Question

What is the molar solubility of silver bromide in each of the following solutions? a. pure...

What is the molar solubility of silver bromide in each of the following solutions?

a. pure water

b. 0.0550 M sodium bromide

c. 0.250 M silver nitrate

Homework Answers

Answer #1

a) In pure water
The Ksp of AgBr =5.2*10^-13
AgBr <-----> Ag+ + Br-
let x = moles/L of AgBr that dissolves : this will give us x mol/L of Ag+ and x mol/L of Br-

[Ag+] = x
[Br-] = x
Ksp = [Ag+][Br-] = (x) ( x )=5.2 x 10^-13
x^2=5.2 * 10^-13
molar solubility x = 7.21*10^-7 M

b. 0.0550 M sodium bromide
AgBr <-----> Ag+ + Br-
let x = moles/L of AgBr that dissolves : this will give us x mol/L of Ag+ and x mol/L of Br-
But we have arleady 0.0550 M of Br- so
[Ag+] = x
[Br-] = x + 0.0550

Ksp = [Ag+][Br-] = (x) ( x+0.0550) = 5.2 x 10^-13

x = molar solubility =9.45*10^-12 M

c. 0.250 M silver nitrate
AgBr <-----> Ag+ + Br-
let x = moles/L of AgBr that dissolves : this will give us x mol/L of Ag+ and x mol/L of Br-
But we have arleady 0.0250 M of Ag+ so
[Ag+] = x + 0.250
[Br-] = x

Ksp = [Ag+][Br-] = (x+0.250) ( x) = 5.2*10^-13

x = molar solubility =2.08*10^-12 M

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