Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Matt17 on September 30, 2013, 04:05:57 PM
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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 + H2O ::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.
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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 ::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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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 ::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?
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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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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.
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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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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.
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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.
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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.
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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.
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Acetate- + H2O ::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 (http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory)). Calculate concentration of OH-, use it to find pOH, convert to pH (pH and pOH dependence (http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product)). Done.
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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?
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No, that's not the correct way of approaching the problem. Google for "ICE table", or study the pH calculation lectures (http://www.chembuddy.com/?left=pH-calculation&right=toc) (ICE tables are not covered there, as the lectures are based on a different, much more flexible approach).
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I hope now will be correct. And thank you for being so patient!
CH3COO- + H20 ::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.
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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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Yes I've made a mistake in calculations, Kb was wrong.
Kb = 6,25 * 10-10
So I'm counting x again.
[tex]6,25*10^{-10} = \frac{x^2}{0,05-x}[/tex]
[tex]x^2 + 6,25*10^{-10}x - 3,125*10^{-11} = 0[/tex]
[tex]x = [OH^{-}] = 5,59*10^{-6} \frac{mol}{dm^3} [/tex]
And of course I know that it's not the end:
[tex]pOH = - log 5,59*10^{-6}[/tex]
[tex]pOH = 5,25 [/tex]
[tex]pH = 14 - 5,25 = 8,75 [/tex]
Which means that the solution will be basic.
Is that correct?
I also have 2 more questions:
1. Hypothtically if I included ioinic strength in calculations, would the error be small enough not to affect the final answer - that the solution is basic? And if not, than in my answer can I simply write that the influence of NaCl is so small that it can be omitted in calculations?
2. I changed the original question a little bit because I didn't know how to translate one phrase to English but I'll try to do it so that I can be sure that everything is correct: The original qustion didn't say that there were 100 cm3 of both solutions but said that: 'Concentrations of all electrolyte solutions are 0,1 mol/dm3. Solutions of CH3COONa and NaCl are mixed in the volume ratio of 1:1 (by volume ratio I mean that both solutions must have the same size eg. 100 cm3, 1 dm3 or whatever). So did I use the concentration of 0,05 in calculations correctly?
Sorry for such a long post.
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[tex]pH = 14 - 5,25 = 8,75 [/tex]
Which means that the solution will be basic.
Is that correct?
Without checking numbers - yes, it is definitely in the right ballpark now.
1. Hypothtically if I included ioinic strength in calculations, would the error be small enough not to affect the final answer - that the solution is basic? And if not, than in my answer can I simply write that the influence of NaCl is so small that it can be omitted in calculations?
In this particular case ionic strength effect would be that the pH is about 0.1 pH unit lower.
2. I changed the original question a little bit because I didn't know how to translate one phrase to English but I'll try to do it so that I can be sure that everything is correct: The original qustion didn't say that there were 100 cm3 of both solutions but said that: 'Concentrations of all electrolyte solutions are 0,1 mol/dm3. Solutions of CH3COONa and NaCl are mixed in the volume ratio of 1:1 (by volume ratio I mean that both solutions must have the same size eg. 100 cm3, 1 dm3 or whatever). So did I use the concentration of 0,05 in calculations correctly?
Why don't you just quote the original wording? :P
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Why don't you just quote the original wording? :P
Because it's not in English. But as far as I can see you are Polish too so I'll write it in Polish (I hope I won't be punished for this). 'Stężenia roztworów elektrolitów wynoszą 0,1 mol/dm3. Roztwory CH3COONa i NaCl są zmieszane w stosunku objętościowym 1:1.
That's it.
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Whoever worded the problem, wanted you to be confused.
But yes, that would mean you start with 0.1 M solutions and you mix them 1:1, ending with 0.05 M.
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You are very helpful! Thank you very much! I've learnt a lot!
