ningcomibiscmsmodibisviewphpid5047047 Sapling Learning Map T

ning.com/ibiscms/mod/ibis/view.php?id-5047047 Sapling Learning Map The pks values for the dibasic base 8 are pkbi 2.10 and pki2 -7.62. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.85 M B(aq) with 0.85 M HCI(aq). Number (a) before addition of any HC Number (b) after addition of 25.0 mL of HCI ! Number (c) after addition of 50.0 mL of HCl Number (d) after addition of 75.0 mL of HCl Number (e) after addition of 1000mL of HCI Hint Previous Give Up & View Solution Check Answer 0 Next Ext about usi Driva-cv oolicv terms ofuse careers contact us helo

Solution

Titration

dibasic base B + 2HCl

pKb1 = -log[Kb1] = 2.10

Kb1 = 8 x 10^-3

pKa1 = 14 - pKb2 = 14 - 7.82 = 6.18

pKa2 = 14 - pKb1 = 14 - 2.10 = 11.9

Ka1 = 6.61 x 10^-7

(a) before addition of any HCl

[B] = 0.85 M

B + H2O <==> BH+ + OH-

let x amount of base reacted

Kb1 = [BH+][OH-]/[B]

8 x 10^-3 = x^2/0.85

x = [OH-] = 0.0825 M

pOH = -log[OH-] = 1.084

pH = 14 - pOH = 12.916

(b) after 25 ml HCl added

initial moles B = 0.85 M x 50 ml = 42.50 mmol

moles HCl added = 0.85 M x 25 ml = 21.25 mmol

half moles B neutralized to first equivalence point

[BH+] formed = [B] remained

first 1/2 equivalence point

pH = pKa2 + log(B/BH+)

pH = pKa2

pKa = 14 - pKb1 = 14 - 2.10 = 11.9

So,

pH = 11.9

(c) after 50 ml HCl added

all of B neutralized to first equivalence point

pH = 1/2(pKa1 + pKa2)

      = 1/2(6.18 + 11.9)

      = 9.04

(d) after 75 ml HCl added

initial moles B = 0.85 M x 50 ml = 42.50 mmol

moles HCl added = 0.85 M x 75 ml = 63.75 mmol

1/2 moles of BH+ neutralized to second equivalence point

[BH2^2+] formed = [BH+] remained

pH = pKa1 + log(BH+/BH2^2+)

      = 6.18

(e) after 100 ml HCl added

all of B forms BH2^2+

molarity of [BH2^2+] = 0.85 M x 50 ml/150 ml = 0.283 M

BH2^2+ + H2O <==> BH+ + H3O+

let x amount has reacted

Ka1 = [BH+][H3O+]/[BH2^2+]

6.61 x 10^-7 = x^2/0.283

x = [H3O+] = 4.32 x 10^-4 M

pH = -log[H3O+] = 3.364