Consider the diprotic acid h2a with k1 100 x 10 4 and k2 100...

Consider the diprotic acid H2A with K1 = 1.00 X 10-4 and K2 = 1.00 X 10-8. Find the pH and the concentrations of H2A, HA- and A2- in a) 0.1000 M H2A b) 0.1000 M NaHA c) 0.1000 M Na2A solutions.

Transcript

0:00 Hello.

00:01 So in the first part, we have this h2a which can be dissociated into ha negative plus h positive.

00:12 So initially the concentration, it is 0 .100 or simply we can write it as 0 .1 and this is 0 and 0 and at equilibrium it is 0 .1 minus x it is there and this is x and x and x it is there.

00:30 So, ka is equal to ha negative and h positive it is there divided by h2a concentration.

00:45 So therefore on putting the value here that is 1 multiply by 10 raised to the power minus 4, it is equal to x squared divided by 0 .1 minus x.

00:56 So therefore on solving this, the concentration, that is x which is equal to h positive and this will become 0 .00316m.

01:13 Now we can calculate the value of ph.

01:17 So therefore, ph is equal to minus log h positive and this will become minus log on putting the value here that is 0 .0033.

01:29 And hence on solving this we will have p h is equal to 2 .50.

01:37 So now the concentration that is h2a, it is equal to 0 .1 minus 0 .00316 and on solving this we will have 0 .0968m and the concentration of ha negative which is equal to x and therefore it is a is 0 .00316m and also this a minus 2 this is equal to 1 multiply by 10 days to the power minus 8 m now so here we are given the concentration that is 0 .1 000 m of n a h a so here this piece p .h is equal to p.

02:39 K .a1 plus p .k .a .2 divided by 2 and this will become 4 plus 8 divided by 2 and hence the value will be here, that is 6.

02:52 Now, this n .aha, it will dissociated into naa negative plus h positive.

03:03 So therefore this ka1 is equal to h.

03:12 This n a n a negative plus multiply by h positive divided by n a h a it is there so we need to find out the concentration of this so therefore this will become on putting the value here one multiply by 10 days to the power minus 4 it is equal to for this it is 0 .1 multiply by for h it is 1 multiply by 10 is to the power minus 6 divided by n .a .h .a.

03:49 So therefore on solving this, this n -aha concentration, it is equal to or simply this is nothing but this is h2a.

04:03 So this will become 0 .001m for h2a.

04:08 Now for the next one we have that is a negative, a minus 2 when it is 0 .001m for h2.

04:18 Get dissociated into h a negative plus oh negative so at initially the concentration here and that is 0 .1 0 and this is 0 and 0 and at equilibrium it will become 0 .1 0 .00 minus x and this is x and x it is there so now k b1 it is equal to x squared divided by 0 .1 minus minus x will be there...

Question

Consider the diprotic acid H2A with K1 = 1.00 X 10-4 and K2 = 1.00 X 10-8. Find the pH and the concentrations of H2A, HA- and A2- in a) 0.1000 M H2A b) 0.1000 M NaHA c) 0.1000 M Na2A solutions.

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