00:01
Hello students in this question we have been given that 20 m l of 0 .5 molar acetic acid is titrated with 30 ml of 0 .2 molar an a.
00:34
1 .8 multiplied by 10 to the power minus 5.
00:40
So let's calculate number of moles first.
00:44
So number of moles of ch3, c -o -o -h would be equal to molarity, mularity multiplied by volume in litter.
01:04
So we will write 0 .5 multiplied by 0 .02 litters and calculating this we will get 0 .01 more.
01:22
Similarly, number of moles of n -a -o -h is equal to 0 .2 multiplied by 0 .03.
01:41
This comes out to be 0 .006 modes.
01:48
Now our reaction is ch3, coo -o -h reacts with naoh to give ch3 coo -na plus water.
02:13
Now the initial concentration, writing initial with i was 0 .01 and 0 .006 and c .ht .c .o.
02:29
N .a.
02:30
Was not formed.
02:32
Now change in concentration would be minus 0 .006 and 0 .006 c .h3 c .c .o .n.
02:45
Will be formed...