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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 7

Lesson 8: Solubility equilibria# Worked example: Calculating solubility from Kₛₚ

A compound's molar solubility in water can be calculated from its K

*ₛₚ*value at 25°C. To do so, first prepare an ICE (Initial, Change, and Equilibrium) table showing the equilibrium concentrations of the ions in terms of*x*, the molar solubility of the compound. Then, plug these expressions into the solubility-product expression for the compound and solve for the value of*x*. Created by Jay.## Want to join the conversation?

- At3:42why do you raise the concentration of the reactants to the power of the coefficient? In what video is that explained?(3 votes)
- What would you do if you were asked to find the ppm of the cu2+ ion or the OH- ion?(2 votes)
- Ppm means: "how many in a million?" as in, "How many grams of Cu²⁺ in a million grams of solution"?

We know that [Cu²⁺] = 1.8 × 10⁻⁵ g/L.

We also know that 1 L of solution = 1000 mL, and the density of water

is 1 g/mL, so we have 1000 g of solution.

∴ [Cu²⁺] = 1.8 × 10⁻⁵ g/1 L = 1.8 × 10⁻⁵ g/10³ g

We want to know how many grams of Cu²⁺ in 10⁶ g of solution, so we multiply the denominator by 1000 to get 10⁶ and the numerator by 1000 to keep the fraction constant.

[Cu²⁺] = (1.8 × 10⁻⁵ g/10³ g) × (10³/10³) = 1.8 × 10⁻² g/10⁶ g = 1.8 × 10⁻² ppm or 0.18 ppm or 180 ppb.(4 votes)

- 4:57how did we get x times 4x squared, and then 4x cubed?(2 votes)
- Jay misspoke, he should have said x times 2x squared which results in 4x cubed. Small math error on his part.

Hope that clears it up.(3 votes)

- At around4:53, why do you times the 2x by 2 to get 4x? Thanks(2 votes)
- You aren't multiplying, you're squaring. 2 times 2 is 4 and x times x is x^2, so 2x times 2x equals 4x^2. Then, multiplying that by x equals 4x^3.(2 votes)

- At5:46, is there some reason not to type (5.5*10^-21)^(1/3) instead of scrolling through the catalog?(2 votes)
- He is using a calculator simulator, so it might be a bit different from a normal graphing calculator(2 votes)

- At4:55while he was solving the problem, he mentioned that we should multiply the solubility "X" by 2 "mole ratio" to get the solubility right. I wonder why we don't do so while calculating the equilibrium expressions generally, and the Ksp specifically, using the molarity "concentration".(2 votes)
- You actually would use the coefficients when solving for equilibrium expressions. Looking back over my notes that I took over the Khanacademy MCAT prep videos I don't see any examples with this, but doing just a little research you can confirm that the coefficients are incorporated when determining any equilibrium expression (even if it is just 1).(2 votes)

- What is the difference between the solubility and the solubility product constant?(2 votes)
- How do you know what values to put into an ICE table?(1 vote)
- Ice table stands for:

Initial concentrations (M)

Change

Equilibrium concentrations (M)

so whatever values you have that you can put in that table you can.(3 votes)

- How do you know when to make the initial concentration for OH- 0 versus making it 1.0x10^-7? Do you only make it 1.0x10^-7 if the problems states that the compound is already in solution?(2 votes)
- If you have a slightly soluble hydroxide, the initial concentration of OH⁻
**before the hydroxide dissolves**is 0.

At this level of study, you can ignore any OH⁻ produced by the water.(1 vote)

- If they asked for the concentration of the chloride anion during equilibrium would you just multiply the molar solubility by two?(2 votes)
- I assume you mean the hydroxide anion. In that case, yes, because you have 2 moles of hydroxide for every mole of copper hydroxide that dissolves in the solution.(1 vote)

## Video transcript

- [Instructor] Let's calculate the molar solubility of calcium fluoride if the Ksp value for calcium fluoride is 3.9 times 10 to the negative
11th at 25 degrees Celsius. The first step is to write the dissolution
equation for calcium fluoride. So, solid calcium fluoride
will dissolve in solution to form aqueous calcium two
plus ions and fluoride anions. And to balance that out,
we need to make sure and include a two in front
of the fluoride anions. The next step is to set up an ICE table, where I stands for initial concentration, C stands for the change in concentration, and E stands for
equilibrium concentration. Before any of the solid
calcium fluoride dissolves, the initial concentrations
of calcium two plus ions and fluoride anions in solution is zero. So we can go ahead and put a zero in here for the initial concentration
of the ions in solution. Some of the calcium
fluoride will dissolve, and we don't know how much. So I like to represent that by
writing -X on the ICE table, where X is the concentration
of calcium fluoride that dissolves. Looking at the mole ratios,
it's a one-to-one mole ratio between calcium fluoride
and calcium two plus ions. So if we're losing X for the concentration of calcium fluoride, we must be gaining X for the concentration of
calcium two plus ions. And since it's a one-to-two mole ratio for calcium two plus
ions to fluoride anions, if we're gaining +X for calcium two plus, we must gain plus +2X for fluoride anions. So the equilibrium concentration
of calcium two plus ions is zero plus X, or just X, and the equilibrium concentration
of fluoride anions will be zero plus 2X, or just 2X. The next step is to
write the Ksp expression from the balanced equation. So Ksp is equal to the concentration of
calcium two plus ions, and since there's a coefficient of one in the balanced equation, that's the concentration
of calcium two plus ions raised to the first power, times the concentration
of fluoride anions, and since there is a coefficient of two in the balanced equation, it's the concentration of
fluoride anions raised to the second power. Pure solids are not included in equilibrium constant expression. So we're going to leave calcium fluoride out of the Ksp expression. The concentration of ions
in our Ksp expression are equilibrium concentrations. Therefore we can plug in X for the equilibrium
concentration of calcium two plus and 2X for the equilibrium
concentration of fluoride anions. We can also plug in the Ksp
value for calcium fluoride. So that would give us 3.9 times 10 to the
negative 11th is equal to X times 2X squared. Next we need to solve for X. So, 3.9 times 10 to the
negative 11th is equal to X times 2X squared. Well, 2X squared is equal to 4X squared times X is equal to 4X cubed. So to solve for X, we need
to divide both sides by four and then take the cube root of both sides. So we'd take the cube
root of the left side and the cube root of X cubed. That gives us X is equal to 2.1 times 10 to the negative fourth. And looking at our ICE table, X represents the equilibrium concentration
of calcium two plus ions. So 2.1 times 10 to the
negative fourth molar is the equilibrium concentration
of calcium two plus ions. For the fluoride anions, the equilibrium concentration is 2X. So two times 2.1 times 10 to
the negative fourth is 4.2, let me go ahead and write that down here, 4.2 times 10 to the negative fourth molar for the equilibrium
concentration of fluoride anions. Our goal was to calculate the molar solubility of calcium fluoride. And molar solubility refers to the concentration of
our salt that dissolved to form a saturated
solution at equilibrium. So if X refers to the concentration of calcium
two plus ions at equilibrium, looking at our mole ratios, that's also the concentration of calcium
fluoride that dissolved. Therefore, 2.1 times 10 to
the negative fourth molar is also the molar solubility
of calcium fluoride. Technically at a constant
temperature of 25 degrees, the concentration of a
solid doesn't change. And so you'll see most
textbooks not to put in -X on the ICE table. I like
to just put it in though to remind me that X in
this case does refer to the molar solubility.