College

Use the molar solubility, [tex]1.55 \times 10^{-5} \, M[/tex], in pure water to calculate [tex]K_{\text{sp}}[/tex] for [tex]Ag_2SO_3[/tex]. Express your answer using three significant figures.

[tex]K_{\text{sp}} =[/tex]

Answer :

To find the solubility product constant ([tex]\(K_{\text{sp}}\)[/tex]) for [tex]\( \text{Ag}_2\text{SO}_3 \)[/tex] using its molar solubility in pure water, follow these steps:

1. Understand the Dissociation Reaction:
[tex]\(\text{Ag}_2\text{SO}_3\)[/tex] dissociates in water according to the following equation:
[tex]\[
\text{Ag}_2\text{SO}_3(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{SO}_3^{2-}(aq)
\][/tex]

2. Use the Molar Solubility:
Given that the molar solubility of [tex]\(\text{Ag}_2\text{SO}_3\)[/tex] is [tex]\(1.55 \times 10^{-5} \, \text{M}\)[/tex]. Let's denote this molar solubility as [tex]\( s \)[/tex].

3. Determine Ion Concentrations at Equilibrium:
- Concentration of [tex]\(\text{Ag}^+\)[/tex] ions will be [tex]\(2s\)[/tex] because there are 2 moles of [tex]\(\text{Ag}^+\)[/tex] produced for each mole of [tex]\(\text{Ag}_2\text{SO}_3\)[/tex].
- Concentration of [tex]\(\text{SO}_3^{2-}\)[/tex] ions will be [tex]\(s\)[/tex].

4. Write the Expression for [tex]\(K_{\text{sp}}\)[/tex]:
[tex]\[
K_{\text{sp}} = [\text{Ag}^+]^2 \times [\text{SO}_3^{2-}]
\][/tex]

5. Substitute the Equilibrium Concentrations:
[tex]\[
K_{\text{sp}} = (2s)^2 \times (s) = 4s^3
\][/tex]

6. Calculate [tex]\(K_{\text{sp}}\)[/tex]:
[tex]\[
K_{\text{sp}} = 4 \times (1.55 \times 10^{-5})^3
\][/tex]

7. Result:
After performing the calculation, the solubility product constant [tex]\(K_{\text{sp}}\)[/tex] is found to be:
[tex]\[
K_{\text{sp}} = 1.49 \times 10^{-14}
\][/tex]

This is the [tex]\(K_{\text{sp}}\)[/tex] value for [tex]\( \text{Ag}_2\text{SO}_3 \)[/tex] to three significant figures.