What is the pH of the solution created by combining 12.20 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
with 8.00 mL of the 0.10 M HC2H3O2(aq)?
mL NaOH pH w/HCl pH w/HC2H3O2
12.20
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my work:
0.000420 moles of NaOH / 0.020 litres = 0.0210 Molar NaOH
- log of 0.0210 Molar NaOH =
pOH of 1.68
since pOH + pH = 14
pH+ = 12.32
I did this and I got a pH of 12.32 for both NaOH and HC2H3O2.
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Its the second part I'm having trouble with.
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
mL NaOH pH w/HCl pH w/HC2H3O2
12.20
pH=pka +log(base/acid)
pka=4.75
for the base I got =0.06039602
for the acid I got = 0.02079
apparently this is wrong.
Is the pH for NaOH different than the pH of HC2H3O2? cuz for the first part they had the same pH (pH=12.32)
Thanks.
There have been a number of posts lately concerning NaOH/HCl and NaOH/HAc (acetic acid--my abbreviation). However, in all cases, moles NaOH are less than moles HAc and an acetic/acetate buffer is created which behave markedly different than the solution you posted.
I don't see anything wrong with your HCl/NaOH work except that 12.20 mL + 8.00 mL = 20.2 mL (0.0202 L and not 0.020 L). That small difference doesn't change the pOH or the pH.
For the NaOH/HAc part, I think the pH is the same. I looked at the OH^- contribution from the hydrolysis of the Ac^- and it is negligible if I didn't make a mistake in my arithmetic. The second part of the problem is a little confusing to me, too, but I think for different reasons. For the NaOH/HCl portion, you have 0.42 millimoles in 120.2 mL = (OH^-) and you can go from there to pOH and pH. The NaOH/HAc part would be the same thing.
What confuses me? I have worked several problems in the last week or so worded almost the same BUT the NaOH/HAc portions has ALWAYS had less NaOH than HAc which forms a buffer solution of HAc/Ac^- and the part about adding 100 mL as a dilution is easy to answer for that. That idea is that the HCl is changed markedly whereas the HAc/Ac^- is not changed at all with dilution. I wonder if somehow these volumes have become interchanged. For example, one I did yesterday had 2.4 mL NaOH + 8.00 HAc (which makes the buffered solution very well). If you find that the NaOH/HAc is NOT the same, I would appreciate you posting it (and if it is different, why is it different).
See this post just a few above yours.
http://www.jiskha.com/display.cgi?id=1272915437
HI again Dr.Bob
This is what I'm doing
.10M (.0122L)= .00122 moles of NaOH
.00122/(.008+.0122)= .06039M
.10M (.008)= .0008 moles CH3COOH
.0008-.00122=.00042 moles HCl unreacted
.00042/(.008+.0122)=.020792M CH3COOH
pH= pka + log (x/y)
pH= 4.75 + log (.06039/.020792)
pH= 5.213
but its not the right answer. What am I doing wrong?
This is what I'm doing
.10M (.0122L)= .00122 moles of NaOH
.00122/(.008+.0122)= .06039M
I get 0.06040 M NaOH
.10M (.008)= .0008 moles CH3COOH
.0008-.00122=.00042 moles HCl unreacted
.00042/(.008+.0122)=.020792M CH3COOH
The HCl is not in excess and unreacted. The reaction is 0.00122 moles NaOH - 0.0008 moles HCl = 0.00042
moles NaOH unreacted and in excess. Then (NaOH) = mols/L = 0.00042/0.0202 = 0.02079 M = 1.68 pOH and 12.32 pH.
pH= pka + log (x/y)
pH= 4.75 + log (.06039/.020792)
pH= 5.213
but its not the right answer. What am I doing wrong?
The HAc is never in excess. The HCl is never in excess.
moles NaOH = 0.00122
moles acetic acid = 0.0008
moles HCl = 0.0008.
Did you see the link that DOES provide a buffer?
Soo...what am I doing wrong? Sorry I'm a bit confused here.
pH= 4.75 + log (.06040/.02079) ?
Then I get 5.203 which isn't the right answer either.
The basic problem here is that you are trying to use the Henderson-Hasselbalch buffer equation to solve a problem that is not a buffer. Your solution is not a buffered solution problem. There is an excess of NaOH when reacted with HCl and the pH is determined by the excess OH for pH = 12.32. The reaction between NaOH and HAc (I use HAc because it's easier to type than HC2H3O2) again is between more NaOH than HAc; therefore, all of the HAc is used and there is an excess of NaOH. Again, this provides a pH of about 12 something. It is NOT a buffer. There is acetate ion but there is no acetic acid. All of the acetic acid was neutralized by the more than enough NaOH. See the link I gave above for a problem that has some NaOH but enough HAc to provide an excess of HAc. Under those conditions there is a buffer present and the HH equation applies. The HH equation has a base and an acid (in the Bronsted-Lowry sense). In the HAc/Ac^- system, the Ac^- is the base and the HAc is the acid. You have Ac^- and the concn can be calculated but you do NOT have any acid present (no HAc). Instead, you have NaOH and that is what is determining the pH of the solution.
To calculate the pH of the solution when the 8.00 mL of 0.10 M HC2H3O2(aq) is diluted with 100 mL of water, you can use the equation pH = pKa + log(base/acid).
First, let's calculate the moles of HC2H3O2 in the 8.00 mL solution:
moles = volume (L) * concentration (M) = 0.00800 L * 0.10 M = 0.00080 moles HC2H3O2
Now, let's calculate the concentration of the diluted solution:
final volume = initial volume (dilute solution) + volume of water = 0.00800 L + 0.100 L = 0.108 L
concentration = moles/volume = 0.00080 moles / 0.108 L = 0.00741 M HC2H3O2
Now, let's determine the ratio of the base (NaOH) to the acid (HC2H3O2):
moles(base) = moles(acid) = 0.000420 moles NaOH
moles(acid) = 0.000800 moles HC2H3O2
Now, let's calculate the pH:
pH = pKa + log(base/acid)
pKa for HC2H3O2 is approximately 4.75.
For NaOH:
pH = pKa + log(0.000420 moles NaOH / 0.000800 moles HC2H3O2)
pH = 4.75 + log(0.525)
pH = 4.75 + (-0.279)
pH = 4.471
So, the pH of the solution with 12.20 mL of NaOH and 8.00 mL of HC2H3O2 (diluted with water) is approximately 4.471.
To answer your question, the pH of NaOH and HC2H3O2 will be different because they are different substances and have different pKa values. In the first part, the pH was the same because you calculated the pH of the NaOH solution only. But in the second part, you are calculating the pH of the HC2H3O2 solution, which is a weak acid with a different pKa value.