Question

How do you solve the Ksp equations?

Answer 1

They all follow a basic pattern, which I will describe and demonstrate below...

Explanation:

To illustrate the method, I will use a generic insoluble solid `XY_2` which will have a `K_(sp)` value of `4xx106(-15)`

When dissolved in water, our compound dissolves according to

`XY_2 (s) rarr X^(2+)(aq) + 2 Y^-` `(aq)`

The `K_(sp)` expression is `[X^(2+)] [Y^-]^2 = 4xx10^(-15)`

Here's the solution:

First, let the variable `x` represent the solubility of the solid (the amount that dissolves in one litre). What we must do is express both ion concentrations in terms of `x`.

First, since each formula unit of `XY_2` contains one `X^(2+)` ion, the equilibrium concentration of `X^(2+)` will be `x`.

Similarly, each unit of `XY_2` contains two `Y^-` ions, menaing the equilibrium concentration of `Y^-` will be `2x`.

Placing these variables into the `K_(sp)` expression:

`(x)(2x)^2 = 4xx10^(-15)`

To solve this, note that `(2x)^2` = `4x^2`, so our expression becomes

`4x^3=4xx10^(-15)`

Divide each side by 4, then take the cube root:

`x=root(3)(1xx10^(-15)` `= 1xx10^(-5)` mol/L

This means the solubility is `1xx10^(-5)` and the ion concentrations are `[X^(2+)] = 1xx10^(-5) M and [Y^-] = 2xx10^(-5)` at equilibrium.

Answer 2

Ksp is a particular form of Equilibrium in solution. Use an I.C.E. chart (Initial, Change, Equilibrium) to correctly determine the quantities of reactants and products before using th equations.

Explanation:

Some simpler solutions may be calculated directly. For example, to find the concentration of aluminum ion inn equilibrium with hydroxide, given the Ksp.
`[Al^(+3)]` in `Al(OH)_3 " solution with" [OH^-] = 2.9 * 10^(-9)` M

`Al(OH)_3(s) ↔ Al^(3+)(aq) + 3OH^-`(aq)

`Ksp = [Al^(3+)][OH^-]^3 = 1.8 * 10^(-33)`

Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum ion is:

`[Al^(3+)] = (Ksp)/[OH^-]^3 = 1.8 * 10^(-33)`

`= (1.8 * 10^(-33))/(2.9 * 10^(-9))^3`

`[Al^(3+)] = (1.8 * 10^(-33))/(2.9 * 10^(-9))^3`

`= 0.738 * 10^(-7)`

`= 7.38 * 10^(-8) M`

The other way – finding the Ksp from the solubility – is just a rearrangement of the equation.
The molar solubility of `MnCO_3` is `4.2 * 10^(-6)` M. What is the Ksp for this compound?

For `MnCO_3` dissolving, we write

`MnCO_3(s) ↔ Mn^(2+)(aq) + CO_3^(2-)`(aq)

For every mole of `MnCO_3` that dissolves, one mole of `Mn^(2+)` will be produced and one mole of `CO_3^(2-)` will be produced. If the molar solubility of `MnCO_3` is s mol/L, then the concentrations of Mn2+ and `CO_3^(2-)` are:

`[Mn2+] = [CO_3^(2-)] = s = 4.2 * 10^(-6)` M

`Ksp = [Mn^(2+)] [CO_3^(2-)] = s^2`

`= (4.2 * 10^(-6))^2 = 1.8 * 10^(-11)`