The solubilities of AgBr in water, in 0 .01 M CaBr 2 , 0 .01 M KBr and in 0 .05 M AgNO 3 are S 1 ,S 2 ,S 3 and S 4 respectively. Then the order of solubilities is

Step-by-step solution

Solubility of AgBr in 0 .01 M CaBr 2 S 2 ......(i) CaBr 2 is a strong electrolyte and dissociates completely to give common ion Br − CaBr 2 0.01 M → Ca 2+ 0 .01 M + 2Br − 0 .02 M AgBr(s) ⇌ Ag + (aq)+Br − (aq); K sp S 2 0 .02+S 2 ⇒ S 2 × 0.02 = K sp ~ 0.02 or S 2 = 50 K sp Similarly, KBr gives common ion Br − And AgNO 3 gives common ion Ag + and we can calculate S 3 = 100 K sp ......(iii) And S 4 = 20   K sp ......(iv) respectively From (i), (ii), (iii) and (iv) S 1 > S 3 > S 2 > S 4 (Note: K sp > 100 K sp ∵