Solubility Product (Ksp) worksheet with questions on dissociation equations, Ksp expressions, and calculations for compounds like PbSO₄, NiS, and Ag₂S.

Solubility Product, \( K_{sp} \) - Worksheet 1

#### Problem 1: Write the dissociation equation and the solubility product (\( K_{sp} \)) expression for the dissociation of each of the following substances.

a) \( \text{PbSO}_4(s) \)
- Dissociation Equation:
\[
\text{PbSO}_4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}]
\]

b) \( \text{Mg(OH)}_2(s) \)
- Dissociation Equation:
\[
\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2
\]

c) \( \text{Pb(NO}_3)_2(s) \)
- Dissociation Equation:
\[
\text{Pb(NO}_3)_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{NO}_3^-(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Pb}^{2+}][\text{NO}_3^-]^2
\]

d) \( \text{AlCl}_3(s) \)
- Dissociation Equation:
\[
\text{AlCl}_3(s) \rightleftharpoons \text{Al}^{3+}(aq) + 3\text{Cl}^-(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Al}^{3+}][\text{Cl}^-]^3
\]

e) \( \text{Na}_2\text{S}(s) \)
- Dissociation Equation:
\[
\text{Na}_2\text{S}(s) \rightleftharpoons 2\text{Na}^+(aq) + \text{S}^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Na}^+]^2[\text{S}^{2-}]
\]

f) \( \text{Ca}_3(\text{PO}_4)_2(s) \)
- Dissociation Equation:
\[
\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2
\]

---

#### Problem 2: Write the solubility product expression for the dissociation of each of the following solids in water.

a) Sodium chloride (\( \text{NaCl} \))
- Dissociation Equation:
\[
\text{NaCl}(s) \rightleftharpoons \text{Na}^+(aq) + \text{Cl}^-(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Na}^+][\text{Cl}^-]
\]

b) Calcium carbonate (\( \text{CaCO}_3 \))
- Dissociation Equation:
\[
\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}]
\]

c) Silver sulfide (\( \text{Ag}_2\text{S} \))
- Dissociation Equation:
\[
\text{Ag}_2\text{S}(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{S}^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ag}^+]^2[\text{S}^{2-}]
\]

d) Aluminum hydroxide (\( \text{Al(OH)}_3 \))
- Dissociation Equation:
\[
\text{Al(OH)}_3(s) \rightleftharpoons \text{Al}^{3+}(aq) + 3\text{OH}^-(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3
\]

e) Magnesium sulfite (\( \text{MgSO}_3 \))
- Dissociation Equation:
\[
\text{MgSO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{SO}_3^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Mg}^{2+}][\text{SO}_3^{2-}]
\]

f) Barium phosphate (\( \text{Ba}_3(\text{PO}_4)_2 \))
- Dissociation Equation:
\[
\text{Ba}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ba}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ba}^{2+}]^3[\text{PO}_4^{3-}]^2
\]

---

#### Problem 3: The solubility of \( \text{NiS} \) is \( 1.0 \times 10^{-12} \, \text{mol/L} \). Calculate the value of \( K_{sp} \) for this compound.

- Dissociation Equation:
\[
\text{NiS}(s) \rightleftharpoons \text{Ni}^{2+}(aq) + \text{S}^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ni}^{2+}][\text{S}^{2-}]
\]
- Given Solubility:
\[
s = 1.0 \times 10^{-12} \, \text{mol/L}
\]
- Concentrations at Equilibrium:
\[
[\text{Ni}^{2+}] = s = 1.0 \times 10^{-12} \, \text{mol/L}
\]
\[
[\text{S}^{2-}] = s = 1.0 \times 10^{-12} \, \text{mol/L}
\]
- Calculate \( K_{sp} \):
\[
K_{sp} = (1.0 \times 10^{-12})(1.0 \times 10^{-12}) = 1.0 \times 10^{-24}
\]

Answer:
\[
\boxed{1.0 \times 10^{-24}}
\]

---

#### Problem 4: The solubility of calcium hydroxide (\( \text{Ca(OH)}_2 \)) is \( 0.02 \, \text{mol/L} \). Calculate the solubility product constant for this substance.

- Dissociation Equation:
\[
\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2
\]
- Given Solubility:
\[
s = 0.02 \, \text{mol/L}
\]
- Concentrations at Equilibrium:
\[
[\text{Ca}^{2+}] = s = 0.02 \, \text{mol/L}
\]
\[
[\text{OH}^-] = 2s = 2 \times 0.02 = 0.04 \, \text{mol/L}
\]
- Calculate \( K_{sp} \):
\[
K_{sp} = (0.02)(0.04)^2 = (0.02)(0.0016) = 3.2 \times 10^{-5}
\]

Answer:
\[
\boxed{3.2 \times 10^{-5}}
\]

---

#### Problem 5: The solubility of silver sulfide (\( \text{Ag}_2\text{S} \)) is \( 1.3 \times 10^{-6} \, \text{mol/L} \). Calculate the \( K_{sp} \) value.

