Question

The solubility product (K_sp) of tin(II) hydroxide is 5.4 x 10^-17.

^{a. Write the balanced reaction equation for the process of dissolving tin(II) hydroxide.}

^{b. Write the K_spexpression for this reactions.}

c. Calculate the molar and mass solubility of tin (II) hydroxide using the K_sp expression.

d. How will the solubility be changed if tin (II) hydroxide is dissolved into an acidic solution? explain why.

e. How will the solubility be changed if tin (ll) hydroxide is disolved into a basic solution? explain why

Answer 1

a) The balanced reaction equation for the process of dissolving tin(II) hydroxide is:

Sn(OH)₂(s)    Sn²⁺(aq) + 2 OH^-(aq)

b) let molar solubility of Sn(OH)₂ = x mol/L

This will give x mol/L Sn²⁺ and 2x mol/L OH^-

Ksp = [Sn²⁺][OH^-]² = (x)(2x)² = 4x³

c) Ksp = [Sn²⁺][OH^-]² = (x)(2x)² = 4x³

or, 4x³ = 5.4 x 10^-17

or, x =(5.4 x 10^-17/4)^1/3 = 2.38 X 10^-6 M

d) If tin (II) hydroxide is dissolved into an acidic solution, solubility increases. Because the OH^- ions are neutralized by H⁺ furnished by the acid. Hence, more amount of tin (II) hydroxide will ionise.

e) f tin (II) hydroxide is dissolved into an basic solution, solubility decreases. Because the ionisation of tin (II) hydroxide will be supressed due to the OH^- ions furnished by the base. It is due to common ion effect.