Answer:
[tex]\mathbf{K_{sp} = 9.65*10^{-7} \ to \ 3 \ significant \ figures}[/tex]
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 [tex]K_{sp}[/tex] can be calculated by the formula:
[tex]K_{sp} = [Ca^{2+}][IO^-_3]^2[/tex]
[tex]K_{sp} = [0.00341][0.01682]^2[/tex]
[tex]K_{sp} = [0.00341][2.829124*10^{-4}][/tex]
[tex]K_{sp} = 9.64731284*10^{-7}[/tex]
[tex]\mathbf{K_{sp} = 9.65*10^{-7} \ to \ 3 \ significant \ figures}[/tex]
