [0.0153] / [0.0064]

pH = 10.05

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