00:04
In this problem given 0 .1 molar solution of chloroacetic acid and ka for the chloroacetic acid given as 1 .4 times 10 to the power negative 5 molar at 25 degrees centigrade.
00:19
We have to calculate the ph of this solution and the concentration of other species.
00:26
Now the acid dissociation reaction is clch2cwh plus water, h3o plus plus clch2c2cw minus.
00:54
Now let's make the concentration table.
00:58
Initial concentration is 0 .1 molar.
01:04
Change concentration minus x plus x plus x and equilibrium concentration 0 .1 minus x x and x.
01:17
Now let's write the expression for ca, acid dissociation constant.
01:24
Concentration of h3o plus times concentration of clch to cw minus divided by the equilibrium.
01:35
Concentration of chloroacetic acid and the k a value given 1 .4 times 10 to the power negative 3 concentration of h3 plus x times x divided by 0 .1 minus x therefore x square is equal to 1 .4 times 10 to the bar minus 3 times 0 .1 minus x therefore x square plus 1 .4 times 10 to the power negative 3 x minus 0 .1 times 1 .4 times 10 to the power negative 3 is equal to 0.
02:23
And this quadratic equation and here a is 1, b is 1 .4 times 10 to the power minus 3 and c is minus 0 .1 times 1 .1 times 1 .5 .5 .5 .5...