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## Chemistry library

### Course: Chemistry library>Unit 14

Lesson 3: Solubility equilibria

# Solubility and complex ion formation

Calculating the solubility of silver chloride in pure water and 3.0 M ammonia. Calculating new equilibrium constant to account for formation of complex ion.  Created by Jay.

## Want to join the conversation?

• Was the reason why you decided to keep the 3-2x denominator term @ (instead of assuming it to be 3) due to the magnitude of the result? •  Generally if ksp < 1.0*10^-5 you can assume the x is small enough to disregard. In this problem ksp= 2.4*10^-3 so x is too large to disregard it.
• In part (b), why do we need 2 NH3 to interact with 1 Ag+? How can we determine how many NH3 to be used when we are doing the question? • At why do you multiply the Ksp and the Kf to get a new K for the net reaction? What is the reasoning behind this rule? • At , I don't understand when we can or can't assume x is small. If the K value was 2.9x10^-5 instead of 2.9x10^-3 can I assume x is small? What is the guideline for assuming x is small? Thanks. • A good guideline for the "small x approximation" is the 5% rule. It's used by comparing the x value you get, to the approximation you make in the first steps.

Let's take the dissociation of 0.1M acetic acid (I'll call it HX for simplicity sake), a weak acid with Ka = 1.8x10^-5.

HX + H2O = X- + H3O+
0.1M 0 0
-x +x +x
0.1M-x x x

[the formatting messed up here, but it's a standard ICE table with HX dissociating]

Then Ka = 1.8x10^-5 = x^2/(0.1M-x)

Now that 0.1M-x on the denominator of the right hand side is the pain. Go ahead and approximate it to 0.1, then solve for x and you get x = 0.0134

[this should have been 0.00134, please refer to Jo Ha's comment below]

To check if your approximation of 0.1-x = 0.1 is valid, divide x by 0.1 and see if it falls within 5%:

0.0134/0.1 = 0.134 = 13.4%

Because we exceed 5%, our approximation was invalid and we need to return and solve this problem using quadratic. This would result in x = 0.00133, which is about 10x less than what we got for x when we approximated.

And if you analyze the math in this closely, you can conclude that the small x approximation is going to be invalidated when Ka is really large (i.e., a relatively strong weak acid) and the concentration is really small.
• How can one know that the silver ion will react with ammonia? • at part b I don't understand where the 2 is coming from when he writes NH3? Why is there a two in front of NH3? • For example b), if Ag has a +1 ion charge in the Reactants, why does it have a +2 charge in Products? This confuses me because 2 NH3's are bound to Ag in the Products when it should only just bind 1 NH3. I dont understand why the balanced equation is not: Ag^+(aq) + NH3(aq) = Ag(NH3)^+.
- I understand that NH3 is neutral; this meaning it has no charge. But why are there 2 NH3's in the Reactants? • What's the explanation, or at least intuition behind multiplying equilibrium constants when adding equations? Is it about finding a common probability by multiplying probabilities (individual equilibrium constants) of independent events? • In this problem the equations add up to form the third equation, which has an equilibrium equivalent to the product of the first two equations. What if the products and solutions cancelled out by a different factor (for example the second equation was something like 2Ag + NH3 -> some product. Would we need to multiple the Ksp twice to get the new equilibrium constant? • You would have had to square the K_sp.
The equations might have looked something like this:
2AgCl(s) ⇌ 2Ag⁺(aq) + 2Cl⁻(aq); K₁ = K_sp² = 3.2 × 10⁻²⁰
2Ag⁺(aq) + NH₃(aq) ⇌ Ag₂(NH₃)²⁺(aq); K_i = 1.6 × 10⁷
2AgCl(s) + NH₃(aq) ⇌ Ag₂(NH₃)²⁺(aq) + 2Cl⁻(aq);
K = K₁K_i = 3.2×10⁻²⁰ × 1.6×10⁷ = 5.2 × 10⁻¹³ 