The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.43. Calculate

Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.43. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.50 M B(aq) with 0.50 M HCl(aq).

macmi The pKb values for the dibasic base B are pKb 2.10 and pKb2 7.43 Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.50 M B(aq) with 0.50 M HCl(aq) Number before addition of any HCl Number (b) after addition of 25.0 mL of HC Number (c) after addition of 50.0 mL of HCl Number (d) after addition of 75.0 mL of HC Number (e) after addition of 100.0 mL of HC

Explanation / Answer

pKb1 = 2.1

Kb1 = 10-2.1 = 0.0079

pKb2 = 7.43

Kb2 = 10-7.43 = 3.72 x 10-8

a)

Moles of base = 50 mL * 0.5 M = 25 mmoles

B + H2O -> B+ + OH- ; Kb1

(0.5 – x) -> (x – y) + (x + y)

B+ + H2O -> B2+ + OH- ; Kb2

(x – y) -> y + (x + y)

Assuming y << x, we get

y = Kb2 = 3.72 x 10-8

x2 / (0.5 – x) = Kb1 = 0.0079

x = 0.057

[OH-] = 0.057 M

pOH = -log(0.057) = 1.24

pH = 14 – 1.24 = 12.76

b)

HCl moles added = 0.5 M * 25 mL = 12.5 mmoles

Moles of base, B left = 25 – 12.5 = 12.5 mmoles

pOH at half-equivalence point = pKb1

pOH = 2.1

pH = 14 – 2.1 = 11.9

c)

HCl moles added = 0.5 M * 50 mL = 25 mmoles

pOH at first equivalence point = ½ (pKb1 + pKb2)

= ½ (2.1 + 7.43) = 4.765

pH = 14 – 4.765 = 9.235

d)

HCl moles added = 0.5 M * 75 mL = 37.5 mmoles

pOH at one and a half equivalence point = pKb2

pOH = 7.43

pH = 14 – 7.43 = 6.57

e)

HCl moles added = 0.5 M * 100 mL = 50 mmoles

Moles of B2- formed = 50 /2 = 25 mmoles

Total volume = 100 + 50 = 150 mL

Initial concentration of B2- = 25 / 150 = 0.167 M

B2- + H2O -> BOH- + H+

(0.167 – x) -> x + x

Ka2 = Kw/Kb2 = 2.69 x 10-7

Ka2 = x2 / (0.167 – x)

x = 2.1 x 10-4 M

pH = -log(x) = 3.67

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