Use the given molar solubilities in pure water to calculate K sp for each compound. b. PbF 2 ; molar solubility = 5.63⨉10 -3 M c. MgF 2 ; molar solubility = 2.65⨉10 -4 M

Identify the dissolution equation for the compound. For PbF 2 , the dissolution equation is: PbF 2 (s) ⇌ Pb 2+ (aq) + 2F - (aq).. Write the expression for the solubility product constant (K sp ) based on the dissolution equation. For PbF 2 , K sp = [Pb 2+ ][F - ]^2.. Substitute the molar solubility into the K sp expression. Given the molar solubility of PbF 2 is 5.63⨉10 -3 M, [Pb 2+ ] = 5.63⨉10 -3 M and [F - ] = 2(5.63⨉10 -3 ) M.. Calculate the concentration of F - by multiplying the molar solubility of PbF 2 by 2, as there are two fluoride ions for every formula unit of PbF 2 that dissolves.. Substitute these concentrations back into the K sp expression and simplify to find the K sp value for PbF 2 .

Use the given molar solubilities in pure water to calculate K sp for each compound. b. PbF 2 ; molar solubility = 5.63⨉10 -3 M c. MgF 2 ; molar solubility = 2.65⨉10 -4 M

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.

Use the given molar solubilities in pure water to calculate Ksp for each compound. b. PbF2; molar solubility = 5.63⨉10-3 M c. MgF2; molar solubility = 2.65⨉10-4 M

Identify the dissolution equation for the compound. For PbF 2 , the dissolution equation is: PbF 2 (s) ⇌ Pb 2+ (aq) + 2F - (aq). Write the expression for the solubility product constant (K sp ) based on the dissolution equation. For PbF 2 , K sp = [Pb 2+ ][F - ]^2. Substitute the molar solubility into the K sp expression. Given the molar solubility of PbF 2 is 5.63⨉10 -3 M, [Pb 2+ ] = 5.63⨉10 -3 M and [F - ] = 2(5.63⨉10 -3 ) M. Calculate the concentration of F - by multiplying the molar solubility of PbF 2 by 2, as there are two fluoride ions for every formula unit of PbF 2 that dissolves. Substitute these concentrations back into the K sp expression and simplify to find the K sp value for PbF 2 .

Ch.18 - Aqueous Ionic Equilibrium

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Tro 6th Edition

Ch.18 - Aqueous Ionic Equilibrium

Problem 95b

Chapter 18, Problem 95b

Use the given molar solubilities in pure water to calculate K_sp for each compound. b. PbF₂; molar solubility = 5.63⨉10^-3 M c. MgF₂; molar solubility = 2.65⨉10^-4 M

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Identify the dissolution equation for the compound. For PbF₂, the dissolution equation is: PbF₂(s) ⇌ Pb²⁺(aq) + 2F^-(aq).

Write the expression for the solubility product constant (K_sp) based on the dissolution equation. For PbF₂, K_sp = [Pb²⁺][F^-]^2.

Substitute the molar solubility into the K_sp expression. Given the molar solubility of PbF₂ is 5.63⨉10^-3 M, [Pb²⁺] = 5.63⨉10^-3 M and [F^-] = 2(5.63⨉10^-3) M.

Calculate the concentration of F^- by multiplying the molar solubility of PbF₂ by 2, as there are two fluoride ions for every formula unit of PbF₂ that dissolves.

Substitute these concentrations back into the K_sp expression and simplify to find the K_sp value for PbF₂.

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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 is the number of moles of a solute that can dissolve in one liter of solution at a given temperature. It is a crucial measure in determining how much of a compound can be dissolved in water, which directly influences the calculation of the solubility product constant (Ksp). In this case, the molar solubility of PbF2 is given as 5.63 × 10^-3 M.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that reflects the extent to which a sparingly soluble ionic compound dissolves in water. It is calculated from the concentrations of the ions in a saturated solution at equilibrium. For PbF2, Ksp can be derived from its dissociation into Pb²⁺ and F⁻ ions, using the molar solubility to express their concentrations.

Dissociation of Ionic Compounds

Dissociation refers to the process by which an ionic compound separates into its constituent ions when dissolved in water. For PbF2, the dissociation can be represented as PbF2(s) ⇌ Pb²⁺(aq) + 2F⁻(aq). Understanding this process is essential for calculating Ksp, as it allows us to relate the molar solubility to the concentrations of the ions produced in solution.