[Byrne]

Okay, so I have an initial 0.0010 mol/L solution of ethanoic acid (HC2H3O2).  Finding the pH of this solution is simple.  Using the Ka value for ethanoic acid and an ice chart, I determined the pH of this solution to be 3.87, which agrees with the answer provided with the problem.

However, in the next part of the problem, things get a little tricky.

7.2 g of NaC2H3O2 are added to 1.5 L of the solution described above.  Find the pH.

Okay, so what I did was first determine the concentration of the acetate ion, which I determined to be 0.0585 mol/L.

Now I'm not exactly what to do.  What I have done is write down another ICE chart in which the initial concentration for C2H3O2 is 0.0585 mol/L.  This eventually leads to the use of the quadratic formula.  Keeping significant digits in mind when soling for the unknown concentration of hydrogen ions (x), you end up with a concentration of 0, which obviously is not true.  However, if I ignore significant figures and perform all the calculations in my calculator, I end up with a rather small concentration of hydrogen ions, which, when subbed into the pH formula, provides me with a pH that equals what the answer should be.  Is there another way to do this problem?  Are you allowed to ignore significant figures in this case?

[Borek]

You must be doing some error with significant digits on your way to the final result. Show your work.

Also note that SD are to be used in the final answer, all intermediary calculations are always done with full available precision.

[Byrne]

You must be doing some error with significant digits on your way to the final result. Show your work.

Also note that SD are to be used in the final answer, all intermediary calculations are always done with full available precision.

Yeah, that was my question.  The reason I can't really show my work is because it involves the use of the quadratic formula.  So, the junk underneath the root symbol, if rounded to three significant digits, would result in a number that, when added to the b value outside the quadratic formula, equals zero.  However, if I ignore significant figures and allow the junk underneath the root symbol to go to around 9 decimal places, then I end up with a small concentration of hydrogen ions, and when I find the pH, I end up with the correct answer.

My teacher always told me that you cary ONE extra significant digit during intermediate calculations, and I were to do that in this case, I would end up a hydrogen ion concentration of zero, which obviously is not the case.  So it is okay to ignore sig figs until your final answer of that particular step?

[Borek]

So it is okay to ignore sig figs until your final answer of that particular step?

Yes. As I told before - significant digits doesn't apply to intermediary calculations, only to the final result.

[Donaldson Tan]

Ka of Acetic Acid is 1.7E-5

from my calculation, the pH of 0.001M acetic acid solution is 3.91

after adding 7.2g of sodium acetate into 1.5L of the 0.001M acetic acid solution, the final pH is 4.76

Bryne: How does my results compare to your answers?

[Byrne]

Ka of Acetic Acid is 1.7E-5

from my calculation, the pH of 0.001M acetic acid solution is 3.91

after adding 7.2g of sodium acetate into 1.5L of the 0.001M acetic acid solution, the final pH is 4.76

Bryne: How does my results compare to your answers?

The Ka value for acetic acid in my text is 1.8E-5, so I ended up with a pH of 3.87 for the solution.

I ended up with a pH of 6.51 for the solution after 7.2 g of sodium acetate are added.  First, I determined the concentration of sodium acetate, which I then made the initial concentration of the acetate ion for my ICE chart involving the dissociation of acetic acid.  I went through the same process as I did in the first part of the problem, ending up using the quadtratic formula in which my a value was just 1, b value was 0.0585 (0.0585 was my initial concentration of the acetate ion). c value was 1.8E-8 (I used approximation since the Ka value for acetic acid < 10E-4 and this approximation worked in the first part of this problem fine).

[Borek]

If you add acetate to acetic acid you can usually assume there were no dissociation going on, so there no need for ICE table. Enter initial acetic acid and acetate concentrations into Henderson-Hasselbalch equation and you are ready. Nothing more fancy required.

0.0585M and 0.001M are numbers you have to use.

[Donaldson Tan]

OK. I used your Ka data to evaluate.

The pH of 0.001M Acetic Acid is 3.90.
[H+] = [Acetate] = 1.25E-4
[Acetic Acid] = 8.75E-4

Now to solve the second part
Since Acetic acid is monoprotic, we express it as HA: HA <-> H⁺ + A^-

According to Le Chatelier's Principle, add acetate ions into the solution will favour the backward reaction. Let amount of acetate per unit volume converted to acetic acid call x.

initial change in acetate concentration when 7.2g added to 1.5Lsol
= [ 7.2/(molar mass) ] / 1,5L =  0.0585M

	HA		H+	A-
Initial	8.75E-4		1.25E-4	1.25E-4
Change	+x		-x	+0.0585-x
Equilibrium	(8.75E-4 + x)		(1.25E-4 - x)	(1.25E-4 + 0.0585 - x)

upon algebraic manipulation, x = 1.25E-5

[H+] = 3.08E-7
pH = 6.51

ps: i use 10 significant figures to evaluate the pH.

[Borek]

Using H-H:

pH = pKa + log([A-]/[HA])

pH = 4.74 + log(0.0585/0.001) = 6.51

Using ICE table is a waste of time - and source of errors

[Byrne]

Using H-H:

pH = pKa + log([A-]/[HA])

pH = 4.74 + log(0.0585/0.001) = 6.51

Using ICE table is a waste of time - and source of errors

I never realized you could use this equation to solve these types of problems.  Don't know if my teacher would want me doing it this way since he never taught it...

[Borek]

IMHO that's the only correct way of approaching this type of question. You are told that the solution was prepared with acid and conjugated base. If so it's a buffer solution, and if it is a buffer solution, you should use H-H equation.

H-H equation has its limitations, but they are not present here.

[Byrne]

IMHO that's the only correct way of approaching this type of question. You are told that the solution was prepared with acid and conjugated base. If so it's a buffer solution, and if it is a buffer solution, you should use H-H equation.

H-H equation has its limitations, but they are not present here.

Well, we did learn about buffers and the H-H equation, but I didn't realize this was a buffer problem.  Usually, problems that I've come across that involve the use of the H-H equation begin with "Construct a buffer..."

[Borek]

I didn't realize this was a buffer problem

So you realize now

[AWK]

At this concentration of acetic acid its dissociation level is about 25 % and pH about 3.6. I agree with pH of  the buffer solution.

[Borek]

At this concentration of acetic acid its dissociation level is about 25 % and pH about 3.6.

Sorry - but not. 3.91 and 12.4% (for pKa=4.757).