Chapter 18, Problem 71A

Interpretation Introduction

Interpretation: The given compounds, from the most soluble to least soluble, using the data from Table $18.2$ , are ranked.

Concept introduction: The solubility product $\left(K_{\text{sp}}\right)$ is the product of the concentrations of ions in the salt, with each concentration multiplied by the dissociation reaction coefficient for that ion. The lower value of the solubility product constant decreases the compound's solubility.

Answer to Problem 71A

The rank of the given compounds from the most soluble to least soluble is as follows:

1. c. ${\text{SrCO}}_{\text{3}}$
2. b. ${\text{BaSO}}_{\text{4}}$
3. d. $\text{AgI}$
4. a. $\text{CuS}$

Explanation of Solution

The given compounds are as follows:

a. $\text{CuS}$ b. ${\text{BaSO}}_{\text{4}}$ c. ${\text{SrCO}}_{\text{3}}$ d. $\text{AgI}$

The $K_{\text{sp}}$ values from Table $18.2$ are as follows:

1. $K_{\text{sp}}\left(\text{CuS}\right)=8.0\times {10}^{-37}$
2. $K_{\text{sp}}\left({\text{BaSO}}_{\text{4}}\right)=1.1\times {10}^{-10}$
3. $K_{\text{sp}}\left({\text{SrCO}}_{\text{3}}\right)=9.3\times {10}^{-10}$
4. $K_{\text{sp}}\left(\text{AgI}\right)=8.3\times {10}^{-17}$

The order of the solubility product constant is as follows:

  $K_{\text{sp}}\left({\text{SrCO}}_{\text{3}}\right)>K_{\text{sp}}\left({\text{BaSO}}_{\text{4}}\right)>K_{\text{sp}}\left(\text{AgI}\right)>K_{\text{sp}}\left(\text{CuS}\right)$

The compound's solubility decreases as the solubility product constant decreases.

The order of the given compounds from the most soluble to least soluble is as follows:

  ${\text{SrCO}}_{\text{3}}>{\text{BaSO}}_{\text{4}}>\text{AgI}>\text{CuS}$

The rank of the given compounds from the most soluble to least soluble is as follows:

c. ${\text{SrCO}}_{\text{3}}$

b. ${\text{BaSO}}_{\text{4}}$

d. $\text{AgI}$

a. $\text{CuS}$

Conclusion

The rank of the given compounds from the most soluble to least soluble is as follows:

c. ${\text{SrCO}}_{\text{3}}$

b. ${\text{BaSO}}_{\text{4}}$

d. $\text{AgI}$

a. $\text{CuS}$