In this video, we're going to talk about how to calculate the Ksp value given the molar solubility. So in this problem, the molar solubility of nickel carbonate is 3.74 times 10 to the negative 4 moles per liter. Calculate the Ksp value, also known as the solubility product constant for nickel carbonate.
Now the first thing we need to do is write the chemical reaction that's associated with it. Solid nickel carbonate is going to dissociate into the nickel ion and the carbonate ion. So this reaction is in equilibrium.
Now what's important though is the molar ratio. This number that you see here, the molar solubility of nickel carbonate, that is the concentration of nickel carbonate that is actually dissolved in the solution. So that's the amount that is in the aqueous phase. Now because the molar ratio is 1 to 1, the concentration of nickel and carbonate is also equal to the molar solubility, which is basically that value. Now let's write the Ksp expression for this reaction.
To write the equilibrium expression for any K value, it's equal to the products divided by the reactants. Now the products are nickel plus 2 and carbonate. both of these are in the aqueous phase and the reactant is solid nickel carbonate now keep in mind solids and liquids are not included in the equilibrium expression so we're not going to include this so in this example we just got to plug in the concentrations of nickel and carbonate which they're the same as this number they're all equal to 3.7 times 10 to the negative 4 So nickel is going to be the same number, and carbonate is also equal to the same value.
So for this problem, it's 3.74 times 10 to the negative 4 squared. And so the KSB for nickel carbonate is 1.4 times 10 to the minus 7. And therefore, this is the answer. Now let's work on number two. The molar solubility of calcium hydroxide is 6.875 times 10 to the negative 3. Calculate the Ksp value for calcium hydroxide. So first let's write the dissociation reaction.
Solid calcium hydroxide dissociates into calcium and two hydroxide ions and so these are in the aqueous phase now pay attention to the molar ratio so we have a one to one to two molar ratio So with this reaction, go ahead and calculate the Ksp value for this substance. Now let's calculate the concentrations of everything. The concentration of calcium hydroxide in the aqueous phase is given to us.
That's the molar solubility. That's 6.875 times 10 to the minus 3. Now the concentration of calcium...... Notice that these two are in a one-to-one ratio, so therefore they have the same concentration. Now the concentration of hydroxide is going to be twice the concentration of this value, or twice the molar solubility, because there's a 2 here. So for those of you who want to show your work, here's what you can do.
So starting with calcium hydroxide, the molar concentration is 6.875 moles of CaOH2 per liter. So you can start with that. And then use the molar ratio. The molar ratio is 1 to 2. So for every 1 mole of calcium hydroxide that dissolves, you could say that 2 moles of hydroxide ions are produced.
And so you can cancel these units, and now you have the concentration of hydroxide in moles per liter. so it's 6.875 times 10 to the minus 3 multiplied by 2 and so the concentration of hydroxide is.01375 so now that we have the equilibrium concentrations of calcium and hydroxide we can now calculate the KSP value for it And so Ksp, based on this reaction, is going to be equal to the calcium ion concentration multiplied by the hydroxide ion concentration, but that's going to be raised to the second power. Keep in mind the coefficients turn into the exponents in the equilibrium expression. So now we just got to plug in the numbers that we have.
So the concentration for calcium is 6.875 times 10 to the minus 3. And for hydroxide it's 0.01375. And we need to square it. And so this is equal to 1.3 times 10 to the negative 6. So that is the KSB value for calcium hydroxide. So now you know how to calculate the KSB value. given the molar solubility.
Number three, the solubility of calcium phosphate is 5.1 times 10 to the negative 5 grams per liter. Calculate the Ksp value for calcium phosphate. So like before, let's start out with a reaction.
So we have solid calcium phosphate. and it's going to ionize into the calcium plus two ion which is an aqueous phase and also the phosphate ion now we need to balance the reaction So notice that we have a subscript of 3 for calcium, and a subscript for phosphate is a 2. So now it's balanced. And we're given the molar solubility in grams per liter, as opposed to moles per liter. So the first thing we need to do is calculate the molar solubility in moles per liter. So let's start out with this.
We need to find the molar mass of calcium phosphate, so let's go ahead and calculate it. The atomic mass of calcium is 40.08, and there's three of them in this formula unit. And then we have two phosphorus atoms, each with an atomic mass of 30.97, and there's eight oxygen atoms.