But I have a very similar question though it looks easier:
Solutions of CuSO4 and Na2SO4 are mixed together in 1:1 proportions (and concentration of both of them is again 0,1 mol/dm3). The question is the same: is this mixture acidic, basic, neutral or close to nuetral? There aren't any constants given, so I assume that I don't have to count anything.
Na2SO4 only dissociates but doesn't hydrolize. And there aren't any reactions between CuSO4 and Na2SO4.
CuSO4 hydrolizes:
Cu2+ + SO42- + H2O ::equil:: (CuOH)+ + HSO4-
(CuOH)+ + H2O ::equil:: Cu(OH)2 + H+
Which means that the mixture is acidic.
Am I correct?
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CuSO4 solution should be definitely slightly acidic. No idea how much though.
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Thank you! I'd be grateful if you could help me with one more problem.
There are solutions of KF and HCl mixed in proportions of 2:1 (so we can assume that the proportions are the same but concentration of KF is 0,2 mol/dm3 and concentration of HCl is 0,1 mol/dm3). The same question. Constants: Ka(HF) = 6*10-4 .
What I am sure of is that:
KF hydrolizes:
F- + H20 ::equil:: HF + OH-
and
The dissociation of HF is:
HF + H2O ::equil:: H3O+ + F-
And now I don't know if HCl will react with F- or maybe OH-. And then I don't what would be the next step, which ions and reactions to consider in calculations.
I'd be very glad if you could help me with this one, too.
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http://www.chembuddy.com/?left=buffers&right=composition-calculation
See 5th problem (the one about HCl and sodium acetate).
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pKa of hydrated copper(II) is close to 10-8.
Note - if you mix your compound with salt that increase ionic strength then you should correct your pH result for ionic strength, though for 0.05 M CuSO4 + 0.05 M Na2SO4 (I~ 0.3) this is quite difficult.
Try for example aq_solution free program
http://www.acadsoft.co.uk/download/Aq_solutions.zip
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I think I got it:
The reaction:
KF + HCl :rarrow: HF + KCl
And assuming that we have 1 dm3 of 0,2 mol/dm3 KF and 1 dm3 of 0,1 mol/dm3 HCl
[tex][HF] = \frac{0,1}{2} = 0,05[/tex]
[tex][F^-] = \frac{0,2 - 0,1}{2} = 0,05[/tex]
[tex]\frac{[HF]}{[F^-]} = 1[/tex]
Which I think means that
[tex][H^+] = K_a * 1[/tex]
I hope that's correct?
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KF + HCl :rarrow: HF + KCl
Go net ionic, and it is not :rarrow: but ::equil::
And assuming that we have 1 dm3 of 0,2 mol/dm3 KF and 1 dm3 of 0,1 mol/dm3 HCl
Not necessarily follows from the wording of the problem that you posted earlier. Doesn't mean it is wrong.
Which I think means that
[tex][H^+] = K_a * 1[/tex]
Looks OK. For a solution where [A-]= [HA] pH=pKa.
For next problems please start new threads. And an English lesson: you are not counting, you are calculating. In Polish there is no difference, but counting refers to finding number of objects (like you have some eggs and to find out how many are there you count them), while calculating refers to finding a number (to find out how much is 3*5 you calculate). Jest pewna analogia do różnicy między liczbą i ilością - liczba dotyczy obiektów przeliczalnych, jak siedem długopisów (counted), ilość dotyczy obiektów nieprzeliczalnych, na przykład ilość soli w roztworze (którą możesz policzyć - n=C×V - calculate).
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Thank you very much. And thank you for an English lesson - very useful one.
And of course, if I have any more problems with pH, I'll start a new thread.
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I have one more question. Why don't we have to consider hydrolysis of KF in this example? Is that because it is a buffer?
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I have one more question. Why don't we have to consider hydrolysis of KF in this example? Is that because it is a buffer?
In a way - yes. It can be easily shown that the error we make is negligible, compare:
http://www.chembuddy.com/?left=buffers&right=with-ICE-table