Question

Using the solubility of Ca(IO3)2 that you determined in 0.01M KIO3 as a solvent, calculate the value of Ksp using only concentrations, that is assuming that all activity coefficients are 1.0. You should report 3 significant figures.

Answer 1

`\mathbf{K_(sp) = 9.65*10^(-7) \ to \ 3 \ significant \ figures}`

Step-by-step explanation:

Dissociation of Ca(IO₃)₂ is as follows:

Ca(IO₃)₂ ⇄ Ca²⁺ + 2IO₃⁻

This implies that for every mole of Ca(IO₃)₂ that dissolves in water' 1 mole of Ca²⁺ is obtained and 2 moles of IO₃⁻ is obtained.

The solubility in mol/L when Ca(IO₃)₂ is added to KIO₃ = 0.00341

Concentration of Ca²⁺ will be = 0.00341 M ˣ 1 = 0.00341 M ( i.e equal to the solubility of Ca(IO₃)₂)

Concentration of IO₃⁻ will be = 0.00341 M ˣ 2 = 0.00682 M

IO₃⁻ is also obtained from the dissociation of KIO₃ in water. the IO₃⁻ obtained from this process is = 0.01 M

The total concentration of IO₃⁻ now = 0.00682 M + 0.01 M = 0.01682 M

The solubility product
`K_(sp)` can be calculated by the formula:

`K_(sp) = [Ca^(2+)][IO^-_3]^2`

`K_(sp) = [0.00341][0.01682]^2`

`K_(sp) = [0.00341][2.829124*10^(-4)]`

`K_(sp) = 9.64731284*10^(-7)`

`\mathbf{K_(sp) = 9.65*10^(-7) \ to \ 3 \ significant \ figures}`