# How to Calculate Solubility in Presence of Common Ion

Let's take a look at our discussion question this week! We want to determine solubility of iron (II) hydroxide in 0.10 moldm-3 of FeSO4.

Let's consider the dissociation of both salts. Let y be solubility of iron hydroxide, it will dissociate to give y moldm-3 Fe2+ and 2y moldm-3 OH-.

0.10 moldm-3 of FeSO4 will dissociate to give 0.10 moldm-3 of Fe2+ and SO42-.

Notice both salts give Fe2+, so Fe2+ is the common ion.

Based on the common ion effect, we know that solubility of Fe(OH)2 will be suppressed, so we expect solubility in FeSO4 (y) to be lower than solubility in water (x).

Writing the Ksp expression for Fe(OH)2: Concentration of Fe2+ in the system will include contribution of Fe2+ from both salts = (0.10 + y)

Concentration of OH- = 2y

We need to approximate and simplify the calculation as solving for quadratic equation is not in A Level Chemistry syllabus. We assume contribution of Fe2+ from the sparingly soluble salt to be negligible, hence conc of Fe2+ will be 0.10 moldm-3.

We can now calculate solubility (y): This is a smaller value as compared to solubility of Fe(OH)2 in water (previously calculated to be 5.82 x 10-6 moldm-3) which is consistent with the common ion effect.

Topic: Solubility Product, Physical Chemistry, A Level Chemistry, Singapore

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