[nozo]

I have no idea how to solve the ff.. we havent gotten to acid/base lecture yet and our lab tomorrow is about acid/bases ><; so at this point I am purely guessing...

An unknown acid was dissolved into 55 ml of distilled water. Half of this soln was titrated w/ NaOH. The pH of the other half of this soln was measured with a PH meter. The neutralized soln is then added to the orifinal soln and the pH of this combined final soln was measured.

wt. of unknown acid = 1.1342 g
vol of NaOH used = 14.31 mL
conc of NaOH = .2223 M
pH of the orig soln = 2.28
pH of the final acid soln = 3.41

I calculated the ff:

Molecular weight (MW) of the acid used in titration:
n = CV = (.2223) x (.01431) = .003181
MW = 1.1342 g / .003181 = 356.5

Molarity of the unknown acid soln from titration
.005248 (derived from pH of the orig soln)

Ka of unknown acid:
okay, there is a formula on the lab manual, but it doesn't really explain much...

Ka = [H+][A-] / [HA]_eq

I'm assuming that the [H+] would also be .005248 M (from the orig pH soln) and would [A^-] be the salt? How do I solve for the concentration of the salt? It says on the book that [H⁺] = [A^-] if pure acid is dissolved in water... correct? So the concentrations would be the same? And what is [HA]? How do I solve for [HA]?

And finally I need the concentration of undissociated acid from pH measurements & the total concentration of unknown acid from pH measurements. Where do I start? Do I simply get the anti log of the pH of the final acid soln for its concentration?

Please help... my brain is fried 

[Albert]

Given that I'm far from being sure of what the problem is really asking you to calculate, here are my opinions. Check them out:

Molecular weight (MW) of the acid used in titration:
n = CV = (.2223) x (.01431) = .003181
MW = 1.1342 g / .003181 = 356.5

If the initial solution has been divided into two 27.5 mL solutions, I would multiply the moles by two and, then, calculate the MW (1.1342 is the mass of acid in the 55mL solution, isn't it?).

MW is nearby 178.

Molarity of the unknown acid soln from titration
.005248 (derived from pH of the orig soln)

3.18 mmol in 27.5 mL or 6.36 mmol in 55 mL, hence, 0.1157M.

Ka of unknown acid:
okay, there is a formula on the lab manual, but it doesn't really explain much...

Ka = [H+][A-] / [HA]eq

I'm assuming that the [H+] would also be .005248 M (from the orig pH soln) and would [A-] be the salt? How do I solve for the concentration of the salt? It says on the book that [H+] = [A-] if pure acid is dissolved in water... correct? So the concentrations would be the same? And what is [HA]? How do I solve for [HA]?

Ok, here I'm at a loss because I can't understand the question. You have your initial pH (2.28) in the 55mL solution. Am I wrong?
Well, if that's how things are, your concentration of protons is 5.248 mM, while, in the INITIAL solution, the molarity of the acid is 0.1157, right?
Hence, x=Ka=(5.248*10^-3)²/(0.1158 - 5.248*10^-3)= 2.49*10^-4

And finally I need the concentration of undissociated acid from pH measurements & the total concentration of unknown acid from pH measurements.

Here, I apologize, I am really lost on this one. I really don't understand what you are supposed to calculate and, above all, in which solution.
I wish Borek was here.

 

[nozo]

Thanks for the help Albert.. 
Hopefully I can clarify my other questions...

Ok, here I'm at a loss because I can't understand the question. You have your initial pH (2.28) in the 55mL solution. Am I wrong?
Well, if that's how things are, your concentration of protons is 5.248 mM, while, in the INITIAL solution, the molarity of the acid is 0.1157, right?
Hence, x=Ka=(5.248*10^-3)²/(0.1158 - 5.248*10^-3)= 2.49*10^-4

Actually my initial pH 2.28 solution is in the 27.5 ml solution. So to make the corrections, would I simply need to divide 5.248 by 2, and plug in to your equation above?

And finally I need the concentration of undissociated acid from pH measurements & the total concentration of unknown acid from pH measurements.

Here, I apologize, I am really lost on this one. I really don't understand what you are supposed to calculate and, above all, in which solution.
I wish Borek was here.

 

I don't know if I'm explaining this right... but basically the first half of the solution (27.5 ml) was "neutralized" by titration and measured by the pH meter giving a pH of 2.28

Then, the second half of the solution (27.5 ml) was added into the "neutralized" solution (27.5 ml) and measured by the pH meter giving a pH of 3.41

I hope that cleared things up? Or maybe not... 

[Albert]

Actually my initial pH 2.28 solution is in the 27.5 ml solution. So to make the corrections, would I simply need to divide 5.248 by 2, and plug in to your equation above?

No. If I understand, now, BEFORE adding the base, you have a 0.1157M solution (27.5mL) whose pH is 2.28. Right?

Now, using the equation

Ka=(5.248*10^-3)²/(0.1158 - 5.248*10^-3)= 2.49*10^-4

you will get the equilibrium constant for the reaction: HA + H₂O <=> H₃O⁺ + A^-

Ka = [H+][A-] / [HA]eq

I need the concentration of undissociated acid from pH measurements & the total concentration of unknown acid from pH measurements.

What's the difference? I can't see it! That's what puzzles me.             

Can you help me understanding it?

[nozo]

Okay, I spoke with my TA, but he didn't really explain much since he was kind of busy.. but he gave me the answers. I'll try to understand it once I'm done with lab today :p

Anywhoo, for Ka, apparently the answer was the molarity of the pH of the final acid solution. So given the pH 3.41, I simply got the inverse log for that.

Then the eqn you answered

Ka=(5.248*10^-3)²/(0.1158 - 5.248*10^-3)= 2.49*10^-4

was actually the answer for the "concentration of undissociated acid from pH measurements," with slight modifictions...

(5.248*10^-3)²/(3.890*10^-4)= answer

And finally for "total concentration of unknown acid from pH measurements" just add up the molarities for the answer I got above and the molarity for the pH of the original acid solution (inverse log of pH 2.28)