Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sundrops on October 27, 2005, 12:25:35 AM
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I have a 10.00mL solution of 0.2000M CH3COONa is titrated with 0.1000M HCl to the equivalence point.
Write out the equation, calculate the volume HCl required and calculate the pH of the solution at its equivalence point.
Ok, I'm still learning this whole titration thing so bear with me guys, and let me know if I'm doing this right. ::)
I'm not too sure about my Equation...
CH3COONa + H20 + HCl <-> H30+ + CH3COO- + NaCl
but what do i do with my NaCl? do i simply disregard it like I would with the water in my ICE table?
so i used M1V1=M2V2
so, (0.2M)(0.01L)=(0.1M)V2
V2 = 0.02L HCl
and to get my pH, I need my concentration of H+, so I can use pH=-log[H+], however I don't know my Ka value, nor do I know how much [H+] I have because I'm not sure about my equation so I cannot form my ice table either. I'm stuck.
Hopefully there's someone out there who understands titration equations. :'(
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CH3COONa + H20 + HCl <-> H30+ + CH3COO- + NaCl
CH3COONa + HCl ----> CH3COOH + NaCl
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What is ICE table?
Your volume of 0.02L looks right.
Wouldn't the pH be 7 if it is a neutralisation?
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thanks,
the reaction goes to completion? no equilibrium?
in that case is there a Ka value?
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how do you know that it is a neutrlization?
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is it because HCl is such a strong acid that the reaction goes to completion?
an ICE table is where you record the initial, the change, and finally the equilibrium concentrations of your products and reactants.
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I would have thought it would go to completion, how did you measure the endpoint?
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Sorry I think you are right, at equivalence the solution will be slighly acidic.
pKa of actetic acid is about 4.76 I think
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I haven't found the equivalence point yet, these are PRE-lab questions. ::) but in lab we will be using phenolpthalein.
so maybe the equivalence point being when 0.02L HCl has been used.
assuming that that is the case, how do I go about finding pH? I'm not given a pKa or a Ka value, or is there a may to determine the concentration of H+ produced? because I could just do the -log of [H+] and find pH that way...
any ideas?
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How do you know the pKa value?
so could I just go like this:
Ka = 10^(-pKa)
Ka = 1.78E-5
and then I really dont know what to do.
what do i do with the salt? how do i determine my pH?
I'm soooo lost. :'(
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Please, can anyone help me find the pH of this solution?
I have no clue how to go about finding it...
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I don't know how you can find it without knowing the pKa or the concetration of the final solution. Weak base strong acid titrations are more complicated than strong acid base titrations.
Borek would know the answer :-\
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thx anyways... :)
I'll keep trying... :'(
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:'(
I can;t figure it out.... does anyone have any ideas on finding ph?
please? ???
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Ka values you should find in physicochemical tables.
For calculation of the final pH treat the solution as pure diluted acetic acid.
Its concentarion is 0.06667 M (1/3 of staring comcentarion of sodium acetate)
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hey AWK,
I need to determine the pH of the solution at the equivalence point, and then compare my answet to the pH of a 0.0667M CH3COOH solution.
i can determine the pH of the 0.00667M CH3COOH (pH=2.97) but I'm still lost about the pH of my CH3COONa solution.
do you have any ideas on how to go about doing that?
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could I write the equation like:
CH3COOH + H2O <-> CH3COO- + H3O
I 0 0 0.2 0.1
C +x -x -x
E x 0.2-x 0.1-x
Ka = (o.2-x)(0.1-x) / x
using the quadratic formula x=0.1
so [H+] = 0.1-x
so [H+] = 0
so that means that there is no pH. because -log(0) doesn't exist.
so does that mean that the reaction goes to completion? but in that case wouldn't the pH be 7?
I'm a little lost.
help?
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What is ICE table?
Initial, Change, Equlibrium - a way of dealing with equilibrium problems. Sundrops already used it later in the thread.
Wouldn't the pH be 7 if it is a neutralisation?
No, but you have already found that.
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The more I read through the thread the less I understand.
For titration of sodium acetate with hydrochloric acid you cant use phenolopthtaleine, as the final pH will be acidic.
There are three areas of the curve. I will outline very shortly how how to deal with them, check much more detailed explanation on titration curve calculation (http://www.chembuddy.com/?left=pH-calculation&right=titration-curves-calculation) on my site.
pKa value of acetic acid is known and given (as AWK wrote) in physicochemical tables. Consider you know the value, it is about 4.75. The calculations can't be done without knowing Ka.
