Solubility Product Constant (Ksp) - Video Tutorials & Practice Problems

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Solubility Product Constant (Ksp) is associated with any ionic compound, which measures how soluble the compound will be in a solvent.

Ksp

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Solubility Product Constant (Ksp) Concept 1

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now recall that solly ability is a chemical property that deals with the ability of a salute to become dissolved in a solvent. Now connected to this idea of scalability is a new term are so liability product constant, which uses the variable K. S. P. Now this is the equilibrium constant that deals with the sai ability of ionic solids. So for looking at the possibility of ionic solids were dealing with K. S. P. We're gonna stay here. It deals with their soluble itty and we're gonna say soluble itty can also be referred to as concentration or polarity. More clarity uses capital and now the basic idea is the higher your K. S. P. Value than the more soluble your ionic solid and the smaller K. S. P. Value than the less soluble your ionic solid, the less of it will dissolve within a solvent. So just keep this in mind when looking at K. S. P.

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example

Solubility Product Constant (Ksp) Example 1

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here it asks which substance is the most soluble. So here we have silver chloride, magnesium, carbonate, calcium, sulfate and copper to sulfide each of them with their given K. S. P. Value. Remember we want the most soluble of all of these choices. Out of all of these choices. Remember we're gonna say the greater chaos P. Value than the more soluble you are as an ionic solid. So if we look here, we say that the answer has to be seen Since at 7.1 times 10 to the negative, five, it has the smallest negative, so it's the largest value. There it would be the most soluble out of all four choices. So here option C would be the correct answer.

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Solubility Product Constant (Ksp) Concept 2

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now placing an ionic solid within a solvent involves two competing processes. We have the dissolution reaction. So basically where it's being broken up and we also have its reverse. Okay, so dissolution. Looking at the four direction of the reaction, or ionic solid dissolves into its ions. So here we have a two B three solid. So if we go to break this up into its ions, we'd say we have two A's. Remember this three didn't come from B came from A. So this is a three plus ions in solution. Our acquis plus three be the two that A possesses didn't come from A came from B two minus acquis. Remember we know that the first iron is positive and the second one is negative because that's the order in which we write ionic compounds. Now we're going to say that from this equilibrium equation it's equilibrium expression can be determined. Now the equilibrium expression is just the ratio of concentrations of products over react. It's and we're gonna say that recall, the equilibrium expression ignores solids and liquids for its ratios. Alright, so it's important that we know how to break up our ionic solids into their respective ions in order to successfully right your equilibrium expression later on. So keep this in mind in terms of the parameters in which it operates its products overreact ints and it ignores solids and liquids.

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Solubility Product Constant (Ksp) Example 2

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here we need to provide the equilibrium expression for calcium nitrate. So step one says we need to write the equilibrium equation by breaking up the ionic solid into its acquis ions. Remember in solution are ions are in the acquis form? All right so we have calcium nitrate solid. It breaks up into calcium which is in group two ways. So it's two plus that's where this two came from. Not only is that where the two came from but that also means that we have to nitrate ions. So that's two. N. 03 minus one. Remember the charge of nitrate ion acquis. Now that we've done this we're gonna say step to using K. S. P. Right? The equilibrium expression based on the equilibrium equation. Since the reaction is a solid set it equal to one within the equilibrium expression. Remember it's products. Overreact. It's so we'd have calcium in brackets or two plus count. Um See a two plus times N. 03 minus. And remember this to hear whatever the coefficient is that becomes the power. So this would be A two here are reacting as a solid so we just ignore it and replace it with one. But realize here that we have this expression over one, which just translates into those concentrations times each other and nitrates would be squared. Okay, so that would just be are equal or expression for this calcium nitrate solid. So these are the steps you need to take in breaking up your ionic solid into its respective ions and then correctly giving the equilibrium expression for it.

