Question

The solubility rules that you might have memorized in first semester Gen Chem can have some surprising exceptions. For instance, "all" potassium salts are soluble, and "all" perchlorate salts are soluble, so we would expect potassium perchlorate (KClO₄) to be a very soluble salt. But it's not! It is only slightly soluble, with a K_sp value of 1.05 X 10^-2. If 5.0 g of potassium perchlorate are added to 100. mL of a 0.10 M solution of sodium perchlorate (a very soluble salt), what is the concentration of potassium ions once equilibrium has been reached?

Answer 1

Solublization of KClO₄ in aqueous phase is given as,

KClO₄ (aq) <---------> K⁺ (aq) + ClO₄^-(aq)

Expression for solubility product is given as,

Ksp = [K⁺][ClO₄^-]

[K⁺][ClO₄^-] = 1.05 x 10^-2 .(given Ksp value) ---------------- (1)

Let solubility of KClO₄ in water be 'S' moles/L.

Hence, [K⁺] = S moles/L and [ClO₄^-] = S moles/L

Addition of NaClO₄ (aq) will provide extra ClO₄^-(aq) which suppress solubility of KClO₄ (aq).

[Note : Na+ will be spectator ion as said "all" perchlorates are soluble i.e. strong electrolytes except KClO₄]

[NaClO₄] = [ClO₄^-] = 0.10 M / 100 mL = 0.10 x 10 M / 100 x 10 mL= 1.0 M/1000 mL = 1 moles/L.

New, [ClO₄^-] = (S+1) moles/L and [K⁺] = S moles/L.

Using these values in eq.(1),

(S+1)S = 1.05 x 10^-2.

S² + S = 1.05 x 10^-2.

Let us solve this quadratic equation by Perfect square method.

Add (1/2 x Linear coefficient)2 = (1/2 x 1)2 = 1/4 = 0.25 = 25 x 10^-2. to both sides,

S² + S + 0.25 = (1.05 x 10^-2)+ (25 x 10^-2)

(S+0.5)² = 26.05 x 10^-2.

Taking square roots of both sides,

S+0.5 = +0.5104 (-ve root will give -ve answer which is unexpected)

S = 0.5104 - 0.5

S = 0.0104 moles/L

i.e. [K⁺] = 0.01404 moles/L at equilibrium.

is concentration of K⁺ ions.

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