Question

1. Here you have saturated Al(OH)₃ solution in water.(Ksp of Al(OH)₃ is 4.6x10^-33)

a) What is the concentration of OH- in this solution?

b) If you prepare a saturated Al(OH)₃ solution in presence of 0.015 mol of OH- in the solution, what will be the concentration of Al³⁺ and OH-?

Answer 1

Al(OH)3 <-------------> Al+3 + 3OH-

is the solubility equilibrium and the equilibrium constant

Ksp = [Al+3] [Oh-] ³

a) Given Ksp = 4.6x10^-33

Al(OH)3 <-------------> Al+3 + 3OH-

- s 3s where s is the solubility of salt

then Ksp = s (3s)³

=27s⁴

= 4.6x10^-33

Thus s = 1.1424 x10^-7M

Thus the [Oh-] = 3s = 3 x1.1424x10^-7

= 3.427x10^-7 M

b)

Al(OH)3 <-------------> Al+3 + 3OH-

- s 3s where s is the solubility of salt

Now [OH]= 0.015 M

Al(OH)3 <-------------> Al+3 + 3OH-

- s 0.015+s where s is the solubility of salt in the presence of common ion

As value of s is very small compared to 0.015 , 0.015+s = 0.015

Thus Ksp = 4.6x10^-33 = [Al+3] [0.015]³

Thus [Al+3] = 1.36x10^-27 M and

[OH-] = 0.015 M