A 0.576 g sample of a diprotic acid with a molar mass of 255.8 g/mol is dissolve

Question

A 0.576 g sample of a diprotic acid with a molar mass of 255.8 g/mol is dissolved in water to a total volume of 23.0 mL The solution is then titrated with a saturated calcium hydroxide solution. Assuming that the pK_a values for each ionization step are sufficiently different to see two equivalence points, determine the volume of added base for the first and second equivalence points. Enter your answers numerically separated by a comma. The pH after adding 23.0 mL of the base was 3.82. Find the value of pK_a_1. The pH after adding 20.0 mL past the first equivalence point was 8.35. Find the value of pK_a_2.

Explanation / Answer

Ksp of Ca(OH)2 = [Ca2+][OH-]^2 = 5.5 x 10^-6

molarity of solution = cube.rt.(5.5 x 10^-6/4)

                               = 0.011 M

Part A) moles of H2A = 0.576 g/255.8 g/mol = 0.00225 mol

Volume of Ca(OH)2 needed to reach Ist equivalence point = 0.00225 mol x 1000/0.011 x 2 = 102.3 ml

Volume of Ca(OH)2 needed to reach IInd equivalence point = 0.00225 mol x 1000/0.011 = 204.6 ml

Part B) when 23 ml Ca(OH)2 added, pH = 3.82

moles Ca(OH)2 = 0.011 M x 23 ml = 0.253 mmol

moles acid remained = (2.25 - 0.253) = 1.997 mmol

pH = pKa1 = log(base/acid)

pKa1 = 3.82 - log(0.253/1.997) = 4.72

Part C) when 20 ml Ca(OH)2 added past Ist equivalence point, pH = 8.35

moles Ca(OH)2 = 0.011 M x 122.3 ml = 1.35 mmol

moles acid remained = (2.25 - 1.35) = 0.9 mmol

pH = pKa2 = log(base/acid)

pKa2 = 8.35 - log(1.35/0.90) = 8.17

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