[fenor]

Accounting for activities, determine the pH of a 0.01 F CH3COOH solution
containing 0.01 F Ca(NO3)2.

Attempt :
(0.01 CH3COOH)(ActivityCoefiictent)
pOH = -log[(0.01 CH3COOH)(ActivityCoefiictent) ]
pH = 14-pOH

[MrTeo]

pOH = -log[(0.01 CH3COOH)(ActivityCoefiictent) ]

Why pOH? CH₃COOH is a weak acid... Anyway you also need the dissociation constant for acetic acid (K_a=1.8e-5 if i remember well...)

[fenor]

pOH = -log[(0.01 CH3COOH)(ActivityCoefiictent) ]

Why pOH? CH₃COOH is a weak acid... Anyway you also need the dissociation constant for acetic acid (K_a=1.8e-5 if i remember well...)

Yes that was my mistake , but where would I use Ka?

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[fenor]

Ka = [H+] [CH3COO-] / [CH3COOH] = 1.8 x 10^-5
What happens to Ca(NO3)2?

[Borek]

What happens to Ca(NO3)2?

Nothing, it is there just to change ionic strength.

[fenor]

Im really confused here,

Ka = [H+]^2/Ca-[H+]

since CH3COOH = Ca -[H+]

Do I have to find Ca before this?

[Borek]

Ca is ambiguous. But both things that you can mark as Ca are given. And they both equal 0.01M.

Perhaps you should answer the question - what IS Ca? - first.

[fenor]

Ca is equal to HA

Since we need H+

Then H+ = to sqrt ( Ka*Ca)

right?

[Borek]

Ca is equal to HA

No, it equals sum of HA and A^-, that is, analytical concentration of acid.

Since we need H+

Then H+ = to sqrt ( Ka*Ca)

right?

Either right or wrong, depending on the circumstances. Formula you use works only if some condition is meet. You should check it.

[fenor]

That would be right if the 5% rule applies if im not mistaken.

[Borek]

So check if it applies and you will know.

[fenor]

since KA=1.8 *10^-5 is very small then 5% rule applies. So the equation should be right.

[Fenix]

*cough* #2
http://www.chem.ucalgary.ca/courses/f10/chem311/311anmt2f10.pdf
ps. That assignment is mental.