How do you solve the Ksp equations?

2 Answers
Feb 10, 2017

Answer:

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.

Feb 10, 2017

Answer:

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)#