So 3 times 40.08 plus 2 times 30.97 plus 8 times 16, that's equal to a molar mass of 310.18 grams per mole. So we can say that there's 310.18 grams of calcium phosphate per 1 mole of calcium phosphate. So these units will cancel. And so it's going to be 5.1 times 10 to the minus 5 divided by the molar mass, which is 310.18. And so this is going to be 1.644 times 10 to the negative 7 moles per liter.
So that's the molar solubility. That's the first thing we need to do before we calculate the Ksp. now keep in mind this molar solubility represents the concentration of calcium phosphate that's dissolved in solution so now let's calculate the concentration of the calcium plus two ion so notice that it's three times this particular value so it's going to be three times 1.644 times 10 to the minus 7. And so that's going to be 4.932 times 10 to the minus 7, moles per liter or molarity.
Now, the phosphate ion concentration is simply twice the value. And keep in mind, if you want to show your work, you can write it like this. so you need to write that there's two moles of phosphate ions per one mole of calcium phosphate and so that's why you need to multiply by two but for those of you who don't need to show your work for some reason you can just multiply the molar solubility by two and then you'll get the phosphate ion concentration So this is going to be 3.288 times 10 to the minus 7. So now that we have the calcium and the phosphate ion concentrations at equilibrium, we can now calculate the Ksp value. Now granted, the solubility that they give you is always the solubility at equilibrium.
It's not the initial value. We don't need an ice table for this problem. so the KSP value is going to be the calcium plus two ion concentration raised to the third power since we have a coefficient of three and then times the phosphate ion concentration raised to the second power So the concentration of calcium, that's 4.932 times 10 to the negative 7. And for phosphate, it's 3.288 times 10 to the negative 7. And don't forget to raise this to the third power and that to the second power. So go ahead and type in what you see in the calculator. So this is equal to 1.3 times 10 raised to the negative 32. So this is the Ksp value for calcium phosphate.
That's the answer. Number 4. The solubility of the barium plus 2 ion is 2.667 times 10 to the negative 8 moles per liter in the saturated solution of barium phosphate. Calculate the Ksp value for barium phosphate.
So first, let's write the dissolution reaction. Solid barium phosphate. is going to ionize and form the barium plus 2 ion and also the phosphate ion. And looking at the subscripts, it's 3 and 2. So to balance it, we're going to use a 1 on the left, a 3 in front of barium, and a 2 in front of the phosphate ion.
Now, we need to calculate the equilibrium concentrations. these two ions we already have bearing plus 2 which is good the concentration for that is 2.667 times 10 to the minus 8 and that's an insaturated solution so that's the equilibrium concentration so what we need to do is take this value and convert it to the concentration of phosphate and we can use the balance chemical equation to do that so let's start with 2.667 times 10 to minus 8 moles of BA plus 2 per liter and the molar ratio is 3 to 2 so for every three moles of barium that dissolves two moles of phosphate will dissolve along with it so we could cancel these two units so it's going to be 2.667 times 10 to minus 8 multiplied by 2 divided by 3 And so the phosphate ion concentration in this problem is 1.778 times 10 raised to negative 8. Now, let's calculate the Ksp. So let's write the equilibrium expression.
It's going to equal the barium plus 2 concentration times the phosphate concentration. and we need to face this third power and the phosphate concentration to the second power so it's going to be 2.667 times 10 to the minus 8 raised to third power and then 1.778 times 10 to the minus 8 raised to the second power So the answer that I got for this problem is 5.997, which is basically 6, times 10 to the negative 39. So that's the Ksp value for barium phosphate in this problem. And so this is the answer. Now let's work on a different type of problem.
So we have the Ksp value for AgBr. What is the molar solubility of silver bromide? Now the first thing that we need to do is we need to write a balanced chemical equation.
So we've got to write the dissolution reaction for AgBr. So if you were to put solid silver bromide in water, some of it will dissociate into silver cations and bromide anions. Now let's make an ice table. Initially, the concentration of Ag and Br will be 0. As a result, the reaction has to shift to the right. So therefore, the products will increase by x.
The molar ratio is 1 to 1. Now to write the equilibrium expression for this reaction, it's going to be products divided by the reactants. So the products are Ag plus and Br minus. The reactant is AgBr.
However, you cannot include any liquids or solids in the equilibrium expression. So this is the KSB expression for this example. Now, Ksb for AgBr is 5 times 10 to the minus 13. The concentration of Ag plus is X, and for bromide it's X as well.
