Chapter 17.4, Problem 9PPB

 (a)

Interpretation Introduction

Interpretation:

The amount of compounds required to dissolve without precipitate has to be calculated.

Concept introduction:

• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion, raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

To calculate: The amount of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to dissolve ${\text{BaCl}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

The amount of compounds required to dissolve without precipitate has to be calculated.

Concept introduction:

• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion, raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

To calculate: The amount of ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ required to dissolve ${\text{BaCl}}_{\text{2}}$

(c)

Interpretation Introduction

Interpretation:

The amount of compounds required to dissolve without precipitate has to be calculated.

Concept introduction:

• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion, raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

To calculate: The amount of $\text{NaF}$ required to dissolve ${\text{BaCl}}_{\text{2}}$