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# Titration Question

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I am currently studying for my GenChem exam and I am having some difficulty so I thought I may return to gold old pmed to see if any keeners were willing to help.

I am having difficulty in calculating the pH of a solution during a titration of a weak acid (acetic, Ka = 1.8X10^-5) with a strong base (KOH).

The initial concentration of the acid (0mL KOH added) is 0.2M and 100mL. What is the pH of the solution after 50mL 0.1M KOH is added?

I keep getting like 12 or 13.

Since OH is in excess of H+ from the acetic acid, do you just calculate the concentration of the remaining OH and use Kw to calculate pH? That is how I am doing it, and I am getting the wrong answer.

If anyone can help, I would really appreciate it!

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Wow, thanks NewfieMike!

That is pretty much what I was doing, but the answer they are giving is 4.26. I think the answer in the back of the book must be wrong!

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yeah that doesn't make any sense. It's a strong base and a weak acid. you'd have to add a metric **** tonne to get the pH down that low

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You get 4.26 or something if you assume 0.2 in 50 mL is the final concentration of the weak acid from the Ph = pka + base/ acid formula...

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Isn't acetic acid acting as a buffer? Thus when the intial H+ gets consumes the equilibrium of the dissociation of acetic acid shifts and produces more H+? Then I could see how the pH could get that low.

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For one sec I thought that was an interview questions.........and was happy that I didn't get asked that one.

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Isn't acetic acid acting as a buffer? Thus when the intial H+ gets consumes the equilibrium of the dissociation of acetic acid shifts and produces more H+? Then I could see how the pH could get that low.

i think this is far beyond the buffer range for an acid like that.

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HA + OH- --> H2O + A-

HA (0.02 - x) mol

OH (0.005) mol

A- ( x + 0.005) mol

x is very negligible since HA is a weak acid

So:

pH = Pka + log (base/ acid)

pH = -log (1.5 x 10^-5) + log ( 0.005/ 0.02)

pH = 4.82 - 0.60

pH = 4.22

Hmmm....

It is true that the PI of any weak acid + strong base is basic and vice versa but in this case we are not adding enough base to pass the buffering zone (I may be wrong).

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HA + OH- --> H2O + A-

HA (0.02 - x) mol

OH (0.005) mol

A- ( x + 0.005) mol

x is very negligible since HA is a weak acid

So:

pH = Pka + log (base/ acid)

pH = -log (1.5 x 10^-5) + log ( 0.005/ 0.02)

pH = 4.82 - 0.60

pH = 4.22

Hmmm....

It is true that the PI of any weak acid + strong base is basic and vice versa but in this case we are not adding enough base to pass the buffering zone (I may be wrong).

Hey, thanks!

The answer the book is giving is 4.26.

Why, though, is the way that NewfieMike and I were doing it in the first place wrong? Or is it?

Damn....I don't know who to believe!

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You can't look at acid and base dissociation separately in these questions. Dissociation of weak acids/ bases is skewed based on free OH- or H+ ions in the solution. When you look at each separately, you ignore this additional OH- (from strong base) or H+ (from strong acid) effect on their dissociation.

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holy **** did you make that yourself?

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