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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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 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

Verified Solution

Key Concepts

Molar Solubility

Solubility Product Constant (Ksp)

Dissociation of Ionic Compounds

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

Verified Solution

Key Concepts

Molar Solubility

Solubility Product Constant (Ksp)

Dissociation of Ionic Compounds

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