Question

Calculate the molar solubility of Ag₂SO₄ in each solution below. The K_sp of silver sulfate is
1.5 * 10⁻⁵.

(a) 0.26 M AgNO₃:



(b) 0.26 M Na₂SO₄:

Answer 1

Calculate the molar solubility of Ag₂SO₄in each solution below. The K_spof silver sulfate is
1.5 x 10⁻⁵.

(a) 0.26 M AgNO₃:



(b) 0.26 M Na₂SO₄:

Let Solubility of Ag₂SO₄ in AgNO₃ = S mol/L

Partial dissociation of Ag₂SO₄ in aqueous solution is :

..............................Ag₂SO₄ (s) <---------------------> 2 Ag⁺ (aq) ....................+...................SO₄^2- (aq)

................................................................................2S mol/L................................................S mol/L

Complete dissociation of AgNO₃ is :

AgNO₃ (aq) <-----------------------> Ag⁺ (aq)...................+..................NO₃^- (aq)

0.26 M...........................................0.26 M.........................................0.26 M

So, Total concentration of common ion i.e. Ag⁺ = [Ag⁺]_Total = ( 2S + 0.26 ) M

Expression of K_sp for Ag₂SO₄ is :

K_sp = [Ag⁺]²_Total.[SO₄^2-]

1.5 x 10^-5 = (2S+0.26)²S

Since, 2S <<<0.26, So neglect 2S as compare to 0.26.

1.5 x 10^-5 = (0.26)² S

S = 2.2 x 10^-4 M

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In Na₂SO₄

[SO₄^2-]_Total = ( S + 0.26 ) M

K_sp = [Ag⁺]².[SO₄^2-]_Total

1.5 x 10^-5 = (2S)² . (S + 0.26)

Since, S <<<0.26, So neglect S as compare to 0.26.

1.5 x 10^-5 = 4S² x 0.26

S = 3.8 x 10^-3 M