Question

The concentration of [H] and concentration of [OH l of a 0.1 aqueous solution of 2%
acid is [lonic product of water = 1 x 10-14) -
[1] 0.02 x 10-3 M and 5 x 10-11 M
[2] 1 x 10-3 M and 3 x 10-11 M
[3] 2 x 10-3 M and 5 x 10-12 M
[4] 3 x 10-2 M and 4 x 10-13 M​

Answer 1

Solution : 

Given that 

Concentration of solution =1=1 

Degree og ionisation =2%=2100=.02=2%=2100=.02 

Ionic product of water =1×10−14=1×10-14 

Concentration of [H+]=[H+]= Concentration of solution XX 

Degree of ionisation =.1×.2=2×10−3M=.1×.2=2×10-3M 

Concentration of [OH−]=Ionic product of water[H+][OH-]=Ionic product of water[H+] 

=1×10−142×10−3=0.5×10−11=5×10−12M