# How do you solve the Ksp equations?

Feb 10, 2017

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 $X {Y}_{2}$ which will have a ${K}_{s p}$ value of $4 \times 106 \left(- 15\right)$

When dissolved in water, our compound dissolves according to

$X {Y}_{2} \left(s\right) \rightarrow {X}^{2 +} \left(a q\right) + 2 {Y}^{-}$$\left(a q\right)$

The ${K}_{s p}$ expression is $\left[{X}^{2 +}\right] {\left[{Y}^{-}\right]}^{2} = 4 \times {10}^{- 15}$

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 $X {Y}_{2}$ contains one ${X}^{2 +}$ ion, the equilibrium concentration of ${X}^{2 +}$ will be $x$.

Similarly, each unit of $X {Y}_{2}$ contains two ${Y}^{-}$ ions, menaing the equilibrium concentration of ${Y}^{-}$ will be $2 x$.

Placing these variables into the ${K}_{s p}$ expression:

$\left(x\right) {\left(2 x\right)}^{2} = 4 \times {10}^{- 15}$

To solve this, note that ${\left(2 x\right)}^{2}$ = $4 {x}^{2}$, so our expression becomes

$4 {x}^{3} = 4 \times {10}^{- 15}$

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

x=root(3)(1xx10^(-15) $= 1 \times {10}^{- 5}$ mol/L

This means the solubility is $1 \times {10}^{- 5}$ and the ion concentrations are $\left[{X}^{2 +}\right] = 1 \times {10}^{- 5} M \mathmr{and} \left[{Y}^{-}\right] = 2 \times {10}^{- 5}$ at equilibrium.

Feb 10, 2017

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.
$\left[A {l}^{+ 3}\right]$ in $A l {\left(O H\right)}_{3} \text{ solution with} \left[O {H}^{-}\right] = 2.9 \cdot {10}^{- 9}$ M

Al(OH)_3(s) ↔ Al^(3+)(aq) + 3OH^-(aq)

$K s p = \left[A {l}^{3 +}\right] {\left[O {H}^{-}\right]}^{3} = 1.8 \cdot {10}^{- 33}$

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

$\left[A {l}^{3 +}\right] = \frac{K s p}{O {H}^{-}} ^ 3 = 1.8 \cdot {10}^{- 33}$

$= \frac{1.8 \cdot {10}^{- 33}}{2.9 \cdot {10}^{- 9}} ^ 3$

$\left[A {l}^{3 +}\right] = \frac{1.8 \cdot {10}^{- 33}}{2.9 \cdot {10}^{- 9}} ^ 3$

$= 0.738 \cdot {10}^{- 7}$

$= 7.38 \cdot {10}^{- 8} M$

The other way – finding the Ksp from the solubility – is just a rearrangement of the equation.
The molar solubility of $M n C {O}_{3}$ is $4.2 \cdot {10}^{- 6}$ M. What is the Ksp for this compound?

For $M n C {O}_{3}$ dissolving, we write

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

For every mole of $M n C {O}_{3}$ that dissolves, one mole of $M {n}^{2 +}$ will be produced and one mole of $C {O}_{3}^{2 -}$ will be produced. If the molar solubility of $M n C {O}_{3}$ is s mol/L, then the concentrations of Mn2+ and $C {O}_{3}^{2 -}$ are:

$\left[M n 2 +\right] = \left[C {O}_{3}^{2 -}\right] = s = 4.2 \cdot {10}^{- 6}$ M

$K s p = \left[M {n}^{2 +}\right] \left[C {O}_{3}^{2 -}\right] = {s}^{2}$

$= {\left(4.2 \cdot {10}^{- 6}\right)}^{2} = 1.8 \cdot {10}^{- 11}$