# Calculate the solubility of AgBr in 0.005 M NaBr solution. Ksp = 5 x 10–13?

**Answer:**

The equilibrium is

AgBr <> Ag+ + Br-

Ksp = [Ag+][Br-]

Let x = moles/L of AgBr that dissolve . We get x moles/l Ag+ and x moles/L Br-.

But there is arleady some Br- in the solution corresponding to 0.005 M NaBr.

Therefore total Br- concentration at equilibrium will be x + 0.005

5 x 10^-13 = x ( x+0.005)

since x is small compared to 0.05 we can assume

5 x 10^-13 = (x) (x)

x = molar solubility =7.1 x 10^-7 M

the first guy said it correctly and then did the math wrong.

Step 1: When given a Ksp value, set up the chemical reaction that corresponds to the Ksp

AgBr (s) -------> Ag^+1 (aq) + Br^-1 (aq)

Step 2: Make and ICE table, fill in what you have - but remember, in K expressions no solids or liquids appear, which means that we can ignore AgBr (its a solid - that's what Ksp values are for!)

In this case, we are told that AgBr is not in water, it is in a solution which is 0.005 M NaBr. Well, in this compound, NaBr, is there anything that is in common with AgBr? Correct, the Br. This is called a common ion problem. So the question is - if you have 0.005 M NaBr, what [Br^-1] is that?

0.005 moles NaBr/1L * 1 mole Br^-1/1 mol NaBr = 0.005 moles Br^-1/1L

0.005 M NaBr = 0.005 M Br^-1 (because the NaBr and Br in the compound are in a 1:1 mole ratio)

So set up ICE table (the lines are just placeholders)

___AgBr (s) ----> _Ag^+1 (aq) + Br^-1 (aq)

I xxxxxxx_______0_________0.005

C xxxxxx________+x_________+x

E xxxxxx________x________0.005 +x

Ksp = [Br^-1][Ag^+1]

remember that solids do not appear

5 x 10^-13 = [x][0.005 +x]

We will make the assumption that x is small, what this means is that 0.005 + x = 0.005 (that's where the guy above went wrong)

5 x 10^-13 = [x][0.005]

x = 5 x 10^-13/0.005 = 1 x 10^-10

Well, let's check our assumption: change(x)/inital amount * 100

1 x 10^-10/0.005 * 100 = 0.5

This is called the 5% rule. If the change (x) value, divided by the inital concentration that you are comparing x to (0.005) times 100 is less than 5% the assumption that you made is valid and x is indeed small

So: [Ag^+1] = x = 1 x 10^-10 M

[Br^-1] = 0.005 + 1 x 10^-10 = 0.005 M

[AgBr] (molar solubility) = ?

Well, we don't want to use the Br^-1 concentration to figure out the solubility of AgBr, because the Br^-1 in solution is coming predominantly from the NaBr that is in solution.

So the molar solubility of the parent species is always determined based off the ion that you did NOT screw around with, in this case, Ag^+1

1 x 10^-10 moles Ag^+1/1L * 1 mol AgBr/1 mol Ag^+1 = 1 x 10^-10 M

Since Ag^+1 and AgBr have a 1:1 mole ratio, the molar solubility of AgBr is the same as the concentration of Ag^+1 in solution

[AgBr] = 1 x 10^-10 M

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