Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kapital 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?

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.

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.

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.

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?

Show how you do it. n=C*V should combine several unknowns leaving you with just two  volumes of both solutions.

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?? ???

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.

Yes but the problem is that you dont know moles of acetic acid?

Assume you have two unknowns  volumes of both solutions (V_{HAc} and V_{Ac}). 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 V_{HAc} and V_{Ac}  unknowns. You know ratio of these concentrations  from the HendersonHasselbalch equation (http://www.chembuddy.com/?left=pHcalculation&right=pHbuffershendersonhasselbalch). Substitute calculated concentrations into HH equation  that's your second equation.
That yields two equations in two unknowns  V_{HAc} and V_{Ac}. Easy to solve.

Express concentrations of both acetic acid and acetate as a function of initial concentrations (known), final volume (known) and VHAc and VAc  unknowns.
How?

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=dilutionmixing
How many moles of acid in V^{HAc} of acetic acid solution? Number of moles of acetic acid in V^{HAc} 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.