Chemistry worksheet focusing on solubility product calculations for different compounds.

Solubility Product, \( K_{sp} \) - Worksheet 2

#### Problem 1: Calculate the solubility (in mol/L) for each of the following compounds:
- a. FeS (\( K_{sp} = 3.7 \times 10^{-19} \))
- b. Mg(OH)_2 (\( K_{sp} = 9.0 \times 10^{-12} \))
- c. Ag_2S (\( K_{sp} = 1.6 \times 10^{-49} \))
- d. AgCN (\( K_{sp} = 2.0 \times 10^{-12} \))

##### Solution:

The solubility product (\( K_{sp} \)) is related to the solubility (\( s \)) of a compound in terms of its dissociation into ions. The general approach is:

1. Write the dissociation equation for the compound.
2. Express the \( K_{sp} \) in terms of the solubility \( s \).
3. Solve for \( s \).

---

##### a. FeS (\( K_{sp} = 3.7 \times 10^{-19} \))

Dissociation:
\[
\text{FeS} \rightleftharpoons \text{Fe}^{2+} + \text{S}^{2-}
\]

Let the solubility of FeS be \( s \) mol/L. Then:
\[
[\text{Fe}^{2+}] = s \quad \text{and} \quad [\text{S}^{2-}] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Fe}^{2+}][\text{S}^{2-}] = s \cdot s = s^2
\]

Given \( K_{sp} = 3.7 \times 10^{-19} \):
\[
s^2 = 3.7 \times 10^{-19}
\]
\[
s = \sqrt{3.7 \times 10^{-19}} = 1.92 \times 10^{-10} \, \text{mol/L}
\]

Answer:
\[
\boxed{1.92 \times 10^{-10} \, \text{mol/L}}
\]

---

##### b. Mg(OH)_2 (\( K_{sp} = 9.0 \times 10^{-12} \))

Dissociation:
\[
\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-
\]

Let the solubility of Mg(OH)_2 be \( s \) mol/L. Then:
\[
[\text{Mg}^{2+}] = s \quad \text{and} \quad [\text{OH}^-] = 2s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3
\]

Given \( K_{sp} = 9.0 \times 10^{-12} \):
\[
4s^3 = 9.0 \times 10^{-12}
\]
\[
s^3 = \frac{9.0 \times 10^{-12}}{4} = 2.25 \times 10^{-12}
\]
\[
s = \sqrt[3]{2.25 \times 10^{-12}} = 1.31 \times 10^{-4} \, \text{mol/L}
\]

Answer:
\[
\boxed{1.31 \times 10^{-4} \, \text{mol/L}}
\]

---

##### c. Ag_2S (\( K_{sp} = 1.6 \times 10^{-49} \))

Dissociation:
\[
\text{Ag}_2\text{S} \rightleftharpoons 2\text{Ag}^+ + \text{S}^{2-}
\]

Let the solubility of Ag_2S be \( s \) mol/L. Then:
\[
[\text{Ag}^+] = 2s \quad \text{and} \quad [\text{S}^{2-}] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+]^2[\text{S}^{2-}] = (2s)^2 \cdot s = 4s^3
\]

Given \( K_{sp} = 1.6 \times 10^{-49} \):
\[
4s^3 = 1.6 \times 10^{-49}
\]
\[
s^3 = \frac{1.6 \times 10^{-49}}{4} = 4.0 \times 10^{-50}
\]
\[
s = \sqrt[3]{4.0 \times 10^{-50}} = 1.59 \times 10^{-17} \, \text{mol/L}
\]

Answer:
\[
\boxed{1.59 \times 10^{-17} \, \text{mol/L}}
\]

---

##### d. AgCN (\( K_{sp} = 2.0 \times 10^{-12} \))

Dissociation:
\[
\text{AgCN} \rightleftharpoons \text{Ag}^+ + \text{CN}^-
\]

Let the solubility of AgCN be \( s \) mol/L. Then:
\[
[\text{Ag}^+] = s \quad \text{and} \quad [\text{CN}^-] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+][\text{CN}^-] = s \cdot s = s^2
\]

Given \( K_{sp} = 2.0 \times 10^{-12} \):
\[
s^2 = 2.0 \times 10^{-12}
\]
\[
s = \sqrt{2.0 \times 10^{-12}} = 1.41 \times 10^{-6} \, \text{mol/L}
\]

Answer:
\[
\boxed{1.41 \times 10^{-6} \, \text{mol/L}}
\]

---

#### Problem 2: What is the solubility of silver iodide in grams per litre? (\( K_{sp} = 8.3 \times 10^{-17} \))

##### Solution:

Dissociation:
\[
\text{AgI} \rightleftharpoons \text{Ag}^+ + \text{I}^-
\]

