titration is killing me!!!

[inaccensa]

Ok guys, I hope someone can explain this to me. I asked my professor and he was just so mad at me.lol I hope someone here will be nice enuf to clear my misunderstanding.

For acetic acid 10mmol being titrated with NaOH10mmol.
Its a weak acid, lets say the pH is 2. When I initially begin adding NaOH, the weak acid will act as a buffer and give up its protons. These protons will inturn start reacting with the hydroxide ions from the base. At the point, we say that the concentration of acid = concentration of its conjugate base. So in the buffering region we are forming water? Here the pH= pKa. If I had 10mmol of acid, the buffering will last until I add 10mmol? ( Here is the confusion, weak acids don't dissociate completely, but they will once we start adding base, so by the time we leave the buffer solution, won't we have used up all the acid?) When we approach the steep region of the curve, it is said that the concetration of acid = base (this is my confusion..I know I'm not thinking something right here)

Can someone please explain!!

P.S. this is a graph of acetic acid/NaOH
http://biology.hunter.cuny.edu/Goldfarb/Biol710-Fall-2007/Lect1.AAs.Peptides.pH.pK.htm

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[loveoforganic]

You're confused about the buffering region. Buffering does not occur immediately upon addition of your titrant. It occurs when you have 10x as much acid until 10x as much base as acid. Outside of that range, the solution is generally not considered a buffer. This is why you have a steep lip at the beginning of the titration of a weak by a strong.

 

[MadTown ChzHead]

Break down the regions of your titration curve...

1. Before adding any base, pH of the solution is -log[H30+], which depends on the ionizability of acetic acid (a weak acid, so you'll have a mildly acidic solution).

2. Begin adding base. The OH- begins reacting with the available H+ in solution. Eventually you enter your buffering region. Use Henderson/Hasselbalch eqn and mmol of base added, and you can monitor your pH.

3. Continue adding base. You begin to near the midpoint of the buffering region, where pH=pKa. This occurs, according to H/H eqn, when log([conjBASE]/[acid])= 0. When concentration conjugate base = acid, buffer has reached it's maximum buffering capacity. As you add more and more base, you leave the buffering region and soon enter the steep part of the curve, nearing the equivalence point.

4. At the equivalence point, moles of acid and base are equal. pH at equivalence point is determined solely by the hydrolysis of the conjugate base (use Kw=KaKb).

5. Past the equivalence point, you have excess base, so you need to calculate how much extra OH- is floating around to figure out pH.

Does this help at all?

 

[inaccensa]

Before adding any base, pH of the solution is -log[H30+], which depends on the ionizability of acetic acid (a weak acid, so you'll have a mildly acidic solution).

2. Begin adding base. The OH- begins reacting with the available H+ in solution ( so we are forming water and as you add more base, more water will form)Eventually you enter your buffering region. Use Henderson/Hasselbalch eqn and mmol of base added, and you can monitor your pH ( Acid is in equilibrium with its conjugate base, now lets say you have 1mmol of acetic acid,is it when you add 1mmol of NaOH that means you leave the buffering region, I'm guessing not, since at equivalence pt, the mmol base= mmol acid. I'm just confused about the concentration of base in this region, it seems like it will be 0.5mmol)

3. Continue adding base. You begin to near the midpoint of the buffering region, where pH=pKa. This occurs, according to H/H eqn, when log(conjBASE/acid)= 0. When conjugate base = acid, buffer has reached it's maximum buffering capacity. As you add more and more base, you leave the buffering region and soon enter the steep part of the curve, nearing the equivalence point.

4. At the equivalence point, moles of acid and base are equal. pH at equivalence point is determined solely by the hydrolysis of the conjugate base (use Kw=KaKb).

5. Past the equivalence point, you have excess base, so you need to calculate how much extra OH- is floating around to figure out pH.

Does this help at all?[/QUOTE]

 

[zmatrix]

Ok guys, I hope someone can explain this to me. I asked my professor and he was just so mad at me.lol I hope someone here will be nice enuf to clear my misunderstanding.

For acetic acid 10mmol being titrated with NaOH10mmol.
Its a weak acid, lets say the pH is 2. When I initially begin adding NaOH, the weak acid will act as a buffer and give up its protons. These protons will inturn start reacting with the hydroxide ions from the base. At the point, we say that the concentration of acid = concentration of its conjugate base. So in the buffering region we are forming water? Here the pH= pKa. If I had 10mmol of acid, the buffering will last until I add 10mmol? ( Here is the confusion, weak acids don't dissociate completely, but they will once we start adding base, so by the time we leave the buffer solution, won't we have used up all the acid?) When we approach the steep region of the curve, it is said that the concetration of acid = base (this is my confusion..I know I'm not thinking something right here)

Can someone please explain!!

P.S. this is a graph of acetic acid/NaOH
http://biology.hunter.cuny.edu/Goldfarb/Biol710-Fall-2007/Lect1.AAs.Peptides.pH.pK.htm

First... the corrections:

1. Acetic acid = WEAK acid, pKa = 4.76
2. The titration curve found on slide 10 of the linked powerpoint is a titration of HCl (STRONG acid, pKa = -8) with NaOH

-Based on pH=2 alone, a light should go off in your head signaling that this cannot be a weak acid.

Buffering regions are regions where a large addition of titrant results in a small change in pH. In titration graphs (x axis = volume of titrant added, y axis = pH), these would be areas with a flat slope, (close to horizontal)