Question

At 25°C, a saturated solution of cerium(III) hydroxide (Ce(OH)3) contains 5.1 × 10–6 moles of the compound in one liter of solution.

What is the set-up for determining Ksp?

Answer 1

Through four successful titrations we determined the Ksp of calcium hydroxide to be 1.2611 x 10 -5. The accepted Ksp value for this substance is 7.9 x 10 -6.

This resulted in a 59.62% error in the experiment.

When the solution contained some solids, it was known that it is saturated. Since this solution was saturated it was implied that it was at equilibrium. Being at equilibrium the forward reaction had to equal the reverse reaction. This implied that the dissolving and crystallization reactions happened at an equal rate and simultaneously. For these reactions to be equal, some precipitate would have been present, indicating the saturated solution was at equilibrium.

It was necessary to remove all of the solid Ca(OH)2 before titrating because the it would have affected the rate of reaction. When the HCl was titrated with Ca(OH)2, some of the solid would go down with the liquid because of a minuscule amount of filtering.. Since the solids were really concentrated, it would have many reactions with the HCl. Therefore, the amount of Ca(OH)2 had to be only liquid, with no solid pieces going into the burette, which would have neutralized the acid too quickly.

Answer 2

The solubility product of cerium(III) hydroxide in the given solution is
`1.8266* 10^(-20)`.

Step-by-step explanation:

Concentration of cesium hydroxide =
`5.1* 10^(-6)mol/L`

`Ce(OH)_3\rightleftharpoons Ce^(3+)+3OH^-`

S 3S

1 mol of cerium(III) hydroxide gives 1 mol of cerium ion and 3 moles of hydroxide ions.

Concentration of cerium ion =
`[Ce^(3+)] =S`

Concentration of hydroxide ion =
`[OH^-]= 3S`

The solubility prodcut will be given as:

`K_(sp)=[Ce^(3+)]^1[OH^-]^3`

`K_(sp)=S* (3S)^3=27S^4`

`K_(sp)=27* (5.1* 10^(-6)mol/L)^4=1.8266* 10^(-20)`

The solubility product of cerium(III) hydroxide in the given solution is
`1.8266* 10^(-20)`.