Question

а он (15 points) A 0.100 M oxalic acid, HO,CCOH, solution is titrated with 0.100 M KOH. Calculate the pHs when 25.00 mL of oxalic acid solution is titrated with 10, 15, 20, 25, 35, 40, 45, 50 and 55 mL of Nont Kal = 5.4 x 10-2 and K 2 = 5.42 x 10- for oxalic acid. Show all your work and make a graph of pH versus Vo HO, CCO 2H + KOH KC204 + H2Oenwot aunt men (6.0100 M ox ac) (25mL ox ac) = (0,100 M KOH) (VolkoH). 0.100 KOH on or is wer Tore through this 2.5ML KOH @ Equivalence point DHME ecco. H+KOH > Ke204+H2Ooteaedte vorene 01 0.001 100 Molx 0.025 L -0.025 ML TL TL

Answer 1

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