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Question

1.00 L of a buffer that is 0.100 M in HOAc (pKa=4.74) and 0.100 M in NaOAc. What is the pH after the addition of 0.0200 mol NaOH? (log 1.5=0.176)

A
4.92
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B
4.74
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C
8.74
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D
6.74
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Solution

The correct option is A 4.92
Consider the reaction of the strong base with the weak acid first. That reaction goes to completion.
HOAc+OHH2O+OAc

HOAcOHOAcInitial100.0 mmol20.0 mmol100.0 mmolChange20.0 mmol20.0 mmol+20.0 mmolEnd80.0 mmol0120.0 mmol
Using Henderson's equation
pH=pKa+log([A][HA])
pH=4.74+log(120.080.0)
pH=4.92

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