00:01
Hello everyone, so in this question the ph of the solution is 0 .02 molar of an acid the ka for the ha acid is 2 into 10 rastic power minus 12.
00:16
So we have to find out the value of ph.
00:20
So to find out the value of ph we have to find out the concentration of h plus ions and for that what we have to do is so first of all we write the distortion of this reaction.
00:29
So at equilibrium this will be like this that h plus ions are relays here and the a minus ions are so at t equals to zero the concentration of h a is given as 0 .02 and the h a plus ion will be zero and a minus i .m will be zero okay but at t is equal to equilibrium at equilibrium the concentration of h a ion will be 0 .2 and there will be some dissociation of x concentration so, we'll write it out as x and this will be x.
00:59
Now, since the value of ka is equal to is, concentration of h plus ions divided by the concentration of a minus ions, divided by the concentration of h .a.
01:10
Acid.
01:11
So, we can write it out as x multiplied by x divided by 0 .02 minus x.
01:19
And we'll put the value of ka, that is, the value of ka is 2 into 10 .0 .2 minus x.
01:26
And we'll put the value of ka, that is 2 into tandistic power minus 12...