How to calculate the solubility of AgCl in a solution containing NaCl?

[brake4country]

Homework Statement

What mass of AgCl will dissolve in 1 L of water containing 0.0144 moles of NaCl. Ksp = 1.7 x 10^-10

Homework Equations

Ksp = [x][x] and ICE table

The Attempt at a Solution

So, the common ion effect is taking place here and the equilibrium taking place is:
AgCl(s) ↔ Ag+ + Cl-

Since excess Cl- will be present from the NaCl, the Na+ is just a spectator ion.

ICE table:
AgCl(s) ↔ Ag+ + Cl-
x 0 0.0144 M
-x x x
0 x 0.0144 + x

Ksp = [Ag+][Cl-] = 1.7 x 10^-10

(x)(0.0144 + x) = 1.7 x 10^-10

I assume that I can eliminate the x in the parentheses because it would be very small, thus:
0.0144x = 1.7 x 10^-10
x = 1.2 x 10^-8 mol/L
Converting to grams gives me 1.8 x 10^-6 g/L

I was wondering if someone wouldn't mind checking to see if my steps are accurate. Thanks in advance!

 

[Borek]

What have you used for molar mass of AgCl? I got a slightly different result.

 

[brake4country]

I used 144 g/mol. I will recheck my math.

 

[brake4country]

Yes, x = 1.2x10^-8 mol/L
Thus, (1.2x10^-8 mol/L)(144 g/L) = 1.8x10^-6 g/L

 

[Borek]

(1.2x10^-8 mol/L)(144 g/L) = 1.8x10^-6 g/L

No matter how many times you will write it, it won't get correct. Sorry.

 

[brake4country]

Wait! The steps are more important! Can I get any advice on how I am approaching these problems?

 

[Borek]

Your approach is OK, the final result is not.

 

[James Pelezo]

Borek is right, recheck your formula wt.