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: 12

                        

0.1 mol hcl is dissolved in distilled water of volm v then at lim v tend to infinity( ph)soln= ? 2- percentage ionisation of water at certain temp is 3.6 into (10) to the power -7 % find kw and ph of water 3- what concn of ac- ions will reduce h3o+ ion to 2* 10 to the power -4 m in o.4 m soln of hac ka (hac)=1.8* 10 to the power -5

6 years ago

Answers : (1)

Aarti Gupta
askIITians Faculty
300 Points
							

Dear student kindly post your questions one by one.
Sol.1:- 0.1M HCl dissolved in V of H2O.
Since V --------> infinity,thus solution becomes almost neutral.
Thus,pH of neutral sol. is 7.
Sol.2:- 3.6* 10-7% of H2O means 3.6*10-7 m ionizes out of 100m of H2O.
Thus, H2o ---> H+ + OH-
Kw = [H+ ] [ OH- ]
= 3.6 *10-7 * 3.6*10-7
12.96 *10-14
Now for pH = - Log [H+]
= - log(3.6*10-7)
= 6.44
Sol.3:-
CH3COOH --> CH3COO- + H3O+
Ka= [H3O][CH3COO-]/[CH3COOH]
1.8*10-5 = (2*10-4)[CH3COO-] / (0.4)
[CH3COO-] =1.8*10-5 * 0.4 / 2*10-4
=0.036 m

Thanks & regard

Aarti Gupta

askiitians Faculty

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

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