# Chemical Forums

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

Title: Calculating Ksp in an equilibrium reaction, given solubility
Post 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, MgCO3  , 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 : MgCO3   <---> Mg 2+  + CO32-

1.00g MgCO3  x  1 mol MgCO3 / 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][CO3]

I Have no idea what to do next :/   , how can I get the Ksp I am stuck.
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: Borek on March 10, 2009, 08:17:04 PM
1.00g MgCO3

Why 1.00?
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: osmanrulz on March 10, 2009, 08:21:21 PM
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
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: ARGOS++ on March 10, 2009, 08:30:17 PM

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++
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: osmanrulz on March 10, 2009, 08:34:25 PM
ok the assumption is wrong then, I would amount to 1.2 x 10-3   m/L for MgCO3 . How does this fit into the equation. Ksp = [Mg][CO3]    , 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?
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: ARGOS++ on March 10, 2009, 08:40:31 PM

Dear osmanrulz;

Now you forgot the conversion into moles/L, because you have only calculated 0.12g/10L into 0.012g/L (= 1.2 * 10-3), but you used the wrong units for.

Convert now really into moles/L!

Good Luck!
ARGOS++
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: osmanrulz on March 10, 2009, 08:54:58 PM
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][CO3]] = Ksp , so

• = Ksp , [1.4235 x 10-5]2 = Ksp, Ksp = 2.0 x 10-10 , yay I got it, thanks guys! :)
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: AWK on March 12, 2009, 02:05:40 AM
m is the abreviation for meter
mol is for moles (do not use a capital letter)
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: osmanrulz on March 13, 2009, 10:03:32 AM
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.
Title: Re: Calculating Ksp in an equilibrium reaction, given solubility
Post by: Borek on March 13, 2009, 10:38:21 AM
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