When a 0.1M solution of weak acid HA was titrated with a 0.1M NaOH. The pH measured when Volbase=1/2Voleq point was 4.62. Using the activity coeficents, calculate pKa.
fA-=0.854 and fHA=1.00
For this I'm confused once again. I know that using the activity coeficents I can get the activity which would then be used as concentration right?
So..
ax=[A]x * fx
aNaA=[NaA]*fNaA
aNaA= [0.1][0.0854]= 0.0854
aHA=[HA]*fHA
aHA= [0.1][1.0]= 0.1
[H30+]= 10^-pH= 10^-4.62= 2.40e-5
But I'm not sure what equation to use but since it is a buffer I, assuming it would be using a buffer eqzn
BUT since it is a titration wouldn't I have to go and subtract the moles of HA - moles NaOH? This is where I think that the 1/2 vol of equivalence point comes into play but I'm not sure. So do I go and assume that since it is 0.1M that it is cHA= 0.1-0.05 ? but what would the cNaA be then? would it be 0.05moles ? I guess it would but I'm not sure how to apply that here with the activity coeficents since I've only seen it used with the Ksp.
anyways not using that this is what I did..
[H30+]=Ka(cHA/cA-)
2.40e-5 = Ka (0.1/0.0854)
Ka=2.04e-5
pKa= -log(Ka)= 4.69
assuming it was using 1L
.5ml x (0.1mol NaOH/L)= 0.05mol NaOH
aNaA= [0.05][0.854]=0.0427
aHA= [0.05][1.00]= 0.05
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[H30+]=Ka (cHA/cNaA)
2.40e-5=Ka*1.1709
Ka=2.04e-5
pKa= 4.69 ===> the exact same thing!?!
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Yes, you get the same answer because cHA and cNaA cancels; therefore, no matter what numbers you place for those two values, it has no effect on the answer. I suggest, however, that if you want to do it right, you might put in 0.0333. Two or three examples will show you why. Suppose we start with 50 mL of 0.1 M HA and titrate with 0.1 M NaOH. At the halfway point, we have
5 millimols HA to start - 2.5 millimols NaOH added leaving 2.5 millimols HA and forming 2.5 millimols NaA.
(HA) = 2.5/75 = 0.0333 M
(NaA) = 2.5/75 = 0.0333 M.
Start with 100 mL 0.1 M HA.
We have 10 millimols to start - 5 millimols NaOH added leaving 5 millimols HA and forming 5 millimols NaA.
(HA) = 5/150 = 0.0333 M
(NaA) 5/150 = 0.0333 M
Start with 200 mL of 0.1 M HA.
20 millimols HA to start - 10 millimols NaOH leaving 10 millimols HA and forming 10 millimols NaA.
(HA) = 10/300 = 0.0333 M
(NaA) = 10/300 = 0.0333 M
I don't see anything wrong with the way you have worked the problem. We are assuming, I suppose, that the pH of 4.62 measures the ACTIVITY of the H+.
You said that it was supposed to be the same but since the concentration was multiplied by the activities which were both DIFFERENT #'s the ratio WOULD NOT cancel out as I found out....but cNaA and cHA were different #'s and I got as you see above 4.69.
for the pKa...Did you mean that the pKa would be the same or the pH would be the same as the pKa.
I was just thinking about this question since it was 1 of the Q on my final and I got 4.69 when I did the Q over on my test.
You said that it was supposed to be the same but since the concentration was multiplied by the activities which were both DIFFERENT #'s the ratio WOULD NOT cancel out as I found out....but cNaA and cHA were different #'s and I got as you see above 4.69.
for the pKa...Did you mean that the pKa would be the same or the pH would be the same as the pKa.
pKa is the same as pH at the half way point of any titrattion of a weak acid with a strong base just because the base produced (the salt in this case) equals the concentration of the acid remaining. But that is true only when considering concentrations alone (neglecting activity coefficients) or when the acitivity coefficients are the same for both acid and base. You can see that
pH = pKa + log (base/acid).
When base = acid, as it does at the half way point, then base/acid = 1 and log 1 is zero so pH = pKa. Obviously, if the activity coefficients are used and they are not the same then acid/base will not be equal and log of that term will not be zero so pH will not be the same as pKa.
I hope you did well on the test.
Well I did get alot of help from you so...MANY THANKS TO YOU DR.BOB =D
and there will be many more smiles if I find I did well in the class...lab unknowns making me nervous though...=/
Yay I guess that was right then hm?
anyways, it's just like I thought; the activity coeficent would change the ratio and thus the pKa would not equal the pH.
Thanks again for clarifying that
Dr.Bob
From the information given, we know that the pH measured at the halfway point of the titration is 4.62. We need to calculate the pKa using the activity coefficients.
To calculate the pKa, we can use the equation:
[H3O+] = Ka * (cHA / cA-)
But since we are using activity coefficients, we need to apply them to the equation. The activity of a species is given by the product of its concentration and its activity coefficient:
aHA = [HA] * fHA
aA- = [A-] * fA-
In this case, the activity coefficients are given as fHA=1.00 and fA-=0.854.
Now, let's plug in the values we know:
[H3O+] = Ka * (cHA / cA-)
At the halfway point of the titration, the concentration of both HA and A- will be equal, as half of the acid is converted into its conjugate base. So we can set cHA = cA-. Let's call the concentration of both substances "c".
[H3O+] = Ka * (c / c) = Ka
We know that the pH is given as 4.62, which is equal to the negative logarithm of [H3O+]. So let's calculate Ka using the given pH value.
[H3O+] = 10^(-pH) = 10^(-4.62) = 2.40e-5
Now we can solve for Ka:
2.40e-5 = Ka
Finally, calculate the pKa by taking the negative logarithm of Ka:
pKa = -log(Ka) = -log(2.40e-5) = 4.62
Therefore, the calculated pKa using the activity coefficients is 4.62.