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Offline rcole23

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Ksp problem double check
« on: November 30, 2009, 06:42:46 PM »
Just did some math here on a Ksp calculation.

Can someone double check my solution please?


24) Calculate the Ksp for magnesium fluoride at 25 degrees Celsius, given a solubility of 0.00172 g / 100 mL

MaF2   Ma+2 + 2F-

Ksp = [Ma+2]  [2F-]

Calculate the concentration of MgF2

1 mole of Magnesium = 24.31 grams
1 mole of fluoride = 37.99 grams

1 mole of MaF2 = 62.3 grams

[MgF2] = (0.00172 g/100 mL * 1000) / 62.3 g

[MgF2] = 0.028 mol/L (book has to many decimal places, did the math a bunch of times)

Setup a ICE table

                                    MbF2(s) <-> Mg2+ (aq)  +   2F- (aq)
Initial concentration (mol/L)                         0                 0
Change in concentration (mol/L)      +X                +2X
Equilibrium concentration (mol/L)      X                 2x

X = 0.028

Ksp = 0.028 * (2 * 0.028)2

Ksp = 0.028 * (0.056)2
Ksp = 8.8 * 10-5

Offline AWK

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Re: Ksp problem double check
« Reply #1 on: December 01, 2009, 02:10:48 AM »
Concentration of MgF2 (Solubility) = 0.000274 M
AWK

Online Borek

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Re: Ksp problem double check
« Reply #2 on: December 01, 2009, 02:40:31 AM »
Ksp = [Ma+2]  [2F-]

No, not 2F-.

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[MgF2] = 0.028 mol/L (book has to many decimal places, did the math a bunch of times)

No - as AWK wrote, much less. You are doing something worng.

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Setup a ICE table

What for? You don't have to calculate x, x is already given.
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Offline rcole23

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Re: Ksp problem double check
« Reply #3 on: December 06, 2009, 09:47:48 PM »
So my error is in this formula

[MgF2] = (0.00172 g/100 mL * 1000) / 62.3 g

but the book has the following, i just restated what the book had in a different method.

[MgF2] = 0.00172 g/100 mL * 1mol/62.31 g * 1000 mL/L

This is copied exactly as the book has it, which seems to give the same answer.  Can someone give me a tip to help me figure out why it is wrong?  I don't really understand why 1 mol would be divided by the weight of a MgF2 mole.  Any help is very appreciated.

P.S> Borek as per my lesson i use an ICE table to assist me in the KSP calculation, just what I have been taught, and also an ICE table is in the book.

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Re: Ksp problem double check
« Reply #4 on: December 07, 2009, 04:09:47 AM »
So my error is in this formula

[MgF2] = (0.00172 g/100 mL * 1000) / 62.3 g

No, your error is in the way you are evaluting this formula. No idea how you caculate the value of this expressions, but you do it wrong and it is not 0.028.

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I don't really understand why 1 mol would be divided by the weight of a MgF2 mole.

That should in fact read 1 mole of MgCl2/6.3 g - that is just a molar mass.

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Borek as per my lesson i use an ICE table to assist me in the KSP calculation, just what I have been taught, and also an ICE table is in the book.

ICE table is used when you have to calculate equilibrium concentrations, not when they are GIVEN TO YOU. It is not a tool that you have to use every time you see Ksp.
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Offline stewie griffin

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Re: Ksp problem double check
« Reply #5 on: December 07, 2009, 08:29:16 AM »
Yeah the math makes sense for converting .00172g/100mL into molarity, but you must be typing it into your calculator incorrectly. If I type in .00172 divided by 100 I get 1.72 x 10-5. Then multiply that by 1000 to get .0172. Then divide that by 62.3 and I get 2.7 x 10-4, or .00027 mole/L.

Offline rcole23

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Re: Ksp problem double check
« Reply #6 on: December 07, 2009, 09:42:59 AM »
Hey thanks a lot guys.  Know what I was doing?

[MgF2] = (0.00172 g/100 mL * 1000) / 62.3 g

I was taking .00172 g PER 100 mL not [MgF2] = (0.00172 g divided by 100 mL so I was multiplying .00172 by 1000 then dividign by 62.31

You guys may think this is pretty funny to do but this independent study material has a lot of errors and a lack of examples, gets pretty hard for me to learn.  I even approached the school about how inaccurately typed this new curriculum material is and they acknowledged that i was correct, but that’s about it.

When I get a different answer than the book I am never sure if it is because I am correct and the book is wrong or visa versa.

Thanks again very appreciated, only been racking my brain now for a few weeks :o ;D I really try to figure everything out.

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