Question

A solution is prepared by mixing 30.0mL of 0.0600 M SC³⁺ with 40.0mL of 0.0600 M OH^-. After mixing a precipate of Sc(OH)₃ settles out, and the resulting solution is found sprectrophotometrically to be 0.0186 M in Sc³⁺. Determine the K_sp of Sc(OH)₃.

Answer 1

Sc³⁺(aq) + 3(OH^-)(aq) <-------> Sc(OH)₃(s)

moles of Sc³⁺ reacting = molarity*volume of solution in litres = 0.06*0.03 = 0.0018

moles of OH^- reacting = 0.06*0.04 = 0.0024

Now, as per the balanced equation:- Sc³⁺ & (OH^-) reacts in the molar ratio of 1:3

Thus, OH^- is limiting reactant

Hence moles of Sc³⁺ left unreacted = 0.0186*0.07 = 0.0013

moles of Sc³⁺ reacting = 0.0018 - 0.0013 = 0.0005

Thus, moles of [OH^-] reacting = 3*moles of Sc³⁺ = 0.0015

Thus, At eqb.,

[Sc³⁺] = 0.0186 M ; [OH^-] = 0.0009/0.07 = 0.013 M

K_sp = 1/{[Sc³⁺]*[OH^-]³} = 2.45*10⁷