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Determine the molar solubility of CuBr in a solution containing 0.4 M KBr, given that the solubility product constant Ksp for CuBr is 6.3 x 10^-9.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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What is the expression for the solubility product constant (Ksp) for the dissolution of Ba3(PO4)2 in water?
A
Ksp = [Ba2+][PO4^3−]
B
Ksp = [Ba2+]^2[PO4^3−]^3
C
Ksp = [Ba2+]^3[PO4^3−]^2
D
Ksp = [Ba2+]^3[PO4^3−]
Guida verificata passo dopo passo1
Start by writing the balanced chemical equation for the dissolution of Ba3(PO4)2 in water: \( \text{Ba}_3(\text{PO}_4)_2 (s) \rightarrow 3\text{Ba}^{2+} (aq) + 2\text{PO}_4^{3-} (aq) \).
Identify the ions produced in the solution: 3 moles of \( \text{Ba}^{2+} \) ions and 2 moles of \( \text{PO}_4^{3-} \) ions.
The solubility product constant, \( K_{sp} \), is expressed in terms of the concentrations of the ions in the solution. It is the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation.
For the dissolution of \( \text{Ba}_3(\text{PO}_4)_2 \), the expression for \( K_{sp} \) is \( K_{sp} = [\text{Ba}^{2+}]^3[\text{PO}_4^{3-}]^2 \).
This expression reflects the stoichiometry of the dissolution process, where the concentration of \( \text{Ba}^{2+} \) is cubed and the concentration of \( \text{PO}_4^{3-} \) is squared.
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