Answer on Other Chemistry Question for Kathryn
2)a)What is the concentration of SO4 just as BaF2 begins to precipitate when 1.0 M Ba(NO3)2 is added to a solution containing .60 M F and .50 M So4? b)What percent of the sulfate ion remains in solution when the BaF2 begins to preciptiate? c)O Would fractional precipitation work to separate BaF2 and BaSO4? EXPLAIN your answer.
A 0.15 Lsolution of 0.30 M MgCl2 is mixed with 0.15 L of 0.10 M Na2CO3.
Calculate the molar concentration of the carbonate ion left after MgCO3 precipitates. What percentage of the carbonate ion precipitated?
The chemical equation for this reaction is
MgCl2(aq) + Na2CO3(aq) = 2NaCl(aq) + MgCO3(s)
C = n/V
c - concentration (M)
n - amount (moles)
V – volume (L)
n(MgCl2) = C(MgCl2) x V(MgCl2) = 0.30 x 0.15 = 0.045 mol
n(Na2CO3) = C(Na2CO3) x V (Na2CO3) = 0.10 x 0.15 = 0.015 mol
According to the chemical equation:
n(MgCl2) = n(Na2CO3) = 0.015 mol and it’s 0.030 mol of MgCl2 is left in solution after reaction is completed.
Answer: After MgCO3 precipitates it’s no carbonate ion left in the solution. The percentage of the carbonate ion precipitated 100%
a) What is the concentration of SO4 just as BaF2 begins to precipitate when 1.0 M Ba(NO3)2 is added to a solution containing 0.60 M F and 0.50 M SO42-?
b) What percent of the sulfate ion remains in solution when the BaF2 begins
c) Would fractional precipitation work to separate BaF2 and BaSO4? EXPLAIN your answer.
The ion equation for these reactionsare
Ba2+(aq) + SO42-(aq) = BaSO4(s)
Ba2+(aq) + 2F-(aq) = BaF2(s)
From the equations we can see that the amount of Ba2+ is more than it’s needed to complete both of the reactions:
n(Ba2+) = n(SO42-) + n(F-)/2 = 0.50 + 0.60/2 = 0.50 + 0.30 = 0.80 mol, but we have 1.0 mol of Ba2+ ions in the solution.
That’s why if we pour 1 M solution of Ba2+ at once, both of the substances will precipitate almost simultaneously, and it’s very hard to find out when exactly BaF2 starts to precipitate. But it is possible to separate BaF2 and BaSO4 with fractional precipitation, when we are adding a small amount of Ba2+ gradually. As solubility products of BaSO4 and BaF2 differ in 10^4 times
[Ks(BaSO4) = 1.1 x 10^-10; Ks(BaF2) = 1.1 x 10^-6], BaSO4 will precipitate first, it can be collected, and only after this next adding of Ba2+ ions leads to precipitation of BaF2.
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