Here is the tutorial on how to do common types of Ksp problems!

**1. Solid barium sulfate dissolves into its respective ions in water at 25 ^{o}C. If given Ksp is equal to 1.1 x 10^{-10}, calculate the solubility constant.**

To solve this problem, you would need to know that the formula for barium sulfate is BaSO_{4} because barium ion is 2+ and sulfate ion is 2-. Then from your knowledge of the dissolution of solids, you would get:

BaSO_{4} (s) <---> Ba^{2+} (aq) + SO_{4}^{2-} (aq)

The equation is balanced, but make sure to have double arrows, for the reaction does not go to completion. Then using the Ksp equation, you would get:

Ksp = [ Ba^{2+} ] x [ SO_{4}^{2- }]

We are not given any initial concentration of any of these ions so we say that Ba^{2+} have 0 M (moles/liter) and SO_{4}^{2-} have 0 M. However, since the solid barium sulfate dissolves to some extent and we do not know the number (which is what we are trying to figure out), we say that X amount has been added to both ions. So the resulting concentration in equilibrium is [Ba^{2+}] = 0 + X = X and [SO_{4}^{2- }] = 0 + X = X. (Reminder: Brackets around an ion refers to its concentration.) So now you have the concentrations of the two ions and the given Ksp, you can plug into the equation above. You should get:

1.1 x 10^{-10} = (X)(X) = X^{2}

(X^{2})1/2 = (1.1 x 10^{-10})1/2

X = 1.048 x 10^{-5} M = 1.0 x 10^{-5} M = [Ba^{2+}] = [SO_{4}^{2-} ]

**2. Solid magnesium fluoride dissolves into its respective ions in water at 25 ^{o}C. If given that [Mg^{2+}] = 3.0 x 10^{-4} M in equilibrium, calculate the Ksp value.**

To solve this problem, you need to know the formula for magnesium fluoride, which is MgF_{2}. Then as the previous problem, write the chemical equation for the dissolution of MgF_{2}. The formula is:

MgF_{2} (s) ---> Mg^{2+} (aq) + 2F ^{-} (aq)

Did you notice what’s different about this problem compared to the previous one? That’s right. This one has 2 moles of fluorine ion for every 1 mole of magnesium ion. So how do you solve it? It should be easy enough. Just write the Ksp equation and you should get:

Ksp = [Mg^{2+}] x [F ^{-} ]^{2}

If you didn’t notice, the coefficient in front of the fluorine ion became the exponent of the concentration value. The reason for this is because of the mole ratio between the magnesium ion and the fluorine ion. The concentration would multiply by itself for every one other fluorine ion. You are not given any initial concentrations, but you know that [Mg^{2+}] = 3.0 x 10^{-4} M in equilibrium which is also equal to the solubility constant X. You plug that value in for F ^{-} and double it because using the ICE table, F^{ -} would be 2X, and then square it because of the coefficient. You should get:

Ksp = (3.0 x 10^{-4})(6.0 x 10^{-4})^{2} = 1.08 x 10^{-10} = 1.1 x 10^{-10}

**3. In a solution of 0.150 M LiCl at 25 ^{o}C, solid AgCl was placed in. Calculate the solubility constant of AgCl. (Ksp = 1.56 × 10^{-10})**

Set up the chemical equation for the dissolution of AgCl and the Ksp equation as usual. You should get:

AgCl (s) ---> Ag^{+} (aq) + Cl^{-} (aq) Ksp = [Ag^{+}] x [Cl^{- }]

The difference here is that the solid was placed in a solution of lithium chloride, which forms the ions Li^{+} and Cl^{-}. If you haven’t noticed yet, there are Cl^{-} in both the solution and the solid, meaning the solubility constant would be lower than expected since there are more of the ions. So to find the solubility constant, you would need to say that [Cl^{- }] would be 0.150 + X since you start off with 0.150 M of the solution with the Cl^{-} initially. And [Ag^{+}] would just be X. Then plug back into the equation:

1.56 x 10^{-10} = (X)(0.150 + X)

You can take out the X in 0.150 + X, assuming X is so small that it is negligible.

Then you get 1.56 x 10^{-10} = (X)(0.150).

X = 1.04 x 10^{-9} M = [Ag^{+}] which is lower than [Cl^{- }], which is 1.5 x 10^{-8} M. (This is the common ion effect)