Q36 / 102 · NTA Abhyas (JEE Main) 2020 · Chemistry · Ionic Equilibrium

The increasing order of Ag+ ion concentration in

I. Saturated solution of AgCl

II. Saturated solution of Agl

III. 1 M Ag NH 3 2 +   in 0.1 M NH3

IV. 1 M Ag CN 2 -   in 0.1 M KCN

Given :

K sp  of   AgCl = 1 . 0 × 1 0 - 1 0 ,

K sp  of  AgI = 1 . 0 × 1 0 - 1 6

K d  of  Ag NH 3 2 + = 1 . 0 × 1 0 - 8

K d  of  Ag CN 2 - = 1 . 0 × 1 0 - 2 1

  • A
    I < II < III < IV
  • B
    IV < III < II < I
  • Correct answer
    C
    IV < II < III < I
  • D
    IV < II < I < III

Solution

I     Ag + = K sp AgCl = 1 × 1 0 - 1 0 = 1 0 - 5 M

II    Ag + = K sp AgI = 1 × 1 0 - 1 6 = 1 0 - 8 M

III    Ag NH 3 2 + ⇌ Ag + + 2 NH 3

Kd=1×10-8=Ag+NH32Ag NH32aq+=Ag+0.121.0⇒Ag+=1×10-6M

IV   AgCN2(aq)-1⇌Agaq++2CNaq-1

         K CN 0 . 1 M ⟶ K aq + +  CN aq - 1
                                               (0.1M)

K d = 1 × 1 0 - 2 1 = Ag +   CN - 2 Ag  CN 2 - 1 = Ag +   0 . 1 2 1 . 0

[Ag+ ] = 1 x 10- 19 M.

Hence, the correct answer is option: 'C'.

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