What is the molar solubility ofAg2CO3 (Ksp = 4 × 10–13) in 0.1 M Na2CO3 solution ?10–610–72 × 10–62 × 10–7
Question
What is the molar solubility ofAg2CO3 (Ksp = 4 × 10–13) in 0.1 M Na2CO3 solution ?10–610–72 × 10–62 × 10–7
Solution
To solve this problem, we need to understand the concept of molar solubility and the solubility product constant (Ksp).
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Write the balanced chemical equation for the dissolution of Ag2CO3 in water: Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)
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Write the expression for the solubility product constant (Ksp): Ksp = [Ag+]^2[CO3^2-]
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In a 0.1 M Na2CO3 solution, the concentration of CO3^2- ions is 0.1 M. Let's assume the molar solubility of Ag2CO3 is 's'. Therefore, the concentration of Ag+ ions is 2s.
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Substitute these values into the Ksp expression: 4 × 10^-13 = (2s)^2 * 0.1
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Simplify the equation to solve for 's': 4 × 10^-13 = 4s^2 * 0.1 s^2 = 10^-14 s = sqrt(10^-14) s = 10^-7 M
So, the molar solubility of Ag2CO3 in 0.1 M Na2CO3 solution is 10^-7 M.
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