Questions: Consider these compounds: A. PbI2 B. AgI C. CO(OH)2 D. CaCrO4 Complete the following statements by entering the letter corresponding to the correct compound. Without doing any calculations it is possible to determine that silver chromate is more soluble than , and silver chromate is less soluble than It is not possible to determine whether silver chromate is more or less soluble than by simply comparing Ksp values.

Solution Steps

First, identify the compounds given in the question:

• A. \(\mathrm{PbI}_{2}\) (Lead(II) iodide)
• B. \(\mathrm{AgI}\) (Silver iodide)
• C. \(\mathrm{CO(OH)}_{2}\) (Cobalt(II) hydroxide)
• D. \(\mathrm{CaCrO}_{4}\) (Calcium chromate)

The question involves comparing the solubility of silver chromate (\(\mathrm{Ag}_2\mathrm{CrO}_4\)) with the given compounds. Solubility is often compared using the solubility product constant (\(K_{\mathrm{sp}}\)).

• More Soluble than Silver Chromate: Compounds with a higher \(K_{\mathrm{sp}}\) than silver chromate are more soluble.
• Less Soluble than Silver Chromate: Compounds with a lower \(K_{\mathrm{sp}}\) than silver chromate are less soluble.
• Indeterminate Solubility: If \(K_{\mathrm{sp}}\) values are not directly comparable or not provided, solubility cannot be determined.

• More Soluble than Silver Chromate: Without specific \(K_{\mathrm{sp}}\) values, we assume that \(\mathrm{CaCrO}_{4}\) (D) is more soluble because chromates are generally more soluble than iodides.
• Less Soluble than Silver Chromate: \(\mathrm{AgI}\) (B) is typically less soluble than silver chromate.
• Indeterminate Solubility: \(\mathrm{PbI}_{2}\) (A) and \(\mathrm{CO(OH)}_{2}\) (C) cannot be directly compared without specific \(K_{\mathrm{sp}}\) values.

Final Answer