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Topic: Calculating Ksp in an equilibrium reaction, given solubility  (Read 16619 times)

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Offline osmanrulz

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Calculating Ksp in an equilibrium reaction, given solubility
« 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.

Offline Borek

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Offline osmanrulz

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #2 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

Offline ARGOS++

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #3 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++

Offline osmanrulz

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #4 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?

Offline ARGOS++

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #5 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++

Offline osmanrulz

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #6 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! :)

Offline AWK

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #7 on: March 12, 2009, 02:05:40 AM »
m is the abreviation for meter
mol is for moles (do not use a capital letter)
AWK

Offline osmanrulz

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #8 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.

Offline Borek

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Re: Calculating Ksp in an equilibrium reaction, given solubility
« Reply #9 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
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