00:01
Hello, so here in this question we are given with 0 .04 molar na3 po4 and for h3po4 the dissociation constants ka1, ka2, ka3 is given as 7 .11 into 10 to the power negative 3, 6 .32 into 10 to the power negative 8 and 4 .5 into the power negative 13 respectively.
00:24
Okay? what we need to calculate is for the ph.
00:28
So using ka, we can, can calculate for kb.
00:32
By the equation, kb is equal to kw by ka3.
00:39
Now why we have taken ka3 is because we are looking for complete dissociation.
00:45
Okay.
00:46
And the equation corresponding for that is p .o43 minus plus h2o will give h.
00:55
P .o42 minus plus oh minus.
01:00
Okay.
01:01
Now we can substitute the values over here.
01:03
Kw is known to us as 10 to the power negative 40 divided by ka3 is given as 4 .5 into 10 to the power negative 13.
01:12
Okay.
01:13
So solving this we will get kb as 0 .022.
01:18
Okay.
01:19
Now from here we can write the equation for kb as kb is equal to concentration of the products that is hpo4 2 minus a into all...