Question

What is the solubility of AgCl in a solution of 0.10 mole/L NaBr?1.(NaBr completely dissociates.)

Answer 1

AgCl Solubility

in

NaBr = 0.1 M

Ksp of AgCl = 1.8*10^-10

Ksp AgBr = 5*10^-13

then

AgCl <-->Ag+ and Cl-

NaBr ---> Na+ and Br-

then

Ag+ and Br- ---> AgBr(s)

solve for Ag+

Ksp = [ag+][Br-]

5*10^-13 = (S)(S)

S = sqt(5*10^-13) = 7.0710*10^-7 M

Then

[Ag+] = 7.07*10^-7

Therefore

for solubility of AgCl

Ksp = [Ag+][Cl-]

(1.8*10^-10) = (7.07*10^-7)(S)

S = (1.8*10^-10)/(7.07*10^-7) = 0.00025459688 M of AgCl will be in solution

makes sense that solubilty increases since NaBr will precipitate plenty of Ag+