00:02
Hi, let us understand the question.
00:03
We are given 0 .02 molar of cobr2 and ksp of cobr2 is given as 6 .3 multiplied to 10 the power of negative 9.
00:13
We need to calculate the solubility of cubr in pure water and 0 .02 molars of cobr 2.
00:21
So let us see the solution for this.
00:24
See the solution for this.
00:28
So in water, cubr dissociates as c .u.
00:34
And b are negative a.
00:37
Therefore ksp is equal to concentration of cu positive multiplied to concentration of br negative and therefore let the solubility be s so casep value is given is 6 .3 multiplied to 10 xx2 which is equal to s square therefore s is equals to 7 .94 multiplied to 10 plus 0 .95 mol per liter.
01:09
This is the solubility of cubr in pure water.
01:16
Now let us see the solubility of cobr in 0 .0 .02 molar of cobr2.
01:39
Is added to cubr.
01:45
Therefore the concentration of br negative ion changes to 2 multiplied to 0 .0 0 which is equal to 0 .04.
01:56
So total br negative concentration now becomes from cubr and from cobr to so from cubr the concentration was s and from cu br the concentration was s and from cu br the concentration was h and from cobr to the concentration is 0 .4...