(a) I f t he molar solubility of CaF 2 at 35°C i s 1.24 × 10 –3 mol/L, what is K sp at this temperature?

Hello everyone today. We are being asked to calculate K. S. P. When we have a liability of calcium fluoride in pure water that is 3.32 times 10 to the negative fourth. So the first we want to do is we want to draw out calcium fluoride dissociating in water. So we're going to draw this in the solid form and we're going to say that when this dissociates in water it's going to form a one calcium two plus ion in the acquis form as well as to fluoride ions. The two comes from the subscript from the previous product or reacting the next we want to do is we want to set up our ice table here so that we can set our numbers up nice and simple. And so we're not really worried about the solid. So we're going to focus on the right hand side of the arrow. And so our initial concentrations are going to be zero for each of the ions. And so we don't know what these changes going to be. The C. Stands for change but we do know that when there's going to be a change in calcium we can represent that with X. And then fluoride we can say two X. Since we have two of them. And so E. Is going to be basically adding up I. N. C. So we're gonna have X. For calcium and we're gonna have to X. For flooring or fluoride. We're going to say R. K. S. P. Is equal to the concentration of our calcium ions as well multiplied by our fluoride ions. And whenever we have a coefficient that becomes the exponent. And so that's where the two comes from when we simplify this in terms of our values of X, we get the K. S. P. Is equal to calcium which is X. And then fluoride, we're just going to be two X. And we cannot forget that squared. If we wanted to simplify this, this would turn into four x cubed. Where then we can take the K. S. P. And we can equal that to four times X, which is our 3.32 times 10 to the negative fourth. And then we cube that and we get a final que sp of 1.46 times 10 to the negative 10th. And that is our final answer. I hope this helped. And until next time.

Ch.17 - Additional Aspects of Aqueous Equilibria

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Brown 15th Edition

Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 53a

Chapter 17, Problem 53a

(a) I f t he molar solubility of CaF₂ at 35°C i s 1.24 × 10^–3 mol/L, what is K_sp at this temperature?

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Identify the dissolution equation for calcium fluoride: \( \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \).

Write the expression for the solubility product constant \( K_{sp} \) for the dissolution of \( \text{CaF}_2 \): \( K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \).

Recognize that the molar solubility of \( \text{CaF}_2 \) is given as \( 1.24 \times 10^{-3} \) mol/L, which represents the concentration of \( \text{Ca}^{2+} \) ions in solution.

Determine the concentration of \( \text{F}^- \) ions, which is twice the molar solubility of \( \text{CaF}_2 \), because each formula unit of \( \text{CaF}_2 \) produces two \( \text{F}^- \) ions.

Substitute the concentrations of \( \text{Ca}^{2+} \) and \( \text{F}^- \) into the \( K_{sp} \) expression and solve for \( K_{sp} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at a specific temperature, expressed in moles per liter (mol/L). In the case of CaF2, the molar solubility indicates how much calcium fluoride can dissolve in water before reaching saturation.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that quantifies the solubility of a sparingly soluble ionic compound. It is calculated from the concentrations of the ions in a saturated solution, raised to the power of their coefficients in the balanced dissolution equation. For CaF2, Ksp is derived from the concentrations of Ca²⁺ and F⁻ ions.

Dissociation of Ionic Compounds

When ionic compounds like CaF2 dissolve in water, they dissociate into their constituent ions. For CaF2, the dissociation can be represented as CaF2(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). Understanding this dissociation is crucial for calculating Ksp, as it directly relates the molar solubility to the concentrations of the ions produced in solution.