1. The overall K f for the complex ion Ag(NH 3 ) 2 + is 1.7 10 7 . The K sp for

Question

1. The overall Kf for the complex ion Ag(NH3)2+ is 1.7 107. The Ksp for AgI is 1.5 1016. What is the molar solubility of AgI in a solution that is 2.0 M in NH3?

[A] 1.3X10^-3
[B] 8.4X10^-5
[C] 1.0X10^-4
[D] 1.5X10^-9
[E] 5.8X10^-12

2. Silver acetate, AgC2H3O2, is a sparingly soluble salt with Ksp = 1.9 103. Consider a saturated solution in equilibrium with the solid salt. Compare the effects on the solubility of adding to the solution either the acid HNO3 or the base NH3.

[A] Either substance would decrease the solubility.
[B] NH3 would decrease the solubility, but HNO3 would increase it
[C] NH3 would increase the solubility, but HNO3 would have virtually no effect.
[D] Either substance would increase the solubility.
[E] NH3 would increase the solubility, but HNO3 would decrease it.

3. An unknown salt, M2Z, has a Ksp of 3.9 1015. Calculate the solubility in mol/L of M2Z.

Does this problem have enough information?

Explanation / Answer

The complex formation reaction can be written as follows

Ag+ (aq) + 2 NH3 (aq) --------------> Ag(NH3)2+ (aq) ..............Kf = 1.7 x 10^7

The ionization reaction of AgI can be written as

AgI (s) <-----------------> Ag^+ (aq) + I^- (aq) ..................Ksp = 1.5 x 10^-16

Let's add above reactions

AgI (s) + 2 NH3 (aq) <-----------------> Ag(NH3)2^+ (aq) + I^- (aq) Keq = Kf*Ksp ( Ag+ gets cancelled as it is present on both sides)

Keq for the above reaction is

Keq = 1.7 x 10^7 * 1.5 x 10^-16 = 2.55 x 10^-9

Keq expression can be written as

Keq = [ I-] [ Ag(NH3)2^+] / [NH3]^2

2.55 x 10^-9 =  [ I-] [ Ag(NH3)2^+] / [NH3]^2

Let us draw ICE table for the above reaction

Substituting these values in keq equation,

2.55 x 10^-9 = (x) (x) / ( 2 - 2x)^2

2.55 x 10^-9 = x^2 / ( 2 - 2x)^2 ..................take square root on both sides,

5.0498 x 10^-5 = x / (2- 2x)

5.0498 x 10^-5 ( 2-2x) = x

1.00995 x 10^-4 - 0.000100995x = x

1.00995 x 10^-4 = 1.000100995 x

x = 1.00995 x 10^-4 / 1.000100995

x = 1.01 x 10^-4

but x is the concentration of I^-

From ionization equation of AgI we know that equilibrium concentrations of Ag+ and I- are equal

Therefore, we have [Ag+] = 1.0 x 10^-4

The solubility of AgI in 2M ammonia is 1.0 x 10^-4

________________________________________________________________________________________________

The ionization equation for the given salt can be written as follows

AgC2H3O2 <---------------------> Ag+ + C2H3O2^-

If we want to increase the solubility, the equilibrium must shift towards right.

According to Le-Chatelier's principle, if the products are removed, the equilibrium will shift towards right.

Ag^+ ions can be removed by reacting with NH3 , because Ag+ ions form a complex Ag(NH3)2^+ with very high formation constant  ....................................( We have already seen this in previous problem)

Ag+ + 2NH3 --------------> Ag(NH3)2^+ Kf = 1.7 x 10^7

Therefore adding NH3 would increase the solubility .

C2H3O2^- ions are basic in nature and therefore readily react with an acid like HNO3

C2H3O2^- + HNO3 -----------------> HC2H3O2 + NO3^-

Therefore adding an acid like HNO2 would also increase solubility of the salt

Either substance would increase the solubility

Option D is the correct choice

__________________________________________________________________________________________________

Ionization of M2Z can be written as follows

M2Z (s) <----------------------> 2 M^+ (aq) + Z^2- (aq)

Ksp for above reaction can be written as

Ksp = [M^+]^2 [ Z^2-]

Substituting Ksp and equilibrium values we get,

3.9 x 10^-15 = (2x)^2 (x)

3.9 x 10^-15 = 4x^3

x^3 = 9.75 x 10^-16

x = 9.92 x 10^-6 M

x is the solubility for Z^2-

Since mol ratio for M2Z and Z is 1:1 , solubility of M2Z is also same as that of Z^2-

Therefore solubility of M2Z is 9.92 x 10^-6 mol/L

AgI(s) NH3(aq) Ag(NH3)2^+ (aq) I^- (aq) I - 2 M 0 0 C - -2x +x +x E - 2 - 2x x x

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