If a solution is formed with equimolar amounts of [Cd(CN)4]2− (Kf=1.3×107) and [Ag(NH3)2]+ (Kf=1.7×107), what will be true about the system?

Question

If a solution is formed with equimolar amounts of [Cd(CN)4]2− (Kf=1.3×107) and [Ag(NH3)2]+ (Kf=1.7×107), what will be true about the system?

Select the correct answer below:

[CN−]>[NH3]

[CN−]<[NH3]

[CN−]=[NH3]

impossible to tell

Answer 1

Since the concentrations are the same and the formation constants are very similar (dissociation constants are therefore also similar), it seems that the [CN^-] would be greater than the [NH3] because there are 4 moles CN^-/mole of complex vs. 2 moles NH3/mole complex upon dissociation. So, I'd go [CN-] > [NH3].

Answer 2

[Cd(CN)⁴]²⁻ (K_f=1.3×10⁷)

and 

[Ag(NH₃)₂]⁺ (K_f=1.7×10⁷)

(K_f) is the formation constant. It is the equilibrium rate constant for making [Cd(CN)₄]²⁻ or [Ag(NH₃)₂]⁺ from free ions Cd⁺² and CN^-, and Ag⁺ and NH₃, respectively.

Cd⁺² + 4CN^- = [Cd(CN)₄]²⁻ with K_f Cd(CN)₄]²⁻ = 1.3×10⁷

So the reverse reaction is.

[Cd(CN)₄]²⁻ = Cd⁺² + 4CN^- with 1/K_f [Cd(CN)₄]²⁻ = (1.3×10⁷)^-1

= 7.7×10^-8

and we also have

[Ag(NH₃)₂]⁺ = Ag⁺ (aq) + 2NH₃ (aq) with 1/K_f [Ag(NH₃)₂]⁺ = (1.7×10⁷)^-1

= 5.9×10^-8

You can imagine these form a salt with one another like...

[Cd(CN)₄]²⁻ [Ag(NH₃)₂]⁺ = {[Cd(CN)₄][Ag(NH₃)₂]}^-

One could prepare an equimolar mixture of these two by starting with this anionic salt, say of alkali earth metal, i.e., Na⁺, K⁺, etc. Whatever!

So we make a 1M solution of {[Cd(CN)₄][Ag(NH₃)₂]}^-, it would first dissolve into

[Cd(CN)₄][Ag(NH₃)₂]^- = [Cd(CN)₄]²⁻ (aq) + [Ag(NH₃)₂]⁺ (aq)

This undergoes further dissolution into what we are interested in ....

[Cd(CN)₄]²⁻(aq) + [Ag(NH₃)₂]⁺(aq) = Ag⁺(aq) + 2NH₃(aq) + Cd⁺²(aq)+ 4CN^-(aq)

K_fCd⁺²= 7.7×10^-8

K_fAg⁺= 5.9×10^-8

Write the law of mass action for this reaction.

K_fCd⁺²•K_fAg⁺ = 4.52×10^-15 = {[Cd⁺²][Ag⁺][CN^-]⁴[NH₃]²}/{[Cd(CN)₄][Ag(NH₃)₂]}

Make an ICE table. (initial, change, at equilibrium)

_______________[Cd(CN)₄]²⁻ + [Ag(NH₃)₂]⁺ = Ag⁺ + 2NH₃ + Cd⁺² + 4CN^-

Initial_______________1___________1___________0______0_______0______0

Δ_________________ [x]__________[x]__________[x]____[2x]²_____[x]___[4x]⁴

Equilibrium______ 1-[x]________1-[x]________[+x]___[+2x]²____[+x]__[+4x]⁴

Plug the equilibrium concentrations into the equilibrium expression.

4.52×10^-15 = {[x][x][4x]⁴[2x]²}/{[1-x]²} = 1024x⁸/(1-x)²

The below is key.

[NH₃] = [2x]² = 4x²

[CN^-] = [4x]⁴ = 256x⁴

At this point you have from the ice table entries all the necessary relationships.

So there are pre-factors and exponents relating the two. Both come from the stoichiometry. It is clear that x is small from the small 1/k_f values given. so, x² > x⁴, since everyone knows that fractions get smaller when you multiply them together. OK, so how much of a role does the pre-factor play? You can see it doesn't have that much of a contribution. It is only a factor of 64, something that would only change an exponent by at most 2, i.e., on the order of a hundred. That is the mathematical point here, if any. Pre-factors aren't as important as exponents for determining concentration relationships. More importantly, you an make estimates of concentrations by taking (1/n) fractional exponents (n^th root) of the appropriate rate constants. It is all in the law of mass action!

