00:01
Any reaction is given, consider it's hydrolysis.
00:04
That is, reaction is h -co -3 negative in an ex -co's condition, it will react with s2o, it is liquid condition at equilibrium.
00:15
We get here s3o plus, that is, hydrogen ion and here also co3 -2 negative.
00:24
The initial moral concentration is h -c03, that is, 0 .001 molar.
00:34
At initial product formation does not take place, so this is 0.
00:40
After c change here, the concentration will be decreased.
00:45
It will become minus a and here a formation of concentration take place.
00:52
At equilibrium concentration, this will become 0 .001 minus a and here plus a and a.
01:03
These concentrations are take place in a molar form.
01:08
So i will calculate here k equilibrium that will be k equilibrium concentration.
01:16
It is concentration of product upon concentration of reactant.
01:22
Now put the value it will become h3o plus product side and also co3 concentration upon concentration of reactant that is hco3 negative and here h2 in liquid solution.
01:44
So this is our k equilibrium.
01:47
Now solving the k equilibrium we get here 4 .4 multiply 10 raise to the power 7 is equal to a multiply a upon the concentration of hco3 that will be 0 .01...