So, X times X is X squared. Now let's take the square root of both sides. So x...
is equal to 7.07 times 10 to the negative 7 moles per liter. Notice that X represents the concentration of Ag plus and Br minus. Now, notice that everything is in a one-to-one ratio. So, therefore, the concentration of silver bromide that is dissolved in a solution, some of it may be in a solid phase, but the amount that's dissolved in a solution is equal to the concentration of Ag. Ag plus and Br minus due to the one-to-one molar ratio.
So X is also equal to the concentration of the dissolved silver bromide in the solution. So anytime you want to find or calculate the molar solubility of the solid product, the portion that's dissolved in the solution, all you need to do is calculate X. X represents the molar solubility of the substance and so this is the answer.
7.07 times 10 to the minus 7 moles per liter. For the sake of practice, let's try another one. You can work on this example if you want to.
So calculate the molar solubility of lead iodide, or lead 2 iodide. So all we need to do is calculate the value of X. That's going to be the concentration, or the solubility, of lead iodide that's actually dissolved in the solution. so let's start with the substance and it's solid form and it's going to dissociate into PB plus 2 and also iodide now we need to balance the reaction so this is going to be one and this is going to be two Now let's make an ICE table.
So this is going to be 0 and 0. Now this is going to increase by 1x and this is going to increase by 2x based on the molar ratio that you see there. So now let's write the KSB expression for this reaction. So it's going to be PV plus 2 and then times the concentration of iodide. now the coefficients will turn into the exponents in the equilibrium expression now KSP for lead to iodide is 1.4 times 10 to negative 8 and lead plus 2 is equal to X and iodide is equal to 2x 2x times 2x is 4x squared. And x times 4x squared is 4x cubed.
So 1.4 times 10 to the negative 8 is equal to 4x cubed. And keep in mind, our goal is simply to calculate the value of x. So let me rewrite this on top. So at this point, it's all algebra. So what I'm going to do is I'm going to divide both sides by 4 initially.
1.4 times 10 to the negative 8 divided by 4. That's 3.5 times 10 to the negative 9. And that's equal to x cubed. Now the next thing that you want to do is you want to raise... both sides to the 1 3rd power based on the number you see here. So raise it to the reciprocal of that value. When you raise one exponent to another, you need to multiply.
3 times 1 3rd, 3 over 3, that's 1. The 3's cancel. So you get X to the first power is simply X. So it's 3.5 times 10 to the negative 9 raised to the 1 third.
Or you could take the cube root of it. And so X is 1.52 times 10 to the minus 3 moles per liter or molarity. And that's the molar solubility of lead 2 iodide. You just got to calculate the value of X.
The Ksp value for silver phosphate is 1.8 times 10 raised to the negative 18. Calculate the molar solubility of Ag3PO4. So this is similar to the first two problems. Go ahead and try this. So let's begin with the reaction.
So we have Ag3PO4 in the solid phase. And then it's going to dissociate into Ag+, which is in the aqueous phase, and also phosphate, which is also in the aqueous phase. Now, let's go ahead and balance the reaction. So, notice that the subscript is 3. We're going to put a 3 in front of Ag+, and a 1 in front of PO4.
Now, let's make an ice table. So this is going to be 0, 0, plus 3x, plus x, and then 3x and x. So the Ksp expression is going to be Ag plus times phosphate. And then we need to raise this to the third power, and this to the first power.
Now, Ag is equal to 3x, and that's raised to the third power. Phosphate is equal to x. 3 to the third power is 27. So this is 27x cubed times x.
And so this is 27x to the fourth. Now, I'm going to show you a shortcut method that can help you to get this expression. The fact that Ksb is equal to 27x to the fourth.
And then we'll come back to this problem. Now let's start with our first example, which was AgBr. In this substance, notice that the subscripts is 1 and 1. So to calculate the Ksp, for Ag it's going to be 1x raised to the first power, and for Br, 1x raised to the first power. get this expression the KSP is equal to x squared and if you need to calculate the molar solubility just simply calculate X now for the second example it was PBI 2 or PB 1 I 2 so to calculate the KSP expression for PB is going to be 1 X raised to the first power for I 2 2 X raised to the second power and this gave us this expression KSP is equal to 4 X cubed So now for our third example.
This works if you're not dealing with the common ion problems, by the way. If you have a common ion problem, which I'll cover that in another video, you need to actually write out the ICE table. But for a problem like this, if you're given the Ksp value, and you want to calculate the molar solubility, and there's nothing else given to you, then you could use this technique.