[NH₃] = [2x]² = 4x²

[CN^-] = [4x]⁴ = 256x⁴

There really is enough to clearly answer the question given an assumption of the magnitude of x.

The equilibrium constant is small ~1x10^-15, so x is also small <<1. Putting a range on it, x is not smaller than ~(1x10^-15)^1/2 , but not larger than ~(1x10^-15)^1/4, i.e. x10^-7.5 to x10^-3.75 which is around 3x10^-8 to 2x10^-4, respectively.

This comes from the stoichiometric coefficients transferring to the exponents in the law of mass action and their relationship to the exponents in the rate constants. You could determine the relationship between the two, if it helps. Combine the system of equations.

[NH₃]64x² = [CN^-], which would have it that [NH₃]>[CN^-], x is small.

So, we said that x < ~2x10^-4 ( from (1x10^-15)^1/4) and squaring it makes it even smaller ( 64(1x10^-15)^1/2_, 64x10^-7.5 = 2x10^-6) . [CN^-]<[NH₃], [CN^-]<<[NH₃]

Even better...imagine the case where the answer could have been [NH₃] = [CN^-], or where the case is 64x²=1. That is to say, what value of x satisfies that relationship?

x=(1/8). So write the equilibrium expression again with this value of x.

1024x⁸ = 4.52×10^-15 (1 - x)²

K_fCd+2•K_fAg+ = 4.52×10^-15, (this is the value for the K_f's from the problem, so lets solve for them treating them as unknown (K_{f Cd+2}•K_{f Ag+} ) and solving the case of 64x²=1.

K_{f Cd+2}•K_{f Ag+} = 1024x⁸/(1-x)²

(K_{f Cd+2}•K_{f Ag+})^1/2 = 256 x⁴/(1-x)

(K_{f Cd+2}•K_{f Ag+})^1/2 = 256 ((1/8)⁴) / (1-1/8)

(K_{f Cd+2}•K_{f Ag+})^1/2 = 256(1/4096)/(7/8)

(K_{f Cd+2}•K_{f Ag+})^1/2 = (1/16)*(8/7) = (1/2)*(1/7)

You could multiply this out if you wanted. Use it instead for making up a new problem where the answer is [NH₃] = [CN^-], given 1/kf = (1/2)² = 1/4 and 1/kf = (1/7)²=1/49 So, 1/k_f = 2.04x10^-2 and 1/k_f= .25; k_f =2² = 4 and k_f = 7² = 49. Why such a monster change in values? Because of the large exponents for the concentrations in the rate expression.

Let us keep proving the correct answer for the problem posed initially, that with both large and similarly valued k_f values. You see, it isn't only that they are similar in value, they need to be large in magnitude as well. The quick estimate of the magnitude of the concentration is given by raising the rate constant to the 1/n fractional power of the corresponding stoichiometric coefficient as above.

x>0 since concentrations are real, so

1024x⁸ = 4.52×10^-15 (1 - x)², simplify the squared binomial.

32x⁴ = (4.52×10^-15)^1/2 (1 - x)

32x⁴ = 7.72×10^-8 - x*7.72×10^-8

32x⁴ + x*7.72×10^-8 - 7.72×10^-8 = 0

x=7.00x10^-3, so we were right in assuming x<<1, and

[NH₃]64x² = [CN^-]

[NH₃]64(7.00x10^-3)²=[CN^-]

3.14x10^-3[NH₃]=[CN^-]

The answer not only is [NH₃]>[CN^-], but [NH₃] is 318 times larger than [CN^-]. Quite a bit!

Though the question didn't ask it, ammonia will do something in solution.

One should then consider the base hydrolysis of water by ammonia. Ammonia is a gas that dissolves in water to react to form ammonium ion (NH₄⁺) and hydroxyl ion (OH^-), so it would go something like...

NH₃ (g) = NH₃ (aq), the solution phase ammonia gas solubility at a pressure has an equilibrium rate constant k_dissolved = [NH₃(aq)]/[NH₃(g)]

NH₃ (aq) + H₂O = NH₄⁺ (aq) + OH^- , and so does base hydrolysis, k_b .