00:03
Here in this problem, given 0 .4 molar, hf, ka given, we have to find the ph.
00:10
Now, let's write the equation here, hf plus water produces h3o plus and f minus.
00:27
Now let's make the ice table.
00:29
Initial concentration here 0 .4 molar that is given here and initial concentration of the products here 0 and here 0.
00:42
Now change concentration, let's say minus x.
00:45
So here it will be plus x and for f minus plus x.
00:50
So equilibrium concentration 0 .4 minus x molar and here x molar and here x molar and 0 plus x, now, let's write the expression for ka.
01:08
Ka is concentration of h3o plus times concentration of f minus.
01:14
These are the concentration of the products and divided by the equilibrium concentration of hf.
01:23
Now here ka is given 7 .2 times 10 to the power negative 4.
01:29
And concentration of h3o plus is x and concentration of f minus also x here.
01:38
So we write x time x, x square, and concentration of equilibrium concentration of hf, 0 .4 minus x.
01:49
Here we are assuming that x is much, much less than the initial concentration, 0 .4...