Answer :
The solubility product constant (Ksp) of potassium iodate (KIO₃) at a certain temperature, where its solubility is 59.1 g/L, is calculated to be approximately 7.63 x 10⁻².
To calculate the solubility product constant (Ksp) for potassium iodate (KIO₃), we need to first understand the dissociation that occurs when KIO₃ dissolves in water:
KIO₃(s) ⇌ K⁺(aq) + IO₃⁻(aq)
From this, we can see that for each mole of KIO₃ that dissolves, one mole of K⁺ and one mole of IO₃⁻ are produced. Therefore, if the solubility of KIO₃ is 59.1 g/L, we convert this mass to moles using its molar mass (214.00 g/mol for KIO₃).
59.1 g/L ÷ 214.00 g/mol ≈ 0.276 mol/L
Because the stoichiometry of the ions to KIO₃ is 1:1, the concentrations of K⁺ and IO₃⁻ will also be 0.276 mol/L. This means the solubility product constant is:
Ksp = [K⁺][IO₃⁻] = (0.276)(0.276) ≈ 7.63 x 10⁻²
This value represents the solubility product constant for KIO₃ at the given temperature.
