00:01
Hello students, so i'm explaining here that the reaction between ascetication and the noh is will be ch3 c2oh that react with naoh give formation of ch3 c2o positive okay sorry it is a negative and na will be positive okay i'm just corrected okay so this will be the negative and then h2 o so in this cs3 c wh is an acid that give the formation of conjugated base conjugate base formation of because acid give the formation of base so acidic acid is a weak acid we know and the salt sodium acetate that is the sodium acetate is salt okay form due to the reaction between acid and base is the conjugate base of the acetic acid so we know that we acid and it's salt together make the acidic buffer okay so the ph of the acidic buffer can be calculated by using this that ph is equal to pka that is for the acid pka plus log salt upon acid this is for how we calculate the ph of the acid now given value of the acid is we have given acid value point four and strength of acid is strength of acid is 0 .172 mole then the volume of base volume this is the volume of acid and volume of base is given 40 .2 male 40 .2 and the strength of base base is equal to 0 .2 209 mole.
02:12
Okay, so p .k .a of acetic acid is, of acetic acid is 4 .75.
02:20
Okay.
02:21
So amount of acid will be, amount of acid will be, amount of acid will be 108 .4 that multiplied by 0 .172.
02:40
That will 18 .645 .5 .5.
02:43
That will 18 .645.
02:44
Millimole okay and the amount of base amount of base is equal to 40 .2 multiplied by 0 .209 so you get 8 .402 millimole then again here we do that the amount of salt formation reaction salt after reaction after reaction is equal to 8 .40 to mili -mole and the acid remain after the acid -base reaction the amount of acid after reaction then it is 10 .4 10 .24 millimole.
03:51
Then the total volume will be total volume will be equal to just add all these value 10 108 .4 plus 148 .6 these values okay these 10 108 .4 plus 40 .2m.
04:20
This is the total volume okay and now just put the value and get for the ph will be four point this is a pk value 4 .75 plus log 8 .402 upon 148 .6 upon 148 .6 upon 10 .24 upon 148 .6 okay here it is a millymol upon liter.
04:54
And it is a millingol upon later okay so from here you get ph is equal to 4 .664 okay so the ph of the buffer is 4 .664 and the sodium acetate is the conjugate base of acetic acid when which is formed due to the reaction between acidic acid and the noh as is shown in below.
05:19
So after this i am doing part b...