So I just want to highlight that fact. So for this example, we had Ag3PO4. The subscript for PO4 is a 1, because nothing was there. So the Ksp is going to be, for Ag, it's going to be 3x raised to the 3rd power, and for phosphate, 1x raised to the 1st power. And so this is going to be 27x cubed multiplied by x, and so that's how we got the expression Ksp is 27x to the 4th.
So now let's go ahead and finish this problem. So the Ksp for silver phosphate is 1.8 times 10 raised to negative 18 and that's equal to 27x to the 4th power. Now the first thing you want to do is divide by the coefficient 27. That's step 1. So if you divide it by 27, this is going to be 6.67 times 10 to the negative 20. And that's equal to x to the 4th. Now last time, we raised everything to the 1 3rd.
This time, because there's a 4, you want to raise everything to the 1 4th power. Because 4 times 1 4th is 1. you want to take the for fruit of 6.67 times 10 to negative 20 and so the molar solubility is 1.607 times 10 to the negative 5 you can write molarity or you can say moles per liter so that's the molar solubility of silver phosphate Now let's move on to Part B. What is the concentration of Ag plus in a saturated solution of silver phosphate? So how can we find the answer?
I'm going to rewrite the reaction that we had in the beginning. So this was 3 and this was 1. Now keep in mind, X is the molar solubility for this substance. For Ag3PO4. So that's how much silver phosphate is dissolved in a solution.
In a saturated solution. So what we need to do is convert it to this value. Notice that the ratio is 1 to 3. So we simply... have to multiply our answer by 3 but if you want to show your work is we could do so start with what you have 1.6 times 10 to negative 5 moles of silver phosphate that's dissolve in a solution per liter and now for every mole of AG 3 PO 4 that's dissolve there's going to be three moles of a g plus dissolve along with it and so that's why we need to multiply by three so now we're going to have moles over liters for a g plus so the concentration of a g plus in a saturated solution of silver phosphate is going to be 4.821 times 10 to negative 5. And so that's the answer. The Ksp value for lead phosphate is 1 times 10 to negative 54. Calculate the molar solubility of Pb3PO42, lead 2 phosphate.
So, let's use the shortcut method. Let's write an equation that relates Ksp to X. So notice that the subscript for lead is 3, so it's going to be 3X to the 3rd power. And for phosphate, it's 2, so it's going to be 2X raised to the 2nd power. 3 to the 3rd is 27. And 2 squared is 4. 27 times 4 is 108. So KSP is equal to 108 times X to the fifth power.
So let's replace KSP with 1 times 10 to negative 54. And so let's start by dividing both sides by 108. So it's going to be 9.26 times 10 to the negative 57. And that's equal to x raised to the 5th power. So we need to raise both sides to 1 over 5. So on the right side, this is going to equal x. On the left side... this is going to be 6.21 times 10 to the negative 12 moles per liter and so that's the answer for part a now let's move on to Part B what is the concentration of phosphate in a saturated solution of lead to phosphate So let's write the reaction first. So lead-2-phosphate is going to dissociate into the Pb plus 2 ion and also the phosphate ion.
now the coefficients I mean the subscripts are three and two so to balance it needs put a three here and a two in front of phosphate so to calculate the concentration of phosphate notice that it's twice the value of the amount of lead III the lead II phosphate dissolve in the solution it's a one to two ratio so if this represents the solubility of this substance then the concentration of phosphate is simply going to be this number multiplied by 2 so it's gonna be six point twenty one times 10 to negative 12 times two And so the phosphate concentration is 1.242 times 10 to the negative 11. And so that's it. So let's say if we need to calculate the lead to ion concentration, for example. Notice that the molar ratio is 1 to 3, so it's going to be 3 times this number. So 6.21 times 10 to negative 12, multiplied by 3, so that's going to be 1.863 times 10 to minus 11. So once you have the molar solubility, you can easily calculate the concentration of any product ion, just by multiplying by the coefficient that you see here. Now let's move on to part C.
Calculate the solubility of lead 3-phosphate in grams per liter. So we're going to start with this number. 6.21 times 10 to the minus 12 moles of lead 3-phosphate.
per liter and all we need to do is convert moles into grams so we need to calculate the molar mass of lead 2-phosphate so there's three lead atoms two phosphorus atoms and eight oxygen atoms Now the molar mass of lead is 207.2. For phosphorus, it's 30.97. And for oxygen, it's 8 times 16. So let's go ahead and plug in these numbers. So this is 811.54 grams per mole. So we're going to put one mole on the bottom.
