Question

The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.52.

Calculate the pH at each of the points in the titration of 50.0 mL of a 0.70 M B(aq) solution with 0.70 M HCl(aq).

What is the pH before addition of any HCl?

pH=

What is the pH after addition of 25.0 mL HCl?

pH=

What is the pH after addition of 50.0 mL HCl?

pH=

What is the pH after addition of 75.0 mL HCl?

pH=

What is the pH after addition of 100.0 mL HCl?

pH=

Answer 1

Solution : Given that - dibasic Bese titration 500ml of 0.701 pkb =2.10 Ebi = 0.0079 piba = 7.52 kb = 1.39x108 Beau) Solution with #cl (20)070 ml. (@) Before addition of Any HCl B+H20 B + toti Tnitially 0070 0 0 0 At quilibrium 070 xx x FB = [Bht] [ot] [B] 0.0079 = xx u 0.70-1 0.0053 27** se = 01074 Com -] = {@mt 13 }=0.0 J=0.074 M pot Hog 0.074 pott pit = 1.13 = {H - 143

(b) pH affor addition of 2500 ml Hel Milimdes of B- 5080-70 = 35 mmoles, Millimoles of HCl = 25X0.70 = 1745 mmoles. Half way of the Eauilance point : (ok] = kb, = 0.0079. pot Hog 0.0079. sarlo pH = 14-2.10 co) pt after addition of 50.0 ml Hel 2. At pauilvalence point pH all-(pki - pk2) = 14 - (2:10 +7.52) $ 14-481 (d) pit After addition of 75.0 ml Hel... Half way to the first & and Equilance point. [ot ] = tb2 = 4.3axio-om pol. =-|0g (1:32x10-6) : pott = 7.52

pH = H-7.52 TpH = 6+48) (C)pt After addition of 100.0 ml of Hcl. It is gond Fuilvalence point. gond Feuilvance point only acid Buela ka = 10th 1.3axo-8 Ka = 7.57X107 (Bux* ] = 250f 1.o >>{+] is >>(Bts?] =0.933M] BH² + H2O → Initialty. 0.933 Ecuilibrium 0.833 : x BHT + flot 0 0 -x : 2

ka XXX - 0.233- 7.57810* -wa 0233 xxx = 117.6 407 a = 4.12 x10 4 [Hgct] = k12 x10 4 M pH = -log (Hola x104) pH = 3:38.