Question

The lab technician Anna Lytic adds 2.20 mol KOH to 1.00 L of 0.5 M Al(NO3)3. What is the concentration of aluminum ions after the aluminum nitrate has reacted with the potassium hydroxide?  Kf = 3.0 ´ 1033 for Al(OH)4– Answer: 1*10^(-31). Please explain the process thoroughly.

Answer 1

The chemical reactions taking place are -

KOH + Al(NO3)3 -> KNO3 + Al(OH)3

and then Al(OH)3 + KOH -> Al(OH)4- + K+ ------- Eqn #2

If we balance the OH atoms of the first equation by multiplying KOH by 3 and KNO3 by 3 we get

3KOH + Al(NO3)3 -> 3KNO3 + Al(OH)3

Adding this equation with Eqn #2 we get

3KOH + Al(NO3)3 + Al(OH)3 + KOH  -> 3KNO3 + Al(OH)3 + Al(OH)4- + K+

Cancelling out common terms of Al(OH)3 we get

4KOH + Al(NO3)3 -> 3KNO3+ Al(OH)4- + K+ --------- Final equation

Init 2.2 0.5 0 0 0

Comp 0.2 0 1.5 0.5 0.5

So, here we can see for every 1 mole of Al(NO3)3 we need 4 moles of KOH

KOH , Al(NO3)3 and KNO3 will dissociate in ions in the aqeuous medium to from K+ , OH- , Al+3, NO3- ions

So, Final equation can be written as -

4K+ +4OH- + Al+3 + 3NO3- -> 3K+ + 3NO3- + Al(OH)4- + K+

Cancelling out common ions we get -

4OH- + Al+3 -> Al(OH)4- ----- final ionic equation

Initially, we have 1L 0.5M solution of Al(NO3)3 which will contain 0.5*1 = 0.5 moles of Al(NO3)3

as molarity = moles/Vol in L so, moles = molairty*vol.

So, for 0.5 moles of Al(NO3)3 we will need 0.5*4 moles = 2 moles of KOH (from the final equation)

As Kf value is so high the reaction will be almost complete so, at the end of the reaction we will have 2.2 - 2 = 0.2 moles of KOH and 0.5 moles of Al(OH)4- (from the final equation)

Kf is the constant of formation of Al(OH)4- which can be calculated as from the final ionic reaction as

Kf = [Al(OH)₄^-]/[OH^-]⁴[Al⁺³] where [Al(OH)₄^-] , [OH^-], [Al⁺³] are the final concentrations.

After completion concentration of OH- = KOH = 0.2 M and [Al(OH)₄^-] = 0.5 M (here molarity = moles as volume is 1L and molarity = no.of moles / vol in L)

So, 3*10³ = 0.5/(0.2⁴)*[Al⁺³] so, [Al+3] = (0.5/0.2⁴) * 10^-33 = 1.04 * 10^-31 which is roughly 1*10^-31 M