Use the given molar solubilities in pure water to calculate K sp for each compound. b. Ag 2 SO 3 ; molar solubility = 1.55⨉10 -5 M c. Pd(SCN) 2 ; molar solubility = 2.22⨉10 -8 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.

Ch.18 - Aqueous Ionic Equilibrium

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Tro 6th Edition

Ch.18 - Aqueous Ionic Equilibrium

Problem 96b

Chapter 18, Problem 96b

Use the given molar solubilities in pure water to calculate K_sp for each compound. b. Ag₂SO₃; molar solubility = 1.55⨉10^-5 M c. Pd(SCN)₂; molar solubility = 2.22⨉10^-8 M

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Identify the dissolution equation for Ag_2SO_3: Ag_2SO_3(s) \rightleftharpoons 2Ag^+(aq) + SO_3^{2-}(aq).

Determine the relationship between molar solubility and ion concentrations: If the molar solubility of Ag_2SO_3 is s, then [Ag^+] = 2s and [SO_3^{2-}] = s.

Substitute the given molar solubility into the expressions for ion concentrations: [Ag^+] = 2(1.55 \times 10^{-5}) M and [SO_3^{2-}] = 1.55 \times 10^{-5} M.

Write the expression for the solubility product constant, K_{sp}: K_{sp} = [Ag^+]^2[SO_3^{2-}].

Substitute the ion concentrations into the K_{sp} expression and simplify: K_{sp} = (2 \times 1.55 \times 10^{-5})^2 \times (1.55 \times 10^{-5}).

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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 present in a saturated solution. In this context, the molar solubility of Ag₂SO₃ is given as 1.55×10⁻⁵ M, which indicates the maximum concentration of the compound that can dissolve in water.

Solubility Product Constant (Ksp)

The solubility product constant, Ksp, is an equilibrium constant that applies to the solubility of ionic compounds. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. For Ag₂SO₃, Ksp can be calculated using the molar solubility to find the concentrations of Ag⁺ and SO₃²⁻ ions in a saturated solution.

Dissociation of Ionic Compounds

When ionic compounds dissolve in water, they dissociate into their constituent ions. For Ag₂SO₃, the dissociation can be represented as Ag₂SO₃(s) ⇌ 2Ag⁺(aq) + SO₃²⁻(aq). Understanding this dissociation is essential for calculating Ksp, as it allows us to relate the molar solubility to the concentrations of the ions produced in solution.