That's one mole of PB3PO4. And it has a mass of 811.54 grams. So the unit's moles of lead phosphate will cancel.
And so this will give us the solubility in grams per liter. So it's going to be 6.21 times 10 to the negative 12 multiplied by 8, 11.54. And so the solubility is now 5.04 times 10 to the negative 9 grams per liter. So that's the answer. Now let's focus on solving common ion effect problems.
So in this problem, we're given a Ksp value for lead II fluoride. It's 4 times 10 to the negative 8. And our goal is to calculate the molar solubility of solid PbF2 in a solution of sodium fluoride. so first let's write the dissolution reaction for lead fluoride let fluoride will dissociate into lead ions and fluoride ions Now we need to balance the equation.
So this is going to be 1, and then we need a 2 in front of fluoride. Now let's make an ice table. So initially...
the concentration of lead will be 0 the concentration of fluoride is 0.5 because the lead fluoride dissolves in a solution that already contains fluoride sodium fluoride is 0.5 moles per liter Now because the concentration of lead is 0, the reaction has to shift to the right. So therefore, Pb will increase by x, or basically we could say 1x, and fluoride will increase by 2x, due to the 1 to 2 ratio that we see here. So in equilibrium, the concentration of Pb plus 2 is represented by x, and fluoride will be 0.5 plus 2x.
So now let's write the KSB expression. So KSB is going to be products over reactants, and the reactant is a solid, so we cannot include that in the equilibrium expression. So it's just going to be products divided by no reactants in this example.
Now we need to raise this to the second power. So that's the KSB expression for this reaction. Now the KSB is 4 times 10 to the negative 8. And the concentration of the lead 2 plus ion. We said it's X, and the concentration of fluoride is 0.5 plus 2X. Now, notice that Ksp is very small, 10 to the negative 8. That means X is going to be a very small value.
So 0.5 plus 2 times a very small number is still going to be 0.5. So this 2X quantity is negligible. So now we have this expression 4 times 10 to negative 8 is equal to x multiplied by 0.5 squared. So to calculate the value of x, it's simply the KSV value 4 times 10 to negative 8 divided by 0.5 squared.
And so x is equal to 1.6 times 10 to the minus 7. Now, x represents the concentration of the Pb plus 2 ion that's dissolved in the solution. And notice that the Pb plus 2 ion is in a 1 to 1 ratio with lead fluoride. So the amount of lead fluoride that's dissolved in the solution is equal to x in this example. So x represents the molar solubility.
So that's the molar solubility of lead fluoride in... solution of sodium fluoride so that's the answer to Part A now let's focus on Part B what is the molar solubility of lead fluoride in a saturated solution of lead fluoride so in this example we want to calculate the molar solubility if it's not in a sodium fluoride solution So it's just a solution that consists only of PBF2. So how is this problem going to be different than the last problem?
What would you say? In this problem, the concentration of lead is zero, but fluoride will be zero as well. We don't have a common ion in part B. So this is going to increase by 1x, this will increase by 2x.
So the Ksb expression, which is still going to be pb plus 2 times the concentration of fluoride squared. In this case, lead is still x, but fluoride is simply 2x. There's no 0.5 in this example. But Ksb is still going to be 4 times 10 to the minus 8. Now, 2x squared is equal to 4x squared, and 4x squared times x is 4x cubed.
So, our goal is to calculate x, which represents the molar solubility. Now, keep in mind, for the previous answer, it was 1.6 times 10 to the minus 7. Let's see what the new answer is going to be for Part B. So, first, what we need to do is divide both sides by 4. So 4 divided by 4 is 1. So on the left side, it's going to be 1 times 10 to the negative 8, and that's equal to x cubed. Now what we need to do now is take the cube root of both sides, or simply raise both sides to the 1 third power. 3 times 1 third is 1. The cube root of 1 times 10 to negative 8, that's 2.15 times 10 to the minus 3. So this is the molar solubility in a solution of lead fluoride. So now let's focus on part C.
What effect does the presence of a common ion, such as fluoride, have on the molar solubility of lead fluoride? So in part B, if we only had lead fluoride, notice that the solubility, the molar solubility, is 10 to the minus 3. But in the presence of a common ion, when we had the sodium fluoride solution, the molar solubility of PBF2 decreased greatly to 10 to negative 7. So the presence of a common ion reduced the molar solubility of lead fluoride. And so anytime you have a common ion in a solution, the molar solubility of that substance will decrease.