Question

Calculate the molar solubility of Al(OH)₃ in a buffer solution with 0.100 M NH₃ and 0.400 M NH₄+.

a) 0.040 M
b) 0.020 M
c) 0.010 M
d) 0.0050 M

Answer 1

To calculate the molar solubility of Al(OH)₃ in a buffer solution with 0.100 M NH₃ and 0.400 M NH₄+, we can use the common ion effect. The molar solubility will be equal to the concentration of Al³⁺ in the solution at equilibrium.

Step-by-step explanation:

The molar solubility of Al(OH)₃ in the buffer solution can be calculated using the common ion effect. In this case, the common ion is NH₄⁺ from NH₄⁺. The Ksp expression for Al(OH)₃ is given by:

Ksp = [Al³⁺][OH⁻]³

Since Al(OH)₃ dissolves to form one Al³⁺ ion and three OH⁻ ions:

Ksp = [Al³⁺]([OH⁻]³) = [Al³⁺]([NH₄⁺][OH⁻] + [NH₃][OH⁻]³)

Substituting the given concentrations:

Ksp = [Al³⁺][(0.400)([OH⁻]) + (0.100)([OH⁻]³)] = [Al³⁺] + (0.004[OH⁻]³)

The molar solubility of Al(OH)₃ will be equal to the concentration of Al³⁺ in the solution at equilibrium. To find that, we can calculate the concentration of OH⁻ using the formula:

[OH⁻] = sqrt(Ksp / (0.004 - [Al³⁺]))

Finally, substituting the values:

[OH⁻] = sqrt((1.2E-33) / (0.004 - [Al³⁺]))

Solving for [Al³⁺] gives the molar solubility of Al(OH)₃ in the buffer solution.