June 04, 2020, 11:11:19 AM
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### Topic: solubility product  (Read 4716 times)

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#### kapital

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##### solubility product
« on: June 30, 2011, 05:57:17 AM »
What is the solubility( as a molar concentraction of Ca+2 ions) of CaF2 in water solutions with pH = 3)   pKsp=10,42                pKa=3,17

I tried so:

2[Ca+2]=[F-]+[HF ]

3,9*10^-11=[Ca+2]*[F-]^2

6,77*10^-4=(HF)/( [H+][F-])

But I cant get anthyig from this. Can please somene tell me the rigt way?

#### AWK

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##### Re: solubility product
« Reply #1 on: June 30, 2011, 08:11:19 AM »
Quote
6,77*10^-4=(HF)/( [H+][F-])

Something is wrong in this equation.

You have system of 3 equations with 3 variables. It is solvable problem.
AWK

#### kapital

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##### Re: solubility product
« Reply #2 on: July 01, 2011, 04:04:09 PM »
How woud you prapare a 250 ml  of buffer with pH 4,5 from 0,2000 M NaCHCOO3
and 0,300 M CH3COH? (Ka=1,77*10^-4)

I get from the Henderson Hasselbac eqution that the ratio between acetic ion and acetic acid is: 5,6[CH3COOH]=[CH3COO-]

But I dont know how to continue.

#### Borek

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##### Re: solubility product
« Reply #3 on: July 01, 2011, 05:49:59 PM »
Don't post new questions in old thread, start a new one.

Ratio of concentrations gives you one equation, the other one will be sum of volumes. That will yield (with some additional conversion of nCV) two equations in two unknowns, easy to solve.
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#### kapital

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##### Re: solubility product
« Reply #4 on: July 01, 2011, 08:10:00 PM »
Sory for my mistake.

But I stil cant get the answer.

If I writ the equation like that: 250 = V(sodium acetate solution)+V(acetic acid)

I stil have too many unkowns and if I try to convert into moles I get even moore of them.

So can you please tell moore detail?

#### Borek

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##### Re: solubility product
« Reply #5 on: July 02, 2011, 04:28:58 AM »
Show how you do it. n=C*V should combine several unknowns leaving you with just two - volumes of both solutions.
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#### kapital

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##### Re: solubility product
« Reply #6 on: July 02, 2011, 05:25:49 AM »
250 = n(acetic acid)/(0,3)   +     n(sodium acetate)/(0,2)

Unkowns:  n(acetic acid), n(sodium acetate) , [CH3COOH]

[CH3COO-]- is it okay to say that this is abaut 0,2M ??

Well, whey are still three??

#### Borek

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##### Re: solubility product
« Reply #7 on: July 02, 2011, 08:08:40 AM »
I told you twice - use n=C/v (rearranged if necessary) to get rid of either of n or C. If you know number of moles of acetic acid and volume of the solution, concentration of acetic acid is NOT an unknown.
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#### kapital

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##### Re: solubility product
« Reply #8 on: July 03, 2011, 06:50:31 AM »
Yes but the problem is that you dont know moles of acetic acid?

#### Borek

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##### Re: solubility product
« Reply #9 on: July 03, 2011, 07:49:08 AM »
Assume you have two unknowns - volumes of both solutions (VHAc and VAc-). You know sum of volumes - that's your first equation. Express concentrations of both acetic acid and acetate as a function of initial concentrations (known), final volume (known) and VHAc and VAc- - unknowns. You know ratio of these concentrations - from the Henderson-Hasselbalch equation. Substitute calculated concentrations into HH equation - that's your second equation.

That yields two equations in two unknowns - VHAc and VAc-. Easy to solve.
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#### kapital

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##### Re: solubility product
« Reply #10 on: July 04, 2011, 04:23:19 AM »
Express concentrations of both acetic acid and acetate as a function of initial concentrations (known), final volume (known) and VHAc and VAc- - unknowns.

How?

#### Borek

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##### Re: solubility product
« Reply #11 on: July 04, 2011, 04:38:59 PM »
Express concentrations of both acetic acid and acetate as a function of initial concentrations (known), final volume (known) and VHAc and VAc- - unknowns.

How?

This is a simple dilution, one of the first things you should learn, long before trying to solve any buffer questions.

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

How many moles of acid in VHAc of acetic acid solution? Number of moles of acetic acid in VHAc doesn't change during dilution, use C=n/V to calculate concentration in the final buffer solution.

This is only approximation (diluted acid dissociates), but works OK, especially in this case.
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