Solubility Product Constant: Ksp - Video Tutorials & Practice Problems
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concept
Solubility Product Constant (Ksp)
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Now, the solubility product constant uses the variable KSP and represents yet another type of equilibrium constant. Here it measures solubility of solid ionic compounds in a solvent and equilibrium. Here, we're going to say that solubility is just the maximum amount of solid dissolved in a solvent. And it's usually represented as capital m which stands for molar solubility. Now, the magnitude of KSP value determines the degree of solubility. Basically, we're gonna say here, the greater the KSP value than the more soluble your ionic solid will be and the lower your KSP, then the less soluble your ionic solid will be here. We're gonna say this comparison can only be used between compounds that break up into the same number of ions. So for example, you have N AC L which breaks up into two ions and you have uh let's say lithium bromide, which actually breaks up into two ions because they break up into the same number of ions. You can look at their K SPS and directly compare them. So this one has a higher KSP. So it's more soluble. If they broke up into different number of ions, then more calculations would be needed. So for example, if you had barium chloride, this breaks up into three ions. So you couldn't just look at its KSP value and compare it to the other two because it breaks up two different numbers of ions later on. We'll see how to compare them by doing further calculations.
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example
Solubility Product Constant: Ksp Example
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Given the following ionic compounds which have the highest will have the highest concentration of hydroxide ions concentration. Here it says hint which is the most soluble in water. All right. So if we took a look at every single one of these compounds, you will see that they all break up into three ions, they would break up into the respective metal and two hydroxide ions. Now, because they break up into the same number of ions, we can just look at their KSP values and determine which one would produce the most oh minus. Remember, the greater KSP value, the more soluble your ionic solid will be the more ions it'll produce. So if we take a look here negative 17 to the negative 20 to the negative 13 to the negative 27. Here we'd say magnesium hydroxide is our answer. It has the highest KSP value. Therefore, it is the most soluble, meaning that it will break up the most and produce the most oh minus ions as products. This in turn will create the highest hydroxide ion concentration.
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concept
Ksp Calculations
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Now, when it comes to KSP, calculations realize that solubility is an equilibrium process. Hence, calculations will require an ice chart. So if you see KSP within a question, that means you're most likely going to have to use an ice chart. Now, here with KSP, we have a formula we need to remember. So let's say that here we have our equation. Again, KSP deals with the solubility of ionic solids, which means that you're gonna start out with an ionic solid as you're reacted, we look to see how it breaks up into various ions. So let's say that it breaks up into some unknown number of B ions plus some unknown number of C ions. These are ions where they'd have charges. So I'm just gonna write a random chart here for each one. Now A B and C are coefficients, KSP is an equilibrium constant and like all equilibrium constants, it equals products over reactants. However, remember we said KSP deals with solids, remember, solids and liquids are ignored. So although KSP equals products over reactants like other equilibrium constants, the reaction will be a solid and therefore can be ignored. So KSB simplifies suggests equal to products. Now, like other equilibrium constants, we'd have our products within brackets represent concentrations and then remember that their coefficients would become these powers. OK. So that's how we break down the equilibrium expression for a KSP value. OK. So just remember it's equal to products over reactants. The reactant is a solid. So just drop it. So KSP equals products.
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example
Solubility Product Constant: Ksp Example
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Here, we have lead to fluoride is a white solid and has diverse applications in pharmaceuticals, metallurgy and technology. If the concentration of lead to fluoride is 4.2 mo with a KSP of 3.6 times 10 to negative eight, calculate the molar solubility of the solid at 25 °C. All right. So when they say solubility, whether they say calculate solubility or calculate more solubility, they're just asking you to find X and when they're asking for the most liability of the ionic solid, that's just equal to X. Here, we're going to follow the steps in order to solve for this particular question. So step one, we set up an ice chart with solid as the only reactant because we're looking at how it breaks up into ions cross out the reactant side because remember in a nice chart wig no solids and liquids. This is an ice chart. So this is initial change equilibrium using the initial role set products equal to zero. Now remember we lose reactants in order to make products. So using the change row place A plus X for the products, we also have to take into account coefficients. So here this would be plus X, there's a two here. So this would be plus two X using the equilibrium row, we're gonna set up the equilibrium constant expression with KSP and solve for X. So now this is plus X and this is plus two X. Now, here we're going to say that the variable X in the ice chart represents the molar solubility of my ionic solid. All right. So here we're gonna say KSP equals products over reactants. But again, the reactant is a solid. So we ignore it. So it just equals PB two plus or plus two times F minus. Remember the coefficient also pops up here too as a power. All right. So now we're going to uh plug in values that we have. So we're gonna say here KSP is 3.6 times 10 to the negative eight and equilibrium led to his acts fluoride is two X and that's still squared. So we're gonna say this is X times two squared is four, X squared is X squared. So four X squared times X comes out to four X cubed and that's still equal to my KSP. So all we have to do now is solve for X. So divide both sides by four and we'll get X cubed equals 9.0 times 10 to the negative nine. Take the cube root of both sides. When we do that, we get X equals 2.08 times 10 to the minus three and that'll be molarity. So this represents the molar solubility of my ionic solid. So again, if any time they're asking for the molar solubility or solubility of your ionic compound is just equal to X. Now, if they were asking for one of the ions though, it wouldn't just simply be X. What you would need to do is you need to look at the equilibrium row and then look to see what is that ion represent at equilibrium here led to his X. So it would still be this number. But florid ion is two X, which would mean that you take this X answer and plug it into this two X. So it'd be two times X and that would be the true molar solubility of this fluoride ion. So again, be careful with when it comes to the ions for the ions, you have to check the equilibrium row and see what their equation is for the solid. When you solve for X, that is your final answer.
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Problem
Problem
Solubility of Sn(OH)2 was found to be 1.11 x 10-9 M; calculate Ksp of this compound.
A
5.47 × 10−27
B
4.44 × 10−9
C
1.23 × 10−18
D
1.37 × 10−27
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Problem
Problem
If a saturated solution of Ag2CO3 contains 2.56 × 10−4 M of Ag+ ions, determine its solubility product constant.
A
4.30 × 10−15
B
8.39 × 10−12
C
1.68 × 10−11
D
6.55 × 10−8
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Problem
Problem
What is the solubility of CN− ions in a solution of 5.5 M Hg2(CN)2, with a Ksp of 5.0 × 10−40?
A
4.5× 10−20 M
B
5.0 × 10−14 M
C
1.1 × 10−20 M
D
1.0 × 10−13 M
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