Multiple Choice
The concentration of fluoride ions in a saturated solution of barium fluoride (BaF2) is ________ M. The solubility product constant (Ksp) of BaF2 is 1.7 x 10^-6. What is the concentration of fluoride ions?
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
The solubility of Ni₃(PO₄)₂ in water at a particular temperature is 8.0 × 10⁻⁶ M. What is the solubility product constant (Ksp) for Ni₃(PO₄)₂?
A
5.1 × 10⁻³⁶
B
6.4 × 10⁻¹⁷
C
1.0 × 10⁻²⁸
D
2.0 × 10⁻¹⁸
Verified step by step guidance1
Identify the dissociation equation for Ni₃(PO₄)₂ in water: Ni₃(PO₄)₂(s) ⇌ 3Ni²⁺(aq) + 2PO₄³⁻(aq).
Define the solubility (s) of Ni₃(PO₄)₂ as 8.0 × 10⁻⁶ M, which represents the concentration of Ni₃(PO₄)₂ that dissolves in water.
Determine the concentrations of the ions at equilibrium: [Ni²⁺] = 3s and [PO₄³⁻] = 2s, where s is the solubility of Ni₃(PO₄)₂.
Substitute the expressions for the ion concentrations into the solubility product expression: Ksp = [Ni²⁺]³[PO₄³⁻]².
Calculate the Ksp by substituting s = 8.0 × 10⁻⁶ M into the expression: Ksp = (3 × 8.0 × 10⁻⁶)³ × (2 × 8.0 × 10⁻⁶)².
1:47mWatch next
Master Solubility Product Constant with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
Solubility Product Constant: Ksp practice set
Problem sets built by lead tutorsExpert video explanations