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Topic: Counting pH of CH3COONa + NaCl  (Read 18837 times)

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Offline Matt17

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Counting pH of CH3COONa + NaCl
« on: September 30, 2013, 04:05:57 PM »
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: Ka (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 + H2::equil:: 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.
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #1 on: September 30, 2013, 04:31:34 PM »
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 + H2::equil:: CH3COOH + Na+ + OH-

This is almost correct - sodium acetate was already dissociated before the reaction, that makes Na+ a spectator.
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Offline Matt17

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Re: Counting pH of CH3COONa + NaCl
« Reply #2 on: September 30, 2013, 04:57:53 PM »
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- + H2::equil:: 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?
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #3 on: September 30, 2013, 05:37:00 PM »
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.
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Offline Matt17

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Re: Counting pH of CH3COONa + NaCl
« Reply #4 on: October 01, 2013, 01:02:07 AM »
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.
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #5 on: October 01, 2013, 03:20:38 AM »
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.
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Offline AWK

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Re: Counting pH of CH3COONa + NaCl
« Reply #6 on: October 01, 2013, 09:01:29 AM »
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.
AWK

Offline Matt17

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Re: Counting pH of CH3COONa + NaCl
« Reply #7 on: October 01, 2013, 09:21:10 AM »
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 Ka 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.
« Last Edit: October 01, 2013, 09:59:08 AM by Matt17 »
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #8 on: October 01, 2013, 09:57:36 AM »
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.

Quote
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.
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Offline Matt17

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Re: Counting pH of CH3COONa + NaCl
« Reply #9 on: October 01, 2013, 10:01:27 AM »
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.
« Last Edit: October 01, 2013, 10:48:24 AM by Matt17 »
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #10 on: October 01, 2013, 10:13:22 AM »
Acetate- + H2::equil:: HAcetate + OH-

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

If you know Ka, finding Kb 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.
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Offline Matt17

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Re: Counting pH of CH3COONa + NaCl
« Reply #11 on: October 01, 2013, 11:00:05 AM »
Okay. I think I have it but please check if I'm correct.
Ka*Kb = Kw
[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?
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #12 on: October 01, 2013, 01:58:19 PM »
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).
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Offline Matt17

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Re: Counting pH of CH3COONa + NaCl
« Reply #13 on: October 01, 2013, 03:39:30 PM »
I hope now will be correct. And thank you for being so patient!

CH3COO- + H2::equil:: CH3COOH + OH-
Kb = 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.
« Last Edit: October 01, 2013, 03:53:41 PM by Matt17 »
My mother tongue is not English so please forgive me my mistakes. I try to write as clearly as I can.

Offline Borek

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Re: Counting pH of CH3COONa + NaCl
« Reply #14 on: October 01, 2013, 04:36:47 PM »
General approach is OK (although if I would rather use symbol Cb for CH3COO- concentration), but the final result doesn't look OK to me. Check your math. Are you sure about your Kb?

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.
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