How many grams of BaCO3 will dissolve in 3.0 L of water containing a Ba+2 concentration of 0.5 M (Ksp= 5.1E-9)
My work:
BaCO3 <-> Ba+ + CO3-
i 0.5 M
c -s +s +s
e 0.5 +s +s
(s)(0.5+s)= 5.1E-9
s= 1.0E-9mol/L x 3.0 L x 197.336 g/L=
5.92E-6 grams of BaCO3 dissolved
^Is this answer correct?
2 answers
oh my work wasn't entered correctly. the 0.5 M, 2nd +s, and 0.5+s should be under Ba+
The boards don't recognize spaces after the first one.
No, your answer needs work.
BaCO3 ==> Ba^2+ + CO3^2- (note the charges on mine versus yours).
Ba^2+ = 0.5+s
CO3^2- = s
(Ba^2+)(CO3^2-) = 5.1E-9
(0.5+s)(s) = 5.1E-9
s = 5.1E-9/0.5 = 1.02E-8M )note the difference in the exponent)
Then 1.02E-8 x 3.0 L x 197.3 = 6.04E-6 g which I would round to 6.0E-6 g.
I have reconsidered. Your answer is ok but you made errors getting there. You either made two math errors (both exponent errors) or you made a typo twice; however, the final answer you have of 5.9E-6 g is ok if you round to two s.f.
No, your answer needs work.
BaCO3 ==> Ba^2+ + CO3^2- (note the charges on mine versus yours).
Ba^2+ = 0.5+s
CO3^2- = s
(Ba^2+)(CO3^2-) = 5.1E-9
(0.5+s)(s) = 5.1E-9
s = 5.1E-9/0.5 = 1.02E-8M )note the difference in the exponent)
Then 1.02E-8 x 3.0 L x 197.3 = 6.04E-6 g which I would round to 6.0E-6 g.
I have reconsidered. Your answer is ok but you made errors getting there. You either made two math errors (both exponent errors) or you made a typo twice; however, the final answer you have of 5.9E-6 g is ok if you round to two s.f.
1 answer