The solubility of CaCO3 is 7 mg / litre. Calculate the solubility of BaCO3 (in mol/L) from this information and from the fact that when Na2CO3 is added slowly to a solution containing equimolar concentration of Ca+2 and Ba+2, no precipitate of CaCO3 is formed until 90% of Ba+2 has been precipitated as BaCO3 . (Assume no hydrolysis of CO32– ion).
Answers
Explanation:
Answer
MwofCaCO
3
=100gmol
−1
.
[CaCO
3
]=
100×1
7×10
−3
Q
sp
for CaCO
3
=(
100
7×10
−3
)
2
=49×10
−10
When only Ba
2+
is 90% precipitated then only CaCO
3
starts precipitation. Then, let solution contains x M of Ca
2+
and Ba
2+
each.
[Ca
2+
][CO
3
2−
]=49×10
−10
[CO
3
2−
]=
x
49×10
−10
[Ba
2+
]=
100
x×10
(90% Ba
2+
precipitated 10% Ba
2+
left)
K
sp
of BaCO
3
=[Ba
2+
][CO
3
2−
]=
100
x×10
×
x
49×10
−10
=4.9×10
−10
≈5×10
−10
Question:
The solubility of CaCO3 is 7 mg / litre. Calculate the solubility of BaCO3 (in mol/L) from this information and from the fact that when Na2CO3 is added slowly to a solution containing equimolar concentration of Ca+2 and Ba+2, no precipitate of CaCO3 is formed until 90% of Ba+2 has been precipitated as BaCO3 . (Assume no hydrolysis of CO32– ion).
Solution:
Solubility of CaCO3 = [7 * 10^-3]/ 100 mole/L = s
Ksp of CaCO3 = s2 = 49 × 10–10.
When [Ba+2] is 90% precipitated, then only CaCO3 starts precipitation.
If original solution contain 'a' mole/L of Ca+2 & Ba+2, then for CaCO3 precipitation,
[Ca^+2] [CO3^–2] = 49 × 10–10, [CO3 ^–2] = [49*10^-10/a] M
Now for BaCO3, Ksp = [Ba+2] [CO3^–2] = a*10/100 * 49*10^-10 / a = 4.9 × 10^–10