00:01
Of 0 .250 molar lactic acid and ka is given here and in the second portion we have to calculate percent ionization of 2 .0 .250 molar lactic acid in a solution containing 0 .050 molar sodium lactate.
00:21
Okay, let's do this one first.
00:25
The acid plus water in equilibrium with h3o plus plus a minus let's do the ice table here initial concentration 0 .25 more h3o plus 0 here 0 chance concentration say minus x plus x plus x and equilibrium concentration 0 .25 minus x.
01:11
Here it is x and x.
01:14
Now let's write the equilibrium expression equilibrium constant.
01:21
Here is concentration of h3o plus, a minus all this equilibrium concentration and concentration of the lactic acid.
01:32
Ka is given here 1 .4 times 10 to the or negative 4.
01:38
Here, h -3o plus concentration x, a minus also x.
01:44
So x squared divided by 0 .25 minus x.
01:49
Here we are making simplifying an assumption that x is very small compared to 0 .25.
01:57
So we are ignoring x here.
02:00
So x square is 1 .4 times 10 to the core negative 4 times 0.
02:07
So, x will be square root of 1 .4 times 10 to the power minus 4 times 0 .25...