Questions: Which of the following aqueous solutions will not be more soluble under acidic conditions? A Ag2SO4 B ZnF2 C Mg(OH)2 D CuBr

Solution Steps

To determine which compound will not be more soluble under acidic conditions, we need to consider the nature of the anion in each compound. Compounds that contain anions which can react with \( \text{H}^+ \) ions from an acid will generally become more soluble in acidic conditions. This is because the reaction with \( \text{H}^+ \) reduces the concentration of the anion, shifting the equilibrium to dissolve more of the compound.

• A. \( \mathrm{Ag}_{2} \mathrm{SO}_{4} \): The sulfate ion (\( \text{SO}_4^{2-} \)) does not react significantly with \( \text{H}^+ \) ions, so the solubility of silver sulfate is not greatly affected by acidic conditions.

• B. \( \mathrm{ZnF}_{2} \): The fluoride ion (\( \text{F}^- \)) can react with \( \text{H}^+ \) to form \( \text{HF} \), which is a weak acid. This reaction can increase the solubility of zinc fluoride in acidic conditions.

• C. \( \mathrm{Mg}(\mathrm{OH})_{2} \): The hydroxide ion (\( \text{OH}^- \)) reacts with \( \text{H}^+ \) to form water, significantly increasing the solubility of magnesium hydroxide in acidic conditions.

• D. CuBr: The bromide ion (\( \text{Br}^- \)) does not react with \( \text{H}^+ \) ions, so the solubility of copper(I) bromide is not affected by acidic conditions.

From the analysis, both \( \mathrm{Ag}_{2} \mathrm{SO}_{4} \) and CuBr have anions that do not react with \( \text{H}^+ \). However, \( \mathrm{Ag}_{2} \mathrm{SO}_{4} \) is slightly more soluble in acidic conditions due to the complexation of \( \text{Ag}^+ \) with \( \text{H}^+ \) in some cases. CuBr, on the other hand, remains largely unaffected by acidic conditions.

Final Answer