- Dissociation Equation:
\[
\text{Ag}_2\text{S}(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{S}^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Ag}^+]^2[\text{S}^{2-}]
\]
- Given Solubility:
\[
s = 1.3 \times 10^{-6} \, \text{mol/L}
\]
- Concentrations at Equilibrium:
\[
[\text{Ag}^+] = 2s = 2 \times 1.3 \times 10^{-6} = 2.6 \times 10^{-6} \, \text{mol/L}
\]
\[
[\text{S}^{2-}] = s = 1.3 \times 10^{-6} \, \text{mol/L}
\]
- Calculate \( K_{sp} \):
\[
K_{sp} = (2.6 \times 10^{-6})^2(1.3 \times 10^{-6})
\]
\[
K_{sp} = (6.76 \times 10^{-12})(1.3 \times 10^{-6}) = 8.788 \times 10^{-18}
\]

Answer:
\[
\boxed{8.8 \times 10^{-18}}
\]

---

#### Problem 6: Experiments show that \( 0.0059 \, \text{g} \) of \( \text{SrCO}_3 \) will dissolve in \( 1.0 \, \text{L} \) of water. What is the \( K_{sp} \) for strontium carbonate?

- Molar Mass of \( \text{SrCO}_3 \):
\[
\text{Sr} = 87.62 \, \text{g/mol}, \, \text{C} = 12.01 \, \text{g/mol}, \, \text{O} = 16.00 \, \text{g/mol}
\]
\[
\text{Molar Mass} = 87.62 + 12.01 + (3 \times 16.00) = 147.63 \, \text{g/mol}
\]
- Solubility in Moles per Liter:
\[
\text{Moles of } \text{SrCO}_3 = \frac{0.0059 \, \text{g}}{147.63 \, \text{g/mol}} = 4.00 \times 10^{-5} \, \text{mol}
\]
\[
\text{Solubility} = \frac{4.00 \times 10^{-5} \, \text{mol}}{1.0 \, \text{L}} = 4.00 \times 10^{-5} \, \text{mol/L}
\]
- Dissociation Equation:
\[
\text{SrCO}_3(s) \rightleftharpoons \text{Sr}^{2+}(aq) + \text{CO}_3^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Sr}^{2+}][\text{CO}_3^{2-}]
\]
- Concentrations at Equilibrium:
\[
[\text{Sr}^{2+}] = s = 4.00 \times 10^{-5} \, \text{mol/L}
\]
\[
[\text{CO}_3^{2-}] = s = 4.00 \times 10^{-5} \, \text{mol/L}
\]
- Calculate \( K_{sp} \):
\[
K_{sp} = (4.00 \times 10^{-5})(4.00 \times 10^{-5}) = 1.60 \times 10^{-9}
\]

Answer:
\[
\boxed{1.6 \times 10^{-9}}
\]

---

#### Problem 7: The solubility of cobalt (II) sulfide is \( 3.8 \times 10^{-3} \, \text{g/L} \). Calculate the value of \( K_{sp} \).

- Molar Mass of \( \text{CoS} \):
\[
\text{Co} = 58.93 \, \text{g/mol}, \, \text{S} = 32.07 \, \text{g/mol}
\]
\[
\text{Molar Mass} = 58.93 + 32.07 = 91.00 \, \text{g/mol}
\]
- Solubility in Moles per Liter:
\[
\text{Moles of } \text{CoS} = \frac{3.8 \times 10^{-3} \, \text{g}}{91.00 \, \text{g/mol}} = 4.176 \times 10^{-5} \, \text{mol}
\]
\[
\text{Solubility} = \frac{4.176 \times 10^{-5} \, \text{mol}}{1.0 \, \text{L}} = 4.176 \times 10^{-5} \, \text{mol/L}
\]
- Dissociation Equation:
\[
\text{CoS}(s) \rightleftharpoons \text{Co}^{2+}(aq) + \text{S}^{2-}(aq)
\]
- Solubility Product Expression (\( K_{sp} \)):
\[
K_{sp} = [\text{Co}^{2+}][\text{S}^{2-}]
\]
- Concentrations at Equilibrium:
\[
[\text{Co}^{2+}] = s = 4.176 \times 10^{-5} \, \text{mol/L}
\]
\[
[\text{S}^{2-}] = s = 4.176 \times 10^{-5} \, \text{mol/L}
\]
- Calculate \( K_{sp} \):
\[
K_{sp} = (4.176 \times 10^{-5})(4.176 \times 10^{-5}) = 1.743 \times 10^{-9}
\]

Answer:
\[
\boxed{1.7 \times 10^{-9}}
\]

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Final Answers:

1. See detailed dissociation equations and \( K_{sp} \) expressions above.
2. See detailed dissociation equations and \( K_{sp} \) expressions above.
3. \(\boxed{1.0 \times 10^{-24}}\)
4. \(\boxed{3.2 \times 10^{-5}}\)
5. \(\boxed{8.8 \times 10^{-18}}\)
6. \(\boxed{1.6 \times 10^{-9}}\)
7. \(\boxed{1.7 \times 10^{-9}}\)