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Solubility Product Constant (Ksp) Concept 3

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here we can say if the Saudi ability or moller concentrations of ions within ionic solid are known, the K. S. P. Can be calculated if we take a look here we say calculate the K. S. P. Value for silver phosphate which is 83 P. 04 Which has a solid ability of 1.8 times 10 to the - at 25°C. Here we say right the equilibrium equation by breaking up the ionic solids into its acquis ions. Alright so here we have our silver phosphate solid. It's gonna break up into its ions. It's gonna break up into three silver ions. Remember ions are acquis and solution plus our phosphate ion acquis. Yeah. Next we write the equilibrium Russian based on the equilibrium equation. Remember equilibrium expression is K. S. P. Equals products overreacted. It's but your reaction is a solid so we're going to ignore it. So it's just gonna be a G. Plus remember the coefficient here is going to become the power so cubed times P. +043 minus now. Some new stuff. Step three. We're gonna make concentrations of the ions are equal to their coefficients, coefficients multiplied by the X variable. All right. So what do I mean by that? We're going to say this is equal to the coefficient in front of silver ion in our equation as a three. So this equals three X. And it's going to be cute and there's a coefficient of one here. That's invisible. There's gonna be times one X or just X. Next we're gonna substitute in the given Saudi ability value for the X variable and solve for K. S. P. All right. So before we do that we have to work out this algebraic expression here. So three X cubed means that three is going to be cubed and X is going to be cute. three times 3 times three equals X cubed times X. So 27 X cubed times X. Is 27 X to the fourth. So that is what my K. S. P. Is equal to. So now we're gonna do what Step four said it said substitute in the given solid ability value. We're told this liability is this 1.8 times 10 to the negative 18. So K. S. P. Equals 27 times 1.8 times 10 to the negative 18 to the fourth. Yeah. Order of operations. We do what's in the parentheses first so 1.8 times 10 to the -18 to the fourth. When you punch that into, your calculator is going to give you back 1.04976 times 10 to the -71. Then that multi gets multiplied by 27. Yeah When we do that we get as our answer 2. times 10 to the negative 70 as the k. S. P. For silver phosphate. Here, here, answer has two significant figures because the Saudi ability given to us in beginning, also has two significant figures.

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Solubility Product Constant (Ksp) Concept 4

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conversely, the saw the ability of an ionic solid can be determined when it's KsB value is already known. So here it says the KsB value for strontium fluoride srf two is 7.9 times 10 to the negative 10 at 25 degrees Celsius calculate its scalability in polarity. All right. So step one, we're gonna write the equilibrium equation by breaking up the ionic solid into its acquis ions. So that part doesn't change. We have strong team fluoride solid which breaks up into strontium ion plus two four eyed ions. Next we write the equilibrium expression based on the equilibrium equation. So that's K. S. P. Equals our strontium ion times are fluoride ion. And remember the coefficient becomes the power. Next we solve for the cy ability variable X. Based on the given chaos P value. Alright, so remember to do that. We're going to say We have our K sp here which is 79 times 10 to the negative 10 and that equals X. Which stands in for the strontium ion times. Remember the coefficient is going to play a part here, It's gonna get to axe and it's still squared. So we're gonna have 7.9 times 10 to the negative 10 equals X times two squared is two times two. So that's four X squared. So we're just solving this like a math problem. So it's gonna be 7.9 times 10 to the negative 10 equals four X cubed. We're going to divide both sides now by four in order to isolate our X cubed. So when we do that we're going to get here X cubed equals 1.975 Times 10 to the -10. And here we need to take the cube root of both sides in order to just isolate my ex here. So we're going to say here when we take the cube root of both sides, X equals at the end 5.8 times 10 to the negative four moller as my final concentration for this strontium fluoride solid

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Problem

Problem

Determine the equilibrium expression of the barium nitride solid.

A

B

C

D

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Problem

Problem

Manganese (V) hydroxide has a measured solubility of 3.4×10^{–5} M at 25ºC. Calculate its K_{sp} value.

A

5.7×10^{–9}

B

1.5×10^{–27}

C

7.7×10^{–27}

D

4.8×10^{–24}

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Problem

Problem

The K_{sp} value for strontium fluoride, SrF_{2}, is 7.9×10^{–10} at 25ºC. Calculate its solubility in g/L.

A

5.8×10^{–4} g/L

B

7.3x10^{-2} g/L

C

4.6×10^{–6} g/L

D

0.092 g/L

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