Multiple Choice
Given the solubility product constant (Ksp) for AgI is 8.3 x 10^-17 and the formation constant (Kf) for Ag(CN)₂⁻ is 1.0 x 10^21, calculate the equilibrium constant for the reaction AgI(s) + 2CN⁻(aq) ⇌ Ag(CN)₂⁻(aq) + I⁻(aq).
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
What is the molar solubility of CaCO3 in ocean water at pH 8.3, given that the solubility is pH dependent and the dissociation constants are Ka1(H2CO3) = 4.3x10^-7 and Ka2(H2CO3) = 5.6x10^-11?
A
1.3 x 10^-6 M
B
2.5 x 10^-5 M
C
4.8 x 10^-7 M
D
3.2 x 10^-4 M
Verified step by step guidance1
Identify the relevant chemical equilibrium: The dissolution of calcium carbonate (CaCO3) in water can be represented by the equation: CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq).
Consider the effect of pH on carbonate species: At pH 8.3, the carbonate ion (CO₃²⁻) is in equilibrium with bicarbonate (HCO₃⁻) and carbonic acid (H₂CO₃). The relevant equilibria are: H₂CO₃ ⇌ H⁺ + HCO₃⁻ and HCO₃⁻ ⇌ H⁺ + CO₃²⁻.
Use the given dissociation constants: The dissociation constants Ka1 and Ka2 for carbonic acid are provided. These will help determine the concentrations of carbonate species at the given pH.
Calculate the concentration of carbonate ions: Use the pH to find the concentration of H⁺ ions, then apply the equilibrium expressions and dissociation constants to find the concentration of CO₃²⁻.
Determine the molar solubility of CaCO3: Use the concentration of CO₃²⁻ to find the molar solubility of CaCO3, as the solubility product (Ksp) is related to the concentrations of Ca²⁺ and CO₃²⁻ at equilibrium.
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