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 NH 3 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

Question

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 NH 3 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, I < II < III < IV, IV < III < II < I, IV < II < III < I, IV < II < I < III

MARKS App · Accepted Answer

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 K d = 1 × 1 0 - 8 = Ag + NH 3 2 Ag NH 3 2 aq + = Ag + 0 . 1 2 1 . 0 ⇒ Ag + = 1 × 1 0 - 6 M IV Ag CN 2 (aq) - 1 ⇌ Ag aq + + 2 CN aq - 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.

Search any question & find its solution

Looking for more such questions to practice?