Answer :
Sure, let's solve this step by step.
First, let's carefully understand and define the given parameters:
- At [tex]\( 25^{\circ} C \)[/tex], the solubility of the salt [tex]\( AB \)[/tex] in water is [tex]\( 0.0400 \, \text{mol/L} \)[/tex]. This means that at saturation, [tex]\( 0.0400 \, \text{mol/L} \)[/tex] of [tex]\( AB \)[/tex] dissolves in water.
Now, write the dissociation reaction for the salt [tex]\( AB \)[/tex]:
[tex]\[ \text{AB}(s) \rightleftharpoons \text{A}^+(aq) + \text{B}^-(aq) \][/tex]
Since the salt [tex]\( AB \)[/tex] dissociates into its constituent ions [tex]\( \text{A}^+ \)[/tex] and [tex]\( \text{B}^- \)[/tex], the concentrations of the ions in a saturated solution are both equal to the solubility of [tex]\( AB \)[/tex]. Therefore:
[tex]\[ [\text{A}^+] = 0.0400 \, \text{mol/L} \][/tex]
[tex]\[ [\text{B}^-] = 0.0400 \, \text{mol/L} \][/tex]
The solubility product constant [tex]\( K_{sp} \)[/tex] is calculated by multiplying the molar concentrations of the ions in the saturated solution:
[tex]\[ K_{sp} = [\text{A}^+][\text{B}^-] \][/tex]
Substitute the concentrations into the equation:
[tex]\[ K_{sp} = (0.0400 \, \text{mol/L}) \cdot (0.0400 \, \text{mol/L}) \][/tex]
[tex]\[ K_{sp} = 0.0400^2 \][/tex]
[tex]\[ K_{sp} = 0.0016 \][/tex]
Therefore, the solubility product constant [tex]\( K_{sp} \)[/tex] of the salt [tex]\( AB \)[/tex] at [tex]\( 25^{\circ}C \)[/tex] is:
[tex]\[ K_{sp} = 0.0016 \][/tex]
First, let's carefully understand and define the given parameters:
- At [tex]\( 25^{\circ} C \)[/tex], the solubility of the salt [tex]\( AB \)[/tex] in water is [tex]\( 0.0400 \, \text{mol/L} \)[/tex]. This means that at saturation, [tex]\( 0.0400 \, \text{mol/L} \)[/tex] of [tex]\( AB \)[/tex] dissolves in water.
Now, write the dissociation reaction for the salt [tex]\( AB \)[/tex]:
[tex]\[ \text{AB}(s) \rightleftharpoons \text{A}^+(aq) + \text{B}^-(aq) \][/tex]
Since the salt [tex]\( AB \)[/tex] dissociates into its constituent ions [tex]\( \text{A}^+ \)[/tex] and [tex]\( \text{B}^- \)[/tex], the concentrations of the ions in a saturated solution are both equal to the solubility of [tex]\( AB \)[/tex]. Therefore:
[tex]\[ [\text{A}^+] = 0.0400 \, \text{mol/L} \][/tex]
[tex]\[ [\text{B}^-] = 0.0400 \, \text{mol/L} \][/tex]
The solubility product constant [tex]\( K_{sp} \)[/tex] is calculated by multiplying the molar concentrations of the ions in the saturated solution:
[tex]\[ K_{sp} = [\text{A}^+][\text{B}^-] \][/tex]
Substitute the concentrations into the equation:
[tex]\[ K_{sp} = (0.0400 \, \text{mol/L}) \cdot (0.0400 \, \text{mol/L}) \][/tex]
[tex]\[ K_{sp} = 0.0400^2 \][/tex]
[tex]\[ K_{sp} = 0.0016 \][/tex]
Therefore, the solubility product constant [tex]\( K_{sp} \)[/tex] of the salt [tex]\( AB \)[/tex] at [tex]\( 25^{\circ}C \)[/tex] is:
[tex]\[ K_{sp} = 0.0016 \][/tex]
