00:01
So in this problem, we are given a buffer of benzoic acid and sodium benzoid in one liter of water, which has a ph of 4 .55.
00:19
And the freezing point is minus 2 degrees c.
00:23
The density of the solution is 1 .01 gram per ml.
00:28
So we have to calculate the concentration of benzodia acid and sodium benzoid.
00:33
So first we have to do is to calculate the ratio of the concentration of salt.
00:37
Sodium benzoid and benzoic acid using the henderson -hesselbeck equation which is as follows the ph is equal to p -k -a plus log -off base concentration over acid concentration.
00:58
So first we have to find the value of this ratio here.
01:02
So if we just, we know the ph value which is 4 .55 and the p -k -a value for for for for for for for the benzeg acid is 4 .19 for or 4 .187 so you can write the ph we need is 4 .55 pk a for the benzic acid is 4 .187 plus log of base over acid so if we just solve this equation we will find the ratio of the base concentration to acid concentration as 2 .306 so this is the ratio of the base concentration of the acid concentration now we have to use the freezing point depression to find out the molality of the on the solution so as you know that that the depression in the freezing point is given by del t f is equal to m times kf so here del t f is the depression in freezing point m is the molality of the solution and kf is the freezing point constant which has a value of 1 .86 degrees c per more m so from here we can find the molality of the solution just by putting the del t f over kf and as you have seen the freezing point depression value is two degrees c and the kf value we have here is 1 .86 degrees c per m so from here we will have a value of 1 .075m which is the molality of the solution and then we have to find out how many mole of acid and base we have here we will use the use the value of the molality to find out that so the sodium benzoyed will be dissociated as in m c7 h5 .02 sodium benzoid and this will be dissociated as n .f equus plus c7h5, c7h5, o2 minus.
04:52
So that means that one moles of this sodium benzoyt is giving in total two moles of the solute, that means one more sodium and one more benzoyon.
05:11
And that means the total moles of solute if we calculate that, it will be like 1 mole benzate acid plus 2 mole of sodium benzoyate so this is the this is equal to the molality of the solution we just calculated so that means that one mole of acid plus 2 mole of base we have here and that is equal to the molalitium of the solution we calculated here and then we have to convert it to all the number of most of acid so we can do this by and then and then converting this sodium to the benzate acid by just multiplying the mole ratio we just calculated it will be 2 times 2 .306 mol of the acid or the benzoic acid hccc.
06:57
H .c7 h502.
07:00
So our this value is coming from here which is our base to acid concentration ratio.
07:09
So you're multiplying this by to convert the base to acid.
07:13
This will be equal to 1 .075 and then that means in total we have 5 .612 mole acid hc7h502 equal to 1 .075...