Question

rank the following compounds labelled a to e, in order of decreasing solubility in moles/liter, where 5 is assigned to the compound with the greatest solubility and 1 to the compound with the least solubility.
cobalt(ii) sulfide ($k_{sp} = 5.0 \times 10^{-22}$)
barium sulfate ($k_{sp} = 1.5 \times 10^{-9}$)
barium carbonate ($k_{sp} = 1.6 \times 10^{-9}$)
lead(ii) sulfate ($k_{sp} = 1.3 \times 10^{-8}$)
silver bromide ($k_{sp} = 5.0 \times 10^{-13}$)
5
3
4
2
1

Explanation:

Step1: Compare \(K_{sp}\) values

The solubility product constant \(K_{sp}\) is related to the solubility of a sparingly - soluble salt. For salts with the same general formula (e.g., \(AB\) type: \(K_{sp}=s\times s = s^{2}\), \(A_{2}B\) or \(AB_{2}\) type: \(K_{sp}=4s^{3}\)), a larger \(K_{sp}\) value generally corresponds to a higher solubility.
Cobalt(II) sulfide (\(CoS\)): \(K_{sp}=5.0\times 10^{-22}\) (very small, low solubility)
Barium sulfate (\(BaSO_{4}\)): \(K_{sp}=1.5\times 10^{-9}\)
Barium carbonate (\(BaCO_{3}\)): \(K_{sp}=1.6\times 10^{-9}\)
Lead(II) sulfate (\(PbSO_{4}\)): \(K_{sp}=1.3\times 10^{-8}\)
Silver bromide (\(AgBr\)): \(K_{sp}=5.0\times 10^{-13}\)

Step2: Rank based on \(K_{sp}\)

Since \(1.3\times 10^{-8}>1.6\times 10^{-9}>1.5\times 10^{-9}>5.0\times 10^{-13}>5.0\times 10^{-22}\)

Answer:

Lead(II) sulfate - 5, Barium carbonate - 4, Barium sulfate - 3, Silver bromide - 2, Cobalt(II) sulfide - 1