Question

At 25 °C only 0.0560 mol of the generic salt AB2 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?

At 25 °C only 0.0560 mol of the generic salt AB2 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? Number

Answer 1

For a general ionization of AmBn having concentration S M/L i.e. moles/L

AmBn < ------ > mAⁿ⁺ + nB^m-.

We can see that 1 mole of AmBn on ionization gives m moles of Aⁿ⁺ and n moles of B^m- ions.

In saturated solution we will have,

[Aⁿ⁺] = m x S and [B^m-] = n x S

Using this we have,

SO the solubility product is defined as

Ksp = [Aⁿ⁺]^m [B^m-]ⁿ.

For given generic salt,

AB₂(s) < ------ > A²⁺(aq.) + 2B^-1(aq.)

Ksp = [A²⁺]¹ [B^-]²

Ksp = (S) x (2S)².

Ksp = 4S³……….(1)

AT 25 ⁰C [AB2] = 0.0560 mol in 1 L

i.e. S = 0.0560 mole/L

Let us put this value in eq.1

So, Ksp = 4 x (0.0560)³ (mole/L)³.

Ksp = 4 x 1.76 x 10^-4 (mole/L)³.

Ksp = 7.02 x 10^-4 mole³.L^-3

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