If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Chemistry library

### Course: Chemistry library>Unit 14

Lesson 3: Solubility equilibria

# Worked example: Calculating solubility from Kₛₚ

AP.Chem:
SPQ‑5 (EU)
,
SPQ‑5.A (LO)
,
SPQ‑5.A.1 (EK)
,
SPQ‑5.A.2 (EK)
A compound's molar solubility in water can be calculated from its Kₛₚ 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?

• At why do you raise the concentration of the reactants to the power of the coefficient? In what video is that explained?
• What would you do if you were asked to find the ppm of the cu2+ ion or the OH- ion?
• 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.
• how did we get x times 4x squared, and then 4x cubed?
• 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.
• At around , why do you times the 2x by 2 to get 4x? Thanks
• 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.
• At , is there some reason not to type (5.5*10^-21)^(1/3) instead of scrolling through the catalog?
• He is using a calculator simulator, so it might be a bit different from a normal graphing calculator
• At while 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".
• 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).
• 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.
• 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?
• 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?
• 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)
• Why is X expressed in Molar and not in moles ?
(1 vote)
• Concentration is what we care about and typically this is measured in Molar (moles/liter).

To see why, you can think about what it would mean if the equation was in terms of moles. That would mean it wouldn't matter whether we were trying to dissolve a pinch of salt in a drop of water, a tea cup, a swimming pool, or a lake! It would also mean that if we took two identical flasks of saturated salt solutions and poured them together into another larger flask half the salt would precipitate ...

Does that help?