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Topic: Acid-Base problem  (Read 7328 times)

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Offline govibe

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Acid-Base problem
« on: January 04, 2008, 06:20:38 PM »
"How many moles of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at each pH?

    (a) pH = pKa

    (b) pH = 3.79

    (c) pH = 5.16
"

I know that NaOH is a strong base and will completely dissociate into Na+ and OH-. HC2H3O2 is a weak base. The ions in the solution are HC2H3O2, Na+, and OH-. However, I'm not sure what the chemical equation that will occur in the problem...

Offline Kryolith

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Re: Acid-Base problem
« Reply #1 on: January 04, 2008, 06:33:43 PM »
It's hard to identify what "HC2H3O2" means. It's better to write "CH3COOH". A buffer is a solution of a weak acid and its salt. Example:

You have 0.5 mol NaOH and 1 mol CH3COOH in 1 litre. What you get is 0.5 mol/l CH3COONa (sodium acetate) and 0.5 mol/l (unreacted) CH3COOH.

To calculate the pH of buffer solutions look at

http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation

Offline govibe

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Re: Acid-Base problem
« Reply #2 on: January 04, 2008, 08:36:42 PM »
so the base dissocation is

CH3COOH + H2O -> OH- + CH3CO-

and you do an ICE (IRE) box? (since you know concentration of OH-)?

Offline Borek

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Re: Acid-Base problem
« Reply #3 on: January 05, 2008, 04:18:08 AM »
so the base dissocation is

CH3COOH + H2O -> OH- + CH3CO-

That's completely off. CH3COOH is on a base, it is an acid.
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Offline Kryolith

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Re: Acid-Base problem
« Reply #4 on: January 05, 2008, 04:56:31 AM »
An acid is a proton donator, a base a proton acceptor. If CH3COOH (acetic acid) donates a proton to the an acceptor (=base) eg. water you have:

CH3COOH + H2O ------> CH3COO- + H3O+

Every acid has a conjugated base and every base a conugated acid. So there are always acid/base pairs. In the reaction above:
CH3COOH/CH3COO- (acid/base)
H3O+/H2O (acid/base)

For an acetate buffer, you have acetic acid and it's corresponding base acetate (which can be formed with the reaction CH3COOH + OH- --->...). To calculate its pH you have to find out the concentration of both, then insert them in the Henderson-Hasselbalch equation.

Try to work it out.

Offline govibe

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Re: Acid-Base problem
« Reply #5 on: January 05, 2008, 08:58:58 PM »
how does the NaOH come into play here... It breaks up into NA+ and OH- ions, neither of which are in the equation

CH3COOH + H2O ------> CH3COO- + H3O+

Offline Borek

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Re: Acid-Base problem
« Reply #6 on: January 06, 2008, 05:00:27 AM »
What happens when you mix acid (acetic) with a base (sodium hydroxide)?
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Offline govibe

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Re: Acid-Base problem
« Reply #7 on: January 06, 2008, 06:15:36 PM »
^it will form water, but what's the equation that will form involving all of these species?

Offline Kryolith

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Re: Acid-Base problem
« Reply #8 on: January 06, 2008, 06:21:27 PM »
general:
acid + base ---> water + ??

acetic acid + sodium hydroxide ---> water + ??

What does ?? stand for?


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