[Matt17]

Hello!
There are two solutions CH₃COONa (100 cm³) and NaCl (100 cm³) mixed together. Both have the molar concentration of 0,1 mol/dm³. Count the pH. Given values: K_a (CH₃COOH) = 1,6*10^-5

I tried to solve it myself but I couldn't finish.
I know that the hydrolysis of CH₃COONa is:
CH₃COONa + H₂O  CH₃COOH + Na⁺ + OH^-
I think (although I might be wrong) that we can skip NaCl because is does nothing in this case.
After the hydrolysis of CH₃COONa we have a buffer of CH₃COOH/CH₃COONa but I don't know how to use it.

I would be grateful if you could help me. Please remember that my assumptions may be completely wrong.

[Borek]

CH₃COO^- is a weak base - it will react with water producing OH^-.

NaCl presence changes ionic strength of the solution - depending on what you know it should be either taken into account, or ignored.

I know that the hydrolysis of CH₃COONa is:
CH₃COONa + H₂O  CH₃COOH + Na⁺ + OH^-

This is almost correct - sodium acetate was already dissociated before the reaction, that makes Na⁺ a spectator.

[Matt17]

Thank you for your quick response.

This is almost correct - sodium acetate was already dissociated before the reaction, that makes Na⁺ a spectator.

So we have:
CH₃COO^- + H₂0  CH₃COOH + OH^-

NaCl presence changes ionic strength of the solution - depending on what you know it should be either taken into account, or ignored.

It should be taken into account for sure. Could you tell me something more about it, please?

[Borek]

It should be taken into account for sure. Could you tell me something more about it, please?

If you know nothing about ionic strength, chances are you are not expected to take it into account. But if you insist - http://www.chembuddy.com/?left=pH-calculation&right=ionic-strength-activity-coefficients.

[Matt17]

Well, to be honest I have no idea if I should use it or not because this problem is a little more complicated from normal school problems because I'm preparing for a competition. Anyway (regardless if I should use it or not) could you show me the next step? What should I do because I don't know.

[Borek]

Many ways of skinning that cat. In general this is an iterative process, but the first step can do.

First of all - calculate ionic strength of the solution assuming NaCl is the only source of ions. Use it to calculate activity coefficients. Then use your favorite method of calculating H⁺ concentration - just remember each concentration should be multiplied by the activity coefficient in the Ka expression (but not in mass balances). And finally use pH definition, again remembering that it is a minus log of activity.

[AWK]

Both solution give an account to ionic strenght - both approximately the same - about 0.05 for sodium acetate and 0.05 for NaCl => total ionic strength is close to 0.1.

[Matt17]

Guys! I'm really lost. I calculated the ionic strength for NaCl: 0,5*(0,1*1² + 0,1*1²) = 0,1 (and the same would be for sodium acetate). So why ionic strength is 0,05 for both solutions? Also this calculation of activity coefficients seems very difficult. Could you show me how should I do it?

I am also starting to doubt if I should use it or not because the main task of this problem is actually not to count pH precisely but to say whether the solution will be basic, acidic, neutral or close to neutral.

I am also wondering how to use K_a properly in this case because
[tex]K_{a}=\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]}[/tex] is wrong I think because CH₃COOH should be in denominator.
Please help me.

[Borek]

I calculated the ionic strength for NaCl: 0,5*(0,1*1² + 0,1*1²) = 0,1

0.1 M is not the correct value of the concentration. By mixing solutions you are diluting them.

I am also starting to doubt if I should use it or not because the main task of this problem is actually not to count pH precisely but to say whether the solution will be basic, acidic, neutral or close to neutral.

That makes it much easier. Could be NaCl presence is there only to make it look more difficult.

AWK is right about including sodium acetate in the ionic strength calculation, somehow while answering I was thinking about solution of acetic acid, not sodium acetate.

[Matt17]

Okay. So what should I do with this:
[tex]K_{a}=\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]}[/tex]
because it's not correct. (I think)

And sorry for the confusion in the qestion of this answer. In the beginnig I thought that count pH/say if it's acidic or basic was the same. But I was completely wrong.

[Borek]

Acetate^- + H₂O  HAcetate + OH^-

[tex]K_b = \frac {[HAcetate][OH^-]}{[Acetate^-]}[/tex]

If you know K_a, finding K_b in water solutions is trivial (acid and base dissociation constants). Calculate concentration of OH^-, use it to find pOH, convert to pH (pH and pOH dependence). Done.

[Matt17]

Okay. I think I have it but please check if I'm correct.
K_a*K_b = K_w
[tex]K_{b}=\frac{10^{-14}}{1,6*10^{-5}}=1,6*10^{-9}[/tex]

[tex]K_{b}=\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]}[/tex]

[tex][OH^-]=K_{b} * \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}[/tex]

And now I'm not sure here but I think that [CH₃COO^-] = 0,1 mol/dm³  and [CH₃COOH] = 0,1 mol/dm³

So:
[OH^-] = 1,6*10^-9 * 1

Is that all correct?

[Borek]

No, that's not the correct way of approaching the problem. Google for "ICE table", or study the pH calculation lectures (ICE tables are not covered there, as the lectures are based on a different, much more flexible approach).

[Matt17]

I hope now will be correct. And thank you for being so patient!

CH3COO^- + H₂0  CH₃COOH + OH^-
K_b = 1,6*10^-9
So (from ICE table):

[tex]K_b=\frac{x^2}{C_a - x}[/tex]
I wasn't sure if I should use concentration of 0,1 or 0,05. But I decided to use 0,05 since after mixing two solutions this will be the concentration (am I right?).
[tex]x^2 + 0,0000000016x = 0,00000000008[/tex]
x = 8,94*10^-6
and now again I'm not sure but I think that: x = [OH^-] = 8,94*10^-6 mol/dm³

I hope it's fine now.

[Borek]

General approach is OK (although if I would rather use symbol C_b for CH₃COO^- concentration), but the final result doesn't look OK to me. Check your math. Are you sure about your K_b?

Plus, what you gave as a final result is not yet a final result. Best to give answer as pH.

Oh, and please don't write things like 0.000000123 - my eyes hurt from counting zeros. 1.23×10^-7 is much easier to read.