(a) Since Kb1 >> Kb2, you can ignore the second ionization. However, you\\\'ll n
ID: 889297 • Letter: #
Question
(a) Since Kb1 >> Kb2, you can ignore the second ionization. However, you'll need to use either the quadratic equation or successive approximations to solve for [H ].
(b) 25.0 mL is half way to the first equivalence point, so pOH = pKb1.
(c) 50.0 mL is the first equivalence point. BH is amphoteric, so use pH = 1/2(pKa1 pKa2), where pKa = 14.00 – pKb.
(d) 75.0 mL is half way to the second equivalence point, so pOH = pKb2.
(e) 100.0 mL is the second equivalence point. The final product, BH22 , is a weak acid with Ka1 = Kw/Kb2.
The pKb values for the dibasic base B are pKb1 2.10 and pKb27.72. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.80 M B(aq) with 0.80 M HCI(aq). Number (a) before addition of any HCI Number (b) after addition of 25.0 mL of HCI I0 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 HCIExplanation / Answer
(a) before any additon of any HCl
pOH = 1/2 [pKb-logC]
= 1/2 [2.10 - log(0.8) ]
= 1.098
pH + pOH = 14
pH = 12.90
(b) after addition of 25 ml HCl
after adding acid to base it is first half equivalence point . at first half equivalence point
pKb1 = pOH
pOH = 2.10
pH + pOH = 14
pH = 11.9
(c) after addition of 50 ml HCl
it is first equivalence point
B + H+ ------------------> BH+
0.8 x 50 0.8 x 50 0 ---------------------> initial
0 0 40 ----------------> after reaction
BH+ molarity = 40 / total volume = 40 / 100 = 0.4 M
BH+ is salt of strong acid and weak base
so
pH form salt hydrolysis pH = 7- 1/2 [pkb1 + logC]
pH = 7 - 1/2 (2.10 + log 0.4)
pH = 5.75
(d) after addition of 75 ml HCl
here second half of the equiavalence pointt
pOH = pkb2
pOH = 7.72
pH = 6.26
(e) after addition of 100 ml HCl
it is second equivalence point . calculate like first equivalence point
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