Please answer all parts, thanks! :) i always upvote for correctness! 0 4/2/2018
ID: 1038132 • Letter: P
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Please answer all parts, thanks! :) i always upvote for correctness!
0 4/2/2018 1 1:59 PM O 7.3/10 O 4/3/2018 08:19 PM - Print?Calculator-Periodic Table Question 2 of 5 Gradebook Mapr Sapling Learning macmillan learning The pKb values for the dibasic base B are pKb1 2.10 and pKb2 7.71. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.70 M B(aq) with 0.70 M HCl(aq) Number (a) before addition of any HCI Number (b) after addition of 25.0 mL of HCl Number (c) after addition of 50.0 mL of HCI Number (d) after addition of 75.0 mL of HCI Number (e) after addition of 100.0 mL of HCl PreviousGive Up & View Solution Check AnswerNextExit HintExplanation / Answer
dibasic base = pKb1 = 2.10
pKb2 = 7.71
millimoles of base = 50 x 0.7 = 35
(a) before addition of any HCl
pOH = 1/2 [pKb1- logC]
pOH = 1/2 [pKb1 - log 0.7]
= 1/2 (2.10 - log 0.7)
= 1.13
pH + pOH = 14
pH = 14 - pOH
= 12.87
pH = 12.87
(b) after addition of 25.0 mL of HCl
it is half equivalence point. so
pOH = pKb1
pOH = 2.10
pH = 11.9
(c) after addition of 50.0 mL of HCl
it is equivalence point . at equivalence point
pOH = (pKb1 + pKb2 )/ 2
= 2.10 + 7.71 / 2
= 4.91
pH = 9.10
(d) after addition of 75.0 mL of HCl
it is second half equivalence point . so
pOH = pKb2
pOH = 7.71
pH = 6.29
(e) after addition of 100.0 mL of HCl
it is second equivalence point.here it is only BH2+ remains.so its concentration
BH2+ millimoles = 35
BH2+ concentration = 35 / total volume
= 35 / (100 + 50)
= 0.233 M
BH2+ + H2O ------------------> BH+ + H3O+
0.233 0 0
0.233 - x x x
Ka2 = x^2 / (0.33 -x)
5.13 x 10^-7 = x^2 / (0.233 -x)
x = 3.46 x 10^-4
[H3O+] = x = 3.46 x 10^-4 M
pH = -log[H3O+] = -log (3.46 x 10^-4)
= 3.46
pH = 3.46
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