Question

At a certain temperature the K_sp of Ag₂SO₄ is 1.54 x 10^-5. What is the molar soliubility of silver sulfate in water in units of M (mol/liter).

In a titration of a 100.0mL 1.00M NH₃ solution with 1.00M HCl, what is the pH of the solution after the addition of 38.5 mL of HCl? K_b = 1.8 x 10^-5 for NH₃

Answer 1

1)

Ag2SO4 ----------------------> 2Ag+   + SO₄^-2

                                             2S            S

Ksp = [Ag+]^2 [SO₄^-2]

1.54 x 10^-5 = (2S)^2 (S)

1.54 x 10^-5   = 4S^3

S = 0.0157

S = 1.57 x 10^-2 M

molar solubility = 1.57 x 10^-2 M

2)

millimoles of NH3 = 100 x 1 = 100

millimoles of HCl = 38.5 x 1= 38.5

NH3 + HCl ---------------------> NH4Cl

100       38.5                          0 ----------------------------> initial

61.5        0                             38.5 ------------------------> after reaction

in the solution base NH3 and salt NH4Cl remained . so it can form basic buffer

K_b = 1.8 x 10^-5 for NH₃

pKb = -log Kb = -log(1.8 x 10^-5)

pKb = 4.74

pOH = pKb + log [salt/base]

pOH = 4.74 + log [38.5/61.5]

pOH = 4.54

pH + pOH = 14

pH = 14-pOH

pH = 9.46