Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: osmanrulz on March 10, 2009, 08:08:15 PM

Hey Guys, I need help with this question>
When 10.0 L of a saturated solution of magnesium carbonate, MgCO_{3} , is evaporated to dryness, 0.012g of solid magnesium carbonate is obtained. Calculate the Ksp of magnesium carbonate.
Ok, I've tried everything, and I still could not get the answer (it is 2.0 x 10^{10}) .
This is what I've done.
The equation I made is : MgCO_{3} <> Mg ^{2+} + CO_{3}^{2}
1.00g MgCO_{3} x 1 mol MgCO_{3} / 84.3 g/mol = 0.01168 M / 10 L = 1.87 x 10^{3}
Ice Table:
I (ignored) 0 0
C (ignored) + x +x
E 1.86 x 10^{3} +x +x
Ksp = [Mg][CO_{3}]
I Have no idea what to do next :/ , how can I get the Ksp I am stuck.

1.00g MgCO_{3}
Why 1.00?

I just assume 1.00g, that is ideally what you are suppose to do right? (I think?) I am just converting the g / l to M/L

Dear osmanrulz;
It's a wrong assumption!
You know the correct concentration: 0.012g/10L.
Yes you have to convert g/L into moles/L.
Good Luck!
ARGOS^{++}

ok the assumption is wrong then, I would amount to 1.2 x 10^{3} m/L for MgCO_{3} . How does this fit into the equation. Ksp = [Mg][CO_{3}] , but I only have the concentration of Magnesium Carbonate together, I don't have the molar concentrations if it magnesium and carbonate seperately, so How does the molarity i got, fit into the equation, to solve for KsP?

Dear osmanrulz;
Now you forgot the conversion into moles/L, because you have only calculated 0.12g/10L into 0.012g/L (= 1.2 * 103), but you used the wrong units for.
Convert now really into moles/L!
Good Luck!
ARGOS^{++}

Alright so, 1.2 x 10^{3} g / L , is needed to be converted into m / L . Mol / L x Mm (molar mass) = g / L
So I will have to divide 1.2 x 10^{3} g / L by 84.3 g/mol , and I would amount to 1.4235 x 10^{5} mol / L. so the value of x is 1.4235 x 10^{5} mol / L . [Mg][CO_{3}]] = Ksp , so
 = Ksp , [1.4235 x 10^{5}]^{2} = Ksp, Ksp = 2.0 x 10^{10} , yay I got it, thanks guys! :)

m is the abreviation for meter
mol is for moles (do not use a capital letter)

m is the abreviation for meter
mol is for moles (do not use a capital letter)
Yeah your right, I guess I'll avoid that, however the questions written in the course packs do present, that (some x compound) is 0.1 M , so I figured that notation would be ok to use.

M is short for mol/L and is OK, what AWK aimed at was that you wrote something like m/L  which means meters per liter :D