Question

1. Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.5 M H3PO3(aq) with 1.5 M KOH(aq).

pKa1= 1.30   pKa2= 6.70

a) Before the addition of any kOH

b) after adding 25.0ml KOH

c) after adding 50.0ml KOH

d) after adding 75.0ml KOH

e) after adding 100.0ml KOH

2. The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.80. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.60 M B(aq) with 0.60 M HCl(aq).

a) Before the addition of any HCl

b) after the addition of 25ml HCl

c) after 50ml HCl

d) after 75ml HCl

e) after 100ml HCl

3. Copper(I) ions in aqueous solution react with NH3(aq) according to

Cu^+ (aq) + 2NH3 (aq) -------> Cu(NH3)2 ^+ (aq)     Kf= 6.3x10^10

Calculate the solubility (in g·L–1) of CuBr(s) (Ksp = 6.3× 10–9) in 0.61 M NH3(aq).

4.

Given the equation

Ag^+ (aq) + 2NH3 (aq) yields (Ag(NH3)2)^+ (aq)                Kf=2.00x10^7

determine the concentration of NH3(aq) that is required to dissolve 701 mg of AgCl(s) in 100.0 mL of solution. The Ksp of AgCl is 1.77× 10^–10

Answer 1

H3PO3 millimoles = 50 x 1.5 = 75

a) Before any addition of KOH :

H3PO3 ---------------------> H+ +   H2PO3-

1.5                                       0             0

1.5- x                                   x              x

Ka1 = [H+][H2PO3-] / [H3PO3]

0.05 = x^2 / 1.5 - x

x = 0.25

[H+] = 0.25 M

pH = -log [H+] = -log [0.25]

       = 0.6

pH = 0.6

b) after addition of 25.0 mL KOH

it is first equivaelce point

pH = pKa1 = 1.30

pH = 1.30

c) addition of 50.0 mL KOH

it is first equivalence point

pH = 1/2 (pKa1 + pKa2)

pH =1/2 (1.30 + 6.70)

pH = 4.0

3) 75.0 mL KOH

it is seond half equivalece point

pH = pKa2

pH = 6.70

4) 100.0 mL KOH

it is second equivalece point

HPO3^-2 millimoles = 100 x 1.5 = 150

HPO3^-2 molarity = 150 / (50 +100) = 1.0 M

HPO3^-2 + H2O ------------------> H2PO4- + OH-

1 -x                                         x                x

Kb2 = x^2 / 1-x

5.01 x 10^-8 = x^2 / 1.0-x

x^2 + 5.01 x 10^-8x - 5.01 x 10^-8 = 0

x = 2.24 x 10^-4

[OH-] = 2.24 x 10^-4 M

pOH = -log[OH-] = -log (2.24 x 10^-4 )

pOH = 3.65

pH + pOH = 14

pH = 10.35

2)

dibasic base = pKb1 = 2.10

pKb2 = 7.80

millimoles of base = 50 x 0.6 = 30

(a) before addition of any HCl

pOH = 1/2 [pKb1- logC]

pOH = 1/2 [pKb1 - log 0.6]

          = 1/2 (2.10 - log 0.6)

          = 1.16

pH + pOH = 14

pH = 14 - pOH

      = 12.84

pH = 12.84

(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 = 11.9

(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 + 7.80 / 2

         = 4.95

pH = 9.05
(d) after addition of 75.0 mL of HCl

it is second half equivalence point . so

pOH = pKb2

pOH = 7.80

pH = 6.2

(e) after addition of 100.0 mL of HCl

millimoles of HCl = 100 x 0.6 = 60

it is second equivalence point.here it is only BH2+ remains.so its concentration

BH2+ millimoles = 60

BH2+ concentration = 60 / total volume

                                 = 60 / (100 + 50)

                                 = 0.4 M

BH2+ + H2O ------------------> BH+   + H3O+

0.4                                              0             0

0.4 - x                                         x              x

Ka2 = x^2 / (0.33 -x)

6.31 x 10^-7 = x^2 / (0.4 -x)

x = 5.0 x 10^-4

[H3O+] = x = 5.0 x 10^-4 M

pH = -log[H3O+] = -log (5.0 x 10^-4)

       = 3.3

pH = 3.3