00:01
Here in this problem we have to find out the ph of a solution prepared by dissolving 12 .2 grams of benzoic acid in enough water to produce a 500 milliliter solution.
00:13
First we'll find out number of moles of benzoic acid and that will be the mass of benzoic acid given here 12 .2 grams divided by the molar mass of benzoic acid which is 122 .12 grams per mole and we get 0 .0999 moles.
00:45
Now here the volume of the solution given 500 milliliters if we divide by 1000 we get 0 .500 liters because 1 liter is 1000 milliliters.
00:56
Now we can calculate the molarity of this solution.
01:01
Molarity that is moles of the solute divided by the volume of solution in liters.
01:07
Moles of the solute this one 0 .0999 moles of benzoic acid divided by the volume of solution in liters and we get 0 .1998 moles per liter that is molar.
01:24
So this is the molarity of this benzoic acid solution.
01:27
Now let's write the equation for the dissociation reaction of benzoic acid.
01:33
Here is benzoic acid c6h5 cooh plus h2o and here c6h5coo minus plus hydronium ions h3o plus.
01:50
Let's set up ice table.
01:52
Initial concentration 0 .1998 molar benzoic acid.
01:57
Initial concentrations of the products zero.
02:00
Change concentration positive x positive x equilibrium concentration and here 0 plus x x x...