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erie PLAY High Wastex l H&R; Block Budget Cha g.com/ibiscms/mod/ibis/view.php?id-5047047 X · H&R; Block Budget Cho XCopyofCopy of Cashe Jump to... 3/11/2018 11:55 PM A5/103/6/2018 09:01 PM -Prin! Caluator Periodic Table Gradebook The pKb values for the dibasic base B are pin#2.10 and pK62-786. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.55 M B(aq) with 0.55 M HCaq Number ajbefore addition of any HC Number (b) after addition of 25.0 mL ot HC I 0 (c) after addition of 50 0 mL of HCl 110 11 . (d) after addition ot 75.0 ml of CO (e) after addition of 1000mL of HClExplanation / Answer
dibasic base = pKb1 = 2.10
pKb2 = 7.86
millimoles of base = 50 x 0.55 = 27.5
(a) before addition of any HCl
pOH = 1/2 [pKb1- logC]
pOH = 1/2 [pKb1 - log 0.55]
= 1/2 (2.10 - log 0.55)
= 1.18
pH + pOH = 14
pH = 14 - pOH
= 12.8
pH = 12.82
(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 = 11.90
(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 + 7.86 / 2
= 4.98
pH = 9.02
(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 = 7.86
pH = 6.14
(e) after addition of 100.0 mL of HCl
millimoles of HCl = 100 x 0.55 = 55
it is second equivalence point.here it is only BH2+ remains.so its concentration
BH2+ millimoles = 55
BH2+ concentration = 55 / total volume
= 55 / (100 + 50)
= 0.367 M
BH2+ + H2O ------------------> BH+ + H3O+
0.367 0 0
0.367 - x x x
Ka2 = x^2 / (0.367 -x)
7.24 x 10^-7 = x^2 / (0.367 -x)
x = 5.15 x 10^-4
[H3O+] = x = 5.15 x 10^-4 M
pH = -log[H3O+] = -log (5.15 x 10^-4)
= 3.29
pH = 3.29
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