Question 14 of 15 Available From Not Set Due Date Points Possible: 100 Grade Cat
ID: 1034652 • Letter: Q
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Question 14 of 15 Available From Not Set Due Date Points Possible: 100 Grade Category Graded Description 3/28/2018 The pK, values for the dibasic base B are pKoi 2.10 and pk627.46. 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). (a) before addition of any HCI You can dheck your answers You can view solutions when give up on any question You can keep trying to answe until you get it right or give u You lose 5% ofthe points avai answer in your question for e Nu (b) after addition of 25.0 mL of HCl Nu (c) after addition of 50.0 mL of HCl (d) after addition of 75.0 mL of HCl O Help With This Topic (e) after addition of 100.0 mL of HCD { Ex1 Next Previous ? , Aviess So io. O Check Answer HintExplanation / Answer
dibasic base = pKb1 = 2.10
pKb2 = 7.46
millimoles of base = 50 x 0.5 = 25
(a) before addition of any HCl
pOH = 1/2 [pKb1- logC]
pOH = 1/2 [pKb1 - log 0.5]
= 1/2 (2.10 - log 0.5)
= 1.2
pH + pOH = 14
pH = 14 - pOH
= 12.8
pH = 12.8
(b) after addition of 25.0 mL of HCl
millimoles of HCl = 15 x 0.5 = 12.5
it is half equivalence point. so
pOH = pKb1
pOH = 2.10
pH = 11.9
(c) after addition of 50.0 mL of HCl
millimoles of HCl = 25
it is equivalence point . at equivalence point
pOH = (pKb1 + pKb2 )/ 2
= 2.10 + 7.46 / 2
= 4.78
pH = 9.22
(d) after addition of 75.0 mL of HCl
moles of HCl = 75 x 0.5 = 37.5
it is second half equivalence point . so
pOH = pKb2
pOH = 7.46
pH = 6.54
(e) after addition of 100.0 mL of HCl
millimoles of HCl = 100 x 0.5 = 50
it is second equivalence point.here it is only BH2+ remains.so its concentration
BH2+ millimoles = 50
BH2+ concentration = 50 / total volume
= 50 / (100 + 50)
= 0.33 M
BH2+ + H2O ------------------> BH+ + H3O+
0.33 0 0
0.33 - x x x
Ka2 = x^2 / (0.33 -x)
2.88 x 10^-7 = x^2 / (0.33 -x)
x = 3.08 x 10^-4
[H3O+] = x = 3.08 x 10^-4 M
pH = -log[H3O+] = -log (3.08 x 10^-4)
= 3.51
pH = 3.51
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