25cm3 of 0.1moldm-3 MgCl2 was accidentally mixed with 25cm3 of 0.1moldm-3 BaCl2 solutions together.
In order to separate the two metal ions, a stock solution containing 2.50moldm-3 aqueous Na2CO3 was added using a dropper.
[ Given Ksp MgCO3 = 2.0 x 10^-5 Ksp BaCO3 = 2.0 x 10^-9]
How many drops of Na2CO3 is required to precipitate the maximum amount of BaCO3? ( Assume volume of each drop of Na2CO3(aq) is 0.10cm3)
ANS: 10 drops
I am not sure how the answer is taken. I thought to just equate ionic product of BaCO3 to be equal/not more than Ksp of MgCO3, since we want the maximum amount of BaCO3 without precipitating MgCO3.
Qn : How many drops of Na2CO3 is required to precipitate the maximum amount of BaCO3?
Ans : As much Na2CO3 as you have, obviously! The qn didn't specify you're not allowed to precipitate out MgCO3.
Even if the qn did specify such, this question is certainly faulty. You can prove it to yourself by working backwards from the suppossed answer of 10 drops (even 1 drop is too much and will precipitate out MgCO3).
Of course, I could be wrong. If anyone on this forum can obtain the answer of "10 drops", please feel free to share your working here.
Ultima u are right.
I try this question earlier on and I couldn't reach the answer of 10 drops. As the concentration of Na2CO3 is too high, i suppose.
Yes..I thought so too..the answer key says 10 drops but I didnt understand why.
The way I did it, I have 0.08 drops, which is less than a drop!
and yes the question did imply that MgCO3 is not to be precipitated out at all, I did not type out the full question la.. :p