Antimony(III) sulfide (Sb2S3, FW 339.68 g/mole) is a very insoluble, ionic compound with a Kap of 1.7 x 10. Calculate the mass, in grams, of Sb,S; dissolved in Lake Erie (V4.58 x 10 L).
The Correct Answer and Explanation is :
To calculate the mass of antimony(III) sulfide (Sb₂S₃) that dissolves in Lake Erie, we can use the solubility product constant (Ksp), the volume of the lake, and the molar mass of Sb₂S₃.
Step 1: Write the dissociation equation for Sb₂S₃
Antimony(III) sulfide dissolves in water according to the following dissociation equation:
[
\text{Sb}_2\text{S}_3 (s) \rightleftharpoons 2\text{Sb}^{3+} (aq) + 3\text{S}^{2-} (aq)
]
Step 2: Set up the expression for the solubility product constant (Ksp)
The solubility product (Ksp) is given by the equilibrium expression:
[
K_{sp} = [\text{Sb}^{3+}]^2 [\text{S}^{2-}]^3
]
Let the molar solubility of Sb₂S₃ be denoted as ( s ) (in moles per liter). From the dissociation equation, for every mole of Sb₂S₃ that dissolves, 2 moles of Sb³⁺ and 3 moles of S²⁻ are produced. Thus, at equilibrium:
[
[\text{Sb}^{3+}] = 2s \quad \text{and} \quad [\text{S}^{2-}] = 3s
]
Substitute these expressions into the Ksp expression:
[
K_{sp} = (2s)^2 (3s)^3
]
Simplifying this:
[
K_{sp} = 4s^2 \times 27s^3 = 108s^5
]
Given that the Ksp for Sb₂S₃ is ( 1.7 \times 10^{-15} ), we can now solve for ( s ):
[
1.7 \times 10^{-15} = 108s^5
]
Solve for ( s ):
[
s^5 = \frac{1.7 \times 10^{-15}}{108}
]
[
s^5 = 1.57 \times 10^{-17}
]
[
s = \sqrt[5]{1.57 \times 10^{-17}} \approx 1.72 \times 10^{-4} \, \text{mol/L}
]
Step 3: Calculate the moles of Sb₂S₃ dissolved in Lake Erie
The volume of Lake Erie is ( 4.58 \times 10^9 \, \text{L} ). To find the moles of Sb₂S₃ that can dissolve in this volume:
[
\text{moles of Sb}_2\text{S}_3 = s \times \text{volume of Lake Erie}
]
[
\text{moles of Sb}_2\text{S}_3 = (1.72 \times 10^{-4} \, \text{mol/L}) \times (4.58 \times 10^9 \, \text{L})
]
[
\text{moles of Sb}_2\text{S}_3 \approx 7.88 \times 10^5 \, \text{mol}
]
Step 4: Calculate the mass of Sb₂S₃
To find the mass, multiply the moles of Sb₂S₃ by its molar mass (339.68 g/mol):
[
\text{mass of Sb}_2\text{S}_3 = \text{moles of Sb}_2\text{S}_3 \times \text{molar mass of Sb}_2\text{S}_3
]
[
\text{mass of Sb}_2\text{S}_3 = (7.88 \times 10^5 \, \text{mol}) \times (339.68 \, \text{g/mol})
]
[
\text{mass of Sb}_2\text{S}_3 \approx 2.68 \times 10^8 \, \text{g}
]
Final Answer:
The mass of Sb₂S₃ that dissolves in Lake Erie is approximately ( 2.68 \times 10^8 \, \text{g} ).
Explanation:
We calculated the mass of Sb₂S₃ that dissolves by first using the Ksp value to find the molar solubility (( s )) of Sb₂S₃ in water. Using this value, we determined how many moles of Sb₂S₃ would dissolve in the volume of Lake Erie, and finally, we multiplied by the molar mass of Sb₂S₃ to find the total mass.