To calculate the solubility of silver acetate ([tex]$CH_3COOAg$[/tex]) in mol/dm[tex]\(^3\)[/tex] given the solubility product constant ([tex]\(K_{sp}\)[/tex]) of [tex]\(1.9 \times 10^{-3}\)[/tex], we can follow these steps:
### Step-by-Step Solution:
1. Write the Dissociation Equation:
[tex]\[
CH_3COOAg (s) \leftrightarrow CH_3COO^- (aq) + Ag^+ (aq)
\][/tex]
2. Write the Solubility Product Expression ([tex]\(K_{sp}\)[/tex]):
- The solubility product constant is defined as the product of the molar concentrations of the constituent ions, each raised to the power of their coefficients in the balanced equation.
[tex]\[
K_{sp} = [CH_3COO^-][Ag^+]
\][/tex]
3. Express Concentrations in Terms of Solubility (S):
- Let [tex]\(S\)[/tex] be the solubility of [tex]\(CH_3COOAg\)[/tex] in mol/L (or mol/dm[tex]\(^3\)[/tex]). Since one mole of [tex]\(CH_3COOAg\)[/tex] dissociates into one mole of [tex]\(CH_3COO^-\)[/tex] and one mole of [tex]\(Ag^+\)[/tex], the concentrations of both ions in the solution will be equal to [tex]\(S\)[/tex]:
[tex]\[
[CH_3COO^-] = S \quad \text{and} \quad [Ag^+] = S
\][/tex]
4. Substitute These Concentrations into the [tex]\(K_{sp}\)[/tex] Expression:
[tex]\[
K_{sp} = S \times S = S^2
\][/tex]
- Given [tex]\(K_{sp} = 1.9 \times 10^{-3}\)[/tex]:
[tex]\[
1.9 \times 10^{-3} = S^2
\][/tex]
5. Solve for [tex]\(S\)[/tex]:
- Take the square root of both sides to find [tex]\(S\)[/tex]:
[tex]\[
S = \sqrt{1.9 \times 10^{-3}}
\][/tex]
By calculating the square root of [tex]\(1.9 \times 10^{-3}\)[/tex], we find that the solubility [tex]\(S\)[/tex] is approximately:
[tex]\[
S \approx 0.0436 \, \text{mol/dm}^3
\][/tex]
Therefore, the solubility of silver acetate, [tex]\(CH_3COOAg\)[/tex], in mol/dm[tex]\(^3\)[/tex] is [tex]\(\boxed{0.0436}\)[/tex].
