Question

The solubility product constant for La(IO3)3 is 1.00×10-11 at 25 oC. What is the molar concentration of IO3- ions in a saturated solution of La(IO3)3? Assume an ideal solution at 25 oC.

How many grams of La(IO₃)₃ (663.6 g/mol) can be dissolved in 500 mL of pure water at 25 ^oC?

How many grams of La(IO₃)₃ can be dissolved in 500 mL of a 0.100 M KIO₃ solution at 25 ^oC? Use activities for this calculation. 0.964 is the number you use to find the answer

Answer 1

he solubility product constant for La(IO3)3 is 1.00×10-11 at 25 oC. What is the molar concentration of IO3- ions in a saturated solution of La(IO3)3? Assume an ideal solution at 25 oC.

How many grams of La(IO₃)₃ (663.6 g/mol) can be dissolved in 500 mL of pure water at 25 ^oC?

How many grams of La(IO₃)₃ can be dissolved in 500 mL of a 0.100 M KIO₃ solution at 25 ^oC? Use activities for this calculation. 0.964 is the number you use to find the answer

For the salt La (IO₃)₃ at equilibrium,

La (IO₃)₃ < ---------- > La³⁺ + 3 IO₃^3- …………Ksp = 1 x 10^-11 .

Let ‘S’ be the solubility of the salt in moles/L unit at 25 ⁰C.

And hence we have,

[La³⁺] = 1 x S = S moles/L,

[IO₃^3-] = 3 x S = 3S moles/L

Now solubility product (Ksp) is given as,

Ksp = [La³⁺] x [IO₃^3-]³.

Ksp = (S) x (3S)³

Ksp = S x 9S³

Ksp = 9S⁴.

1 x 10^-11 = 9S⁴.

9S⁴ = 1 x 10^-11.

S⁴=(1 x 10^-11) / 9

S⁴=(1.11 x 10^-12)

S=(1 x 10^-11)^1/4……………..(Taking 4^th root of both sides)

S = 1.03 x 10^-3 moles/L.

Hence, [La³⁺] = S moles/L = 1.03 x 10^-3 moles/L.

[IO₃^3-] = 3S moles/L = 3 x 1.03 x 10^-3 moles/L = 3.09 x 10^-3 moles/L.

[IO₃^3-] =3.09 x 10^-3 moles/L.

=========================================

b)Molar mass of La(Io₃)₃ is663.6 g

i.e. 663.6 g in 1000 mL will make 1M La(IO₃)₃

i.e. 663.6/2 g in 500 mL will make 1 M La(IO₃)₃.

So, 331.8 g in 500 mL will make 1 M La(IO₃)₃.

Hence we write,

1 M La(IO₃)₃. = 331.8 g La(IO₃)₃ in 500 mL

1.03 x 10^-3 moles = say ‘A’ g La(IO₃)₃ in 500 mL

A = (331.8 x 1.03 x 10^-3 ) / 1

A = 0.3418 g La(IO₃)₃

0.3418 g La(IO₃)₃ can be dissolved in 500 mL.

==============================

(c) How many grams of La(IO₃)₃ can be dissolved in 500 mL of a 0.100 M KIO₃ solution at 25 ^oC? Use activities for this calculation. 0.964 is the number you use to find the answer

0.1 M KIO3 = 0.1 M IO3^-

Let us find out solubility of La(IO3)3 in this solution

We have Ksp = 1 x 10^-11, [IO3^-]= 0.1 M, [La³⁺]= S = ?

Ksp =[IO3^-] x [La³⁺] = 1 x 10^-11 .

1. X S = 1 x 10^-11 .

S = (1 x 10^-11) / 0.1

S = 1 x 10^-12 moles/L

S = 1 x 10^-12 moles/1000mL…………..(1L = 1000 mL)

S = (1 x 10^-12) / 2 moles/500L…………(divided by 2 as volume halved)

S = 5 x 10^-13 moles/500 mL

Actual concentration = solubilty x activity

So, actual concentration = 5 x 10^-13 x 0.964 = 4.82 x 10^-13 moles/500 mL

We calculated already in (b) that,

1.03 x 10^-3 moles = say 0.3418 g La(IO₃)₃ in 500 mL

4.82 x 10^-13 moles/500 mL = say ‘B’ g La(IO₃)₃ .

B = (0.3418 x 4.82 x 10^-13)/(1.03 x 10^-3)

B = 1.6 x 10^-10 g La(IO₃)₃ can be dissolved in 500 mL of 0.1 M KIO3 solution.