Let the solubility of AgI be \( s \) mol/L. Then:
\[
[\text{Ag}^+] = s \quad \text{and} \quad [\text{I}^-] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+][\text{I}^-] = s \cdot s = s^2
\]

Given \( K_{sp} = 8.3 \times 10^{-17} \):
\[
s^2 = 8.3 \times 10^{-17}
\]
\[
s = \sqrt{8.3 \times 10^{-17}} = 9.11 \times 10^{-9} \, \text{mol/L}
\]

To convert solubility from mol/L to grams per litre, use the molar mass of AgI:
\[
\text{Molar mass of AgI} = 107.87 \, \text{g/mol} + 126.90 \, \text{g/mol} = 234.77 \, \text{g/mol}
\]

Solubility in grams per litre:
\[
\text{Solubility (g/L)} = s \times \text{Molar mass} = 9.11 \times 10^{-9} \, \text{mol/L} \times 234.77 \, \text{g/mol}
\]
\[
\text{Solubility (g/L)} = 2.14 \times 10^{-6} \, \text{g/L}
\]

Answer:
\[
\boxed{2.14 \times 10^{-6} \, \text{g/L}}
\]

---

#### Problem 3: The \( K_{sp} \) value for BaSO_4 is \( 1.5 \times 10^{-9} \). Calculate the number of moles of this salt that will dissolve in 1000 mL of water.

##### Solution:

Dissociation:
\[
\text{BaSO}_4 \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}
\]

Let the solubility of BaSO_4 be \( s \) mol/L. Then:
\[
[\text{Ba}^{2+}] = s \quad \text{and} \quad [\text{SO}_4^{2-}] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \cdot s = s^2
\]

Given \( K_{sp} = 1.5 \times 10^{-9} \):
\[
s^2 = 1.5 \times 10^{-9}
\]
\[
s = \sqrt{1.5 \times 10^{-9}} = 1.22 \times 10^{-5} \, \text{mol/L}
\]

Since the volume is 1000 mL (1 L), the number of moles dissolved is:
\[
\text{Moles} = s \times \text{Volume (L)} = 1.22 \times 10^{-5} \, \text{mol/L} \times 1 \, \text{L} = 1.22 \times 10^{-5} \, \text{mol}
\]

Answer:
\[
\boxed{1.22 \times 10^{-5} \, \text{mol}}
\]

---

#### Problem 4: Calculate the mass of CuS that will dissolve in 1.0 litre of water. (\( K_{sp} = 4.0 \times 10^{-38} \))

##### Solution:

Dissociation:
\[
\text{CuS} \rightleftharpoons \text{Cu}^{2+} + \text{S}^{2-}
\]

Let the solubility of CuS be \( s \) mol/L. Then:
\[
[\text{Cu}^{2+}] = s \quad \text{and} \quad [\text{S}^{2-}] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Cu}^{2+}][\text{S}^{2-}] = s \cdot s = s^2
\]

Given \( K_{sp} = 4.0 \times 10^{-38} \):
\[
s^2 = 4.0 \times 10^{-38}
\]
\[
s = \sqrt{4.0 \times 10^{-38}} = 2.0 \times 10^{-19} \, \text{mol/L}
\]

To find the mass of CuS that dissolves in 1.0 L of water, use the molar mass of CuS:
\[
\text{Molar mass of CuS} = 63.55 \, \text{g/mol} + 32.07 \, \text{g/mol} = 95.62 \, \text{g/mol}
\]

Mass of CuS dissolved:
\[
\text{Mass} = s \times \text{Molar mass} \times \text{Volume (L)} = 2.0 \times 10^{-19} \, \text{mol/L} \times 95.62 \, \text{g/mol} \times 1 \, \text{L}
\]
\[
\text{Mass} = 1.91 \times 10^{-17} \, \text{g}
\]

Answer:
\[
\boxed{1.91 \times 10^{-17} \, \text{g}}
\]

---

#### Problem 5: Calculate the mass of barium carbonate that will dissolve in 100 mL of water. (\( K_{sp} = 8.1 \times 10^{-9} \))

##### Solution:

Dissociation:
\[
\text{BaCO}_3 \rightleftharpoons \text{Ba}^{2+} + \text{CO}_3^{2-}
\]

Let the solubility of BaCO_3 be \( s \) mol/L. Then:
\[
[\text{Ba}^{2+}] = s \quad \text{and} \quad [\text{CO}_3^{2-}] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ba}^{2+}][\text{CO}_3^{2-}] = s \cdot s = s^2
\]

