How to Calculate Solubility in Presence of Common Ion

Let's take a look at our discussion question this week!

230130 question

We want to determine solubility of iron (II) hydroxide in 0.10 moldm-3 of FeSO4.

Let's consider the dissociation of both salts.

230130 dissociation of 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).

Check out this video lesson to learn more about the common ion effect.

Writing the Ksp expression for Fe(OH)2:

230130 Ksp

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.

230130 approximate conc of Fe ii

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):

230130 calculate solubility

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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