Question

Identifying an unkhown diprotic acid.

What is the Ka?

Naoh 0.1008m Un known: 0.212 g PH 2.19 2.66 3.25 3.75 4.70 5.36 6.03 8.63 8.3 / 11.07 8.4 | 11.39 8.5 T 11.44 8.6 11.51 8.7 | 11.60 2.8 11.63 8.9 11.69 111.73 19.1 | 11.77 19. 2 11.87 19.3 11.89 19.4 11.92 19.15 11.96 9.6 12 114 119 12.5 12.68

in 100 ml

Answer 1

The acid graph is made:

Volume NaOH pH 1.95 1.97 2.03 2.19 2.66 8.1 3.25 3.75 4.3 5.36 6.03 8.63 11.07 11.39 11.44 11.51 11.6 11.63 11.69 8.6 8.7 10 15

The equivalence point must occur at 8 mL of NaOH. The moles of acid present are calculated:

n Acid = M * V = 0.1008 M * 0.008 L = 8.1x10 ^ -4 mol

The molar mass is calculated:

MM = g / n = 0.212 / 8.1x10 ^ -4 = 262.9 g / mol

The average equivalence point is 4 mL, at that point:

pH = pKa = 2.66

Ka is calculated:

Ka = 10 ^ -pKa = 10 ^ -2.66 = 2.2x10 ^ -3

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