At first you have solution of sodium acetate only. Acetic anion will react with water producing OH- ions, thus making solution slightly basic. To calculate pH you have to:
1. Calculate pKb of the acetic anion (pKa+pKb=pKw, acetic anion is a conjugated base of acetic acid)
2. Use ICE table to calculate OH- (using Kb and ICE table)
3. Convert pOH to pH.
On the curve you have a mixture of acetic acid and acetic anion - so it is a buffer solution. Assume that all added hydrochloric acid reacted with acetic anion and from simple stoichiometry calculate concentrations of both forms, then use Henderson-Hasselbalch equation (http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch)
At the endpoint all of the acetate was converted to acetic acid, so you have to calculate pH using known concentration of weak acid (don't forget the dilution factor from titration), its Ka value and ICE table.
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thx Borek, that was a HUGE help.
I think it was the initial equation that completely threw me off because I couldn;t formulate an ice table with the HCl and NaCl's floating around.
Ok well here's what I did, (followed your instructions to the letter! :D)
pKa = 4.75
Ka=1.78E-5
Kb = 10E-14 / 1.78E-5
Kb = 5.62E-10
then I made my ICE table;
CH3COO- <-> OH- + CH3COOH
I 0.2 0 0
C -x +x +x
E 0.2-x x x
Kb = [OH-][CH3COOH] / [CH3COO-]
x2 + 5.62E-10x - 1.124E-10 = 0
use quadratic formula and.... x=1.06E-5
therefore [OH-] = 1.06E-5
pOH = -log[OH-]
pOH = 4.97
14-pOH = pH
pH = 9.025
does that look ok?
I feel pretty confident about it, mainly because it makes more sense now ;)
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OK
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wait a second,
wouldn't CH3COO- <-> OH- + CH3COOH work for CH3COOH being titrated with NaOH rather than CH3COONa being titrated with HCL?
so shouldn't the equation be: CH3COOH + H2O <-> CH3COO- + H3O+ ??
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nvmnd - i totally get it now.
thx everyone! ;D
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Titration of acetic acid with NaOH:
CH3COOH + OH- <-> CH3COO- + H2O
Such titration makes sense, as the pH change near end point is large and easy to detect.
Titration of acetate with HCl:
CH3COO- + H+ <-> CH3COOH
Such titration will be not usefull, as the titration curve is very flat (see the picture).
Botth equations describe some equilibria, but they also give you general idea about the direction of the process taking place during titration.
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how do you make graphs like that Borek? It's VERY helpful, is it a downloadable program? its great.
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how do you make graphs like that Borek? It's VERY helpful, is it a downloadable program? its great.
BATE - link in my signature.
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pH = 9.025
Borek, does this mean that the end point of the titration of acetate with HCl is basic?
I don't understand ???
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Borek, does this mean that the end point of the titration of acetate with HCl is basic?
No, it is pH of the sodium acetate solution, BEFORE titration.
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No, it is pH of the sodium acetate solution, BEFORE titration.
Ok this makes more sense :)
Write out the equation, calculate the volume HCl required and calculate the pH of the solution at its equivalence point.
So is there a way to work out the pH at the equivalance point? With the data provided with the question?
Would this mean then that the phenolthalein would have to change from red to colorless for this endpoint?
Sorry for the endless questions, please don't die like the magic goldfish after my three questions ;) ;)
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So is there a way to work out the pH at the equivalance point? With the data provided with the question?
Without Ka - no. With Ka - yes. At the equivalence all acetate have reacted with HCl so it is in form of acetic acid. All you have to do is to calculate pH of acetic acid solution (taking dilution into account).
Would this mean then that the phenolthalein would have to change from red to colorless for this endpoint?
Look at the titration curve I have posted earlier. Phenolophtaleine color will be visible at the very beginning of the titration, when only acetate is present. First drop of HCl will make the solution colorless.
Sorry for the endless questions, please don't die like the magic goldfish after my three questions
I am not going to die, but soon I am going to sleep. It is 2 a.m. here.
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So this is is not really a good question from the start then. If you don't know the Ka and you are only assuming the final volume of the solution (what if you had added water during the titration, for example to wash the sides of the flask etc?)
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So this is is not really a good question from the start then.
In a way. First of all such titration makes no sense. If the question was "check if it makes sense" it will make sense.
I hope you will find sense in the above ;)
If you don't know the Ka and you are only assuming the final volume of the solution (what if you had added water during the titration, for example to wash the sides of the flask etc?)
Lack of Ka is less problematic. I am against giving all Ka values every time they are needed, as in real life you have to judge for yourself what data you need to find answer. The easiest way to paralyze students that are always given necessary constants is to not give the data or to give some excess information. They will spend hours trying to use that excess data.
IMHO instead of listing Ka in the text of the question it should be given in some external data sheet for use throughout the chemistry course. Look for example for chemistry data sheet at Roy Jensen's page (http://www.consol.ca/MacEwan.html).
As for volume - on the general level you are right. But then all titration questions are simplified when compared to the real world so I will not bother myself with such details and I will treat assumption on the volume (sum of sample and titrant) as valid.
On the sidenote, one may prepare some interesting questions on the usability of indicators, depending on the dilution of the sample, and the concentration of titrant used.