Millimoles of oxalic acid = 0.1 mmol/mL * 25 mL = 2.5 mmol
pKa1 = -Log(5.4*10-2) = 1.27
(i) Millimoles of KOH = 10 mL * 0.1 mmol/mL = 1 mmol
According to Henderson-Hasselbulch equation:
pH = pKa1 + Log{[monopotassium oxalate]/[oxalic acid])
i.e. pH = 1.27 + Log{1/(2.5-1)}
i.e. pH = 1.09
(ii) Millimoles of KOH = 15 mL * 0.1 mmol/mL = 1.5 mmol
i.e. pH = 1.27 + Log{1.5/(2.5-1.5)}
i.e. pH = 1.44
(iii) Millimoles of KOH = 20 mL * 0.1 mmol/mL = 2 mmol
i.e. pH = 1.27 + Log{2/(2.5-2)}
i.e. pH = 1.87
(iv) Millimoles of KOH = 25 mL * 0.1 mmol/mL = 2.5 mmol
This is first equivalence-point.
Here, [monopotassium oxalate] = 2.5 mmol/(25+25) mL = 0.05 M
Now, pH = 7 + 1/2 (pKa1 + Log[monopotassium oxalate])
i.e. pH = 7 + 1/2 (1.27 + Log(0.05))
i.e. pH = 6.98
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