Question

Calculate the molar solubility of the following compounds: (a) calcium iodate: \(K_{\mathrm{sp}}=6.5 \times 10^{-6}\) (b) cadmium(III) phosphate: \(K_{\mathrm{sp}}=2.5 \times 10^{-33}\) (c) silver oxalate: \(K_{\mathrm{sp}}=5.4 \times 10^{-12}\)

Short answer

Question: Calculate the molar solubility of the following compounds using the given \(K_{sp}\) values: (a) Calcium iodate, \(K_{sp} = 6.5 \times 10^{-6}\), (b) Cadmium(III) phosphate, \(K_{sp} = 2.5 \times 10^{-33}\), (c) Silver oxalate, \(K_{sp} = 5.4 \times 10^{-12}\). Answer: The molar solubility of the given compounds are: (a) Calcium iodate: \(2.83 \times 10^{-3} \, M\), (b) Cadmium(III) phosphate: \(1.25 \times 10^{-11} \, M\), (c) Silver oxalate: \(3.07 \times 10^{-4} \, M\).

Step-by-step solution

Write dissolution reactions

For each compound, we need to write the balanced dissolution reaction. (a) Calcium iodate: \(Ca(IO_3)_2 \rightarrow Ca^{2+} + 2IO_3^-\) (b) Cadmium(III) phosphate: \(Cd_3(PO_4)_2 \rightarrow 3Cd^{2+} + 2PO_4^{3-}\) (c) Silver oxalate: \(Ag_2C_2O_4 \rightarrow 2Ag^+ + C_2O_4^{2-}\)

Set up the ICE table

We will set up an Initial-Change-Equilibrium (ICE) table for each compound and then we can relate the change in concentration to the molar solubility. (a) Calcium iodate: | | \(Ca^{2+}\) | \(IO_3^-\) | |----------------|-----------|-----------| | Initial | 0 | 0 | | Change | +x | +2x | | Equilibrium | x | 2x | (b) Cadmium(III) phosphate: | | \(Cd^{2+}\) | \(PO_4^{3-}\) | |----------------|-----------|--------------| | Initial | 0 | 0 | | Change | +3x | +2x | | Equilibrium | 3x | 2x | (c) Silver oxalate: | | \(Ag^+\) | \(C_2O_4^{2-}\) | |----------------|--------|---------------| | Initial | 0 | 0 | | Change | +2x | +x | | Equilibrium | 2x | x |

Write expression for \(K_{sp}\) and solve for molar solubility

Now, we will write the expression for \(K_{sp}\) for each compound using the concentrations from the ICE table and solve for the molar solubility, x. (a) Calcium iodate: \(K_{sp} = [Ca^{2+}][IO_3^-]^2 = (x)(2x)^2 = 6.5 \times 10^{-6}\) Solving for x, we get: \(x ≈ 2.83 \times 10^{-3} \, M\) (b) Cadmium(III) phosphate: \(K_{sp} = [Cd^{2+}]^3[PO_4^{3-}]^2 = (3x)^3(2x)^2 = 2.5 \times 10^{-33}\) Solving for x, we get: \(x ≈ 1.25 \times 10^{-11} \, M\) (c) Silver oxalate: \(K_{sp} = [Ag^+]^2[C_2O_4^{2-}] = (2x)^2(x) = 5.4 \times 10^{-12}\) Solving for x, we get: \(x ≈ 3.07 \times 10^{-4} \, M\) So, the molar solubility of the given compounds are: (a) Calcium iodate: \(2.83 \times 10^{-3} \, M\) (b) Cadmium(III) phosphate: \(1.25 \times 10^{-11} \, M\) (c) Silver oxalate: \(3.07 \times 10^{-4} \, M\)