Question

Determine the molar solubility of AgBr in:

Ksp (AgBr) = 7.7 x 10^ -13

a. a solution of pure water

b. a solution containing 0.150 M NaBr

c. A solution containing 0.25 M NaCL

Answer 1

AgBr <-------> Ag+ + Br-
let x = mol/L of AgBr that dissolve
this will give x mol/L Ag+ and x mol/L Br-
Ksp = 7.7 x 10^-13 = [Ag+][Br-] = (x)(x)

Ans a) Ksp = 5 * 10^-13 and solubility in terms of mole per litre = x = 8.77 * 10^-7

Now we have 0.150M Solution of NaBr

thus [Br-] = (0.15 + x)

thus we have Ksp =7.7 * 10 ^-13 = x * (0.15 +x ) thus we get x = 5.1 * 10^-12

Answer : Thus 5.1 * 10 ^-12 moles of AgBr gets dissolved in 0.15M NaBr ( The solubility is suppressed here)

Ans c)

Since the solution of NaCl gives Na + and Cl - ions it does not take part in the equilibrium of solubility product hence the answer remains same as "part a"

Ans c) Dissolved Moles of NaBr = 8.77 * 10^-7

Answer 2

Ksp= 7.7*10^-13
A)
AgBr--->Ag+   + Br-
                x             x
Ksp= [Ag+ ][Br-]
7.7 x 10^-13 =(x)(x)
x = molar solubility = 8.77 x 10^-7 M

b)
NaBr-->Na+ + Br-
              0.15       0.15

AgBr--->Ag+   + Br-
                x             x

7.7 x 10^-13 = (x )(x+0.15)
x^2+0.15x-7.7 x 10^-13=0

x = molar solubility = 5.13 x 10^-13 M

c)
NaCl has no common ion effect

AgBr--->Ag+   + Br-
                x             x
Ksp= [Ag+ ][Br-]
7.7 x 10^-13 =(x)(x)
x = molar solubility = 8.77 x 10^-7 M