MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        Find the solubility of AgCN in a buffer solution having pH=3.0 (given Lap for AgCN = 1.2*10^-15 and Ka For HCN = 4.8*10^-10
9 months ago

Answers : (1)

Arun
16022 Points
							
Dear Piyush
 
AgCN (s) Ag+ (aq) + CN- (aq) ...... Ksp 
H+ (aq) + CN- (aq) HCN (aq) ..... 1/Ka 
Overall: 
AgCN (s) + H+ (aq) Ag+ (aq) + HCN (aq) ....Ko= Ksp/Ka. 
Ko = (6.00x10^-17) / (4.93x10^-10) = 1.22x10^-7, and 
Ko = [Ag+][HCN]/[H+] 
Also, [H+] = -log[H+]; so [H+] = 10^-3.00 = 0.0010; since the solution is buffered to 3.00. 
For every x moles of AgCN that dissolves in a liter of the buffer, x moles of Ag+ and x moles of HCN are in solution (ie., the solubility), so substituting into the Ko expression: 
1.22x10^-7 = x² / 0.001; 
solving, x = 1.10x10^-5 M.
 
 
 
Regards
Arun (askIITians forum expert)
9 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 276 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details