- #1

brake4country

- 216

- 7

## 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!