Given \( K_{sp} = 8.1 \times 10^{-9} \):
\[
s^2 = 8.1 \times 10^{-9}
\]
\[
s = \sqrt{8.1 \times 10^{-9}} = 9.0 \times 10^{-5} \, \text{mol/L}
\]

To find the mass of BaCO_3 that dissolves in 100 mL (0.1 L) of water, use the molar mass of BaCO_3:
\[
\text{Molar mass of BaCO}_3 = 137.33 \, \text{g/mol} + 12.01 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol} = 197.34 \, \text{g/mol}
\]

Mass of BaCO_3 dissolved:
\[
\text{Mass} = s \times \text{Molar mass} \times \text{Volume (L)} = 9.0 \times 10^{-5} \, \text{mol/L} \times 197.34 \, \text{g/mol} \times 0.1 \, \text{L}
\]
\[
\text{Mass} = 1.78 \times 10^{-4} \, \text{g}
\]

Answer:
\[
\boxed{1.78 \times 10^{-4} \, \text{g}}
\]

---

#### Problem 6: Calculate the mass of MnS that will dissolve in 1.0 L of water. (\( K_{sp} = 4.9 \times 10^{-9} \))

##### Solution:

Dissociation:
\[
\text{MnS} \rightleftharpoons \text{Mn}^{2+} + \text{S}^{2-}
\]

Let the solubility of MnS be \( s \) mol/L. Then:
\[
[\text{Mn}^{2+}] = s \quad \text{and} \quad [\text{S}^{2-}] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Mn}^{2+}][\text{S}^{2-}] = s \cdot s = s^2
\]

Given \( K_{sp} = 4.9 \times 10^{-9} \):
\[
s^2 = 4.9 \times 10^{-9}
\]
\[
s = \sqrt{4.9 \times 10^{-9}} = 7.0 \times 10^{-5} \, \text{mol/L}
\]

To find the mass of MnS that dissolves in 1.0 L of water, use the molar mass of MnS:
\[
\text{Molar mass of MnS} = 54.94 \, \text{g/mol} + 32.07 \, \text{g/mol} = 87.01 \, \text{g/mol}
\]

Mass of MnS dissolved:
\[
\text{Mass} = s \times \text{Molar mass} \times \text{Volume (L)} = 7.0 \times 10^{-5} \, \text{mol/L} \times 87.01 \, \text{g/mol} \times 1 \, \text{L}
\]
\[
\text{Mass} = 6.09 \times 10^{-3} \, \text{g}
\]

Answer:
\[
\boxed{6.09 \times 10^{-3} \, \text{g}}
\]

---

#### Problem 7: What volume of water is necessary to dissolve 0.010 mol of silver chloride? (\( K_{sp} = 1.7 \times 10^{-10} \))

##### Solution:

Dissociation:
\[
\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-
\]

Let the solubility of AgCl be \( s \) mol/L. Then:
\[
[\text{Ag}^+] = s \quad \text{and} \quad [\text{Cl}^-] = s
\]

The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \cdot s = s^2
\]

Given \( K_{sp} = 1.7 \times 10^{-10} \):
\[
s^2 = 1.7 \times 10^{-10}
\]
\[
s = \sqrt{1.7 \times 10^{-10}} = 4.12 \times 10^{-5} \, \text{mol/L}
\]

The solubility \( s \) represents the maximum concentration of AgCl that can dissolve in water. To dissolve 0.010 mol of AgCl, we need to calculate the required volume \( V \) using the formula:
\[
\text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}}
\]
\[
s = \frac{0.010 \, \text{mol}}{V}
\]
\[
4.12 \times 10^{-5} \, \text{mol/L} = \frac{0.010 \, \text{mol}}{V}
\]
\[
V = \frac{0.010 \, \text{mol}}{4.12 \times 10^{-5} \, \text{mol/L}} = 242.72 \, \text{L}
\]

Answer:
\[
\boxed{243 \, \text{L}}
\]

---

Final Answers:

1. a. \( \boxed{1.92 \times 10^{-10} \, \text{mol/L}} \)
b. \( \boxed{1.31 \times 10^{-4} \, \text{mol/L}} \)
c. \( \boxed{1.59 \times 10^{-17} \, \text{mol/L}} \)
d. \( \boxed{1.41 \times 10^{-6} \, \text{mol/L}} \)
2. \( \boxed{2.14 \times 10^{-6} \, \text{g/L}} \)
3. \( \boxed{1.22 \times 10^{-5} \, \text{mol}} \)
4. \( \boxed{1.91 \times 10^{-17} \, \text{g}} \)
5. \( \boxed{1.78 \times 10^{-4} \, \text{g}} \)
6. \( \boxed{6.09 \times 10^{-3} \, \text{g}} \)
7. \( \boxed{243 \, \text{L}} \)