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Chapter 16: Problem 41

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a \(0.20 \mathrm{M} \mathrm{KIO}_{3}\) solution is \(4.4 \times\) \(10^{-8} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\).

Short Answer

Expert verified
The solubility product constant, Ksp, for Ce(IO3)3 is approximately \(6.364 \times 10^{-21}\).

Step by step solution

01

Write the balanced chemical equation for Ce(IO3)3 dissolution

Cе(IO3)3(s) ⇌ Ce^3+(aq) + 3 IO3^-(aq)
02

Write down the given concentrations and solubilities

Initial concentration of KIO3: 0.20 M Solubility of Cе(IO3)3 in KIO3 solution: 4.4 x 10^-8 mol/L
03

Set up an ICE table

ICE table for the reaction: | | Ce^3+(aq) | IO3^-(aq) | | ------ | --------- | --------- | | Initial| 0 | 0.20 | | Change | +s | +3s | | Equilibrium| s | 0.20 + 3s | The solubility of Ce(IO3)3 is given as 4.4 x 10^-8 mol/L. So, s = 4.4 x 10^-8 M.
04

Find the equilibrium concentrations

At equilibrium, the concentrations will be: Ce^3+(aq): 4.4 × 10^-8 M IO3^-(aq): 0.20 + 3 × (4.4 × 10^-8) = 0.20 + 1.32 × 10^-7 = 0.200000132 M
05

Write the Ksp expression for the balanced equation

For the balanced equation, Ksp = [Ce^3+][IO3^-]^3
06

Calculate the Ksp value for Ce(IO3)3

Plug the equilibrium concentrations into the Ksp expression: Ksp = (4.4 × 10^-8)(0.200000132)^3 Ksp = 6.364 × 10^-21 The Ksp for Ce(IO3)3 is approximately 6.364 × 10^-21.

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Most popular questions from this chapter

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{~L}\) of a \(1.0 \mathrm{M} \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}=4.0 \times\) \(10^{-11}\) and \(K_{\mathrm{a}}\) for \(\mathrm{HF}=7.2 \times 10^{-4}\). Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\).

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