InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Saccharin (K, 2 x 1012) is a weak acid represented by formula HSaC. A 4 x 10 moleamount of saccharin is dissolved in 200 cc water of pH 3. Assuming no change involume, calculate the concentration of Sac- ions in the resulting solution atequilibrium. |
|
Answer» We have, [HSaC] = mole/litre= (4×10-4)/(200/1000)= 2×10-3MHere dissociation of HSaC takes place in presence of [H+] = 10-3 HSaC ↔ H+ + SaC- Concentration before dissociation 2×10-3 10-3 0In presence of H+the dissociation of HSaC is almost negligible because of commonion effect. Thus at equilibrium[HSaC] = 2×10-3; [H+] = 10-3As, Ka = {[H+][SaC-]}/[HSaC]Therefore, 2×10-12 = {(10-3)[SaC-]}/(2×10-3)Therefore,[SaC-]= 4×10-12M (A) |
|