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?
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.
To determine if the answer is correct, we can calculate the solubility product (Ksp) expression for BaCO3 based on the given concentration of Ba+2. The expression is as follows:
Ksp = [Ba+2][CO3-]
Substituting the given values:
5.1E-9 = (0.5 + s) * s
Now, we can solve for 's':
0.5s + s^2 = 5.1E-9
Rearranging the equation gives us:
s^2 + 0.5s - 5.1E-9 = 0
Since the quadratic equation does not have a simple factorization, we can use the quadratic formula. Solving for 's' gives two possible values: s = -0.500000001 or s ≈ 1.0E-9
Since the solubility cannot be negative, we consider the positive root, s ≈ 1.0E-9 mol/L.
To convert this concentration into grams of BaCO3, we can use the molar mass of BaCO3, which is approximately 197.336 g/mol.
Given that the volume of water is 3.0 L, we can calculate the number of grams of BaCO3 dissolved:
Grams of BaCO3 = solubility (mol/L) * volume (L) * molar mass (g/mol)
= (1.0E-9 mol/L) * (3.0 L) * (197.336 g/mol)
≈ 5.92E-6 grams of BaCO3
Based on the calculations, the correct answer is indeed approximately 5.92E-6 grams of BaCO3 dissolved in 3.0 L of water containing a Ba+2 concentration of 0.5 M.