BR Chapter 5 number 5. acid titration

[UIUCstudent]

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What is the acetate anion concentration in a solution made by mixing 20 mL .30M HOAc with 10 mL .30 M NaOH?

B. .10M H3CCO2-
C. .15M H3CCO2-

So I thought it would be C because you have a half equivalent of a strong base reacting with a weak base. Therefore half of the solution should be conjugate base and the other half would be conjugate acid. The correct answer is B, however. I'm not sure why my thought process was wrong.

 

[jowen88]

look at the BR answer explanation as its a little more clear, but I try to think of these things in term of mmol which is uncomfortable for some.

so you have .3M of 20ml HOAc
.3M of 10ml of NaOH
BUT its kind of like an equilibrium table problem, in that you can't just consider reactivity. I.E. (.3M)(20ml)=6 mmol of HOAc reacts with (.3)(10ml)=3 mmol NaOH. The NaOH reacts to completion being a strong a base with the HOAc, thus it converts 3mmol of that 6mmol HOAc to OAc-. However, you've also diluted your solution by 50% (From an initial 30ml.)

3mmol/30ml(total, new volume)=0.1M

Also, UIUCstudent, I'm at UIUC too, if you're interested, I was going to start an MCAT student organization for kids taking the MCAT to pool resources and provide info to the undergrads who don't know about student doc etc. PM if you're interested.

 

[UIUCstudent]

look at the BR answer explanation as its a little more clear, but I try to think of these things in term of mmol which is uncomfortable for some.

so you have .3M of 20ml HOAc
.3M of 10ml of NaOH
BUT its kind of like an equilibrium table problem, in that you can't just consider reactivity. I.E. (.3M)(20ml)=6 mmol of HOAc reacts with (.3)(10ml)=3 mmol NaOH. The NaOH reacts to completion being a strong a base with the HOAc, thus it converts 3mmol of that 6mmol HOAc to OAc-. However, you've also diluted your solution by 50% (From an initial 30ml.)

3mmol/30ml(total, new volume)=0.1M

Also, UIUCstudent, I'm at UIUC too, if you're interested, I was going to start an MCAT student organization for kids taking the MCAT to pool resources and provide info to the undergrads who don't know about student doc etc. PM if you're interested.

Hmm I understood the math concept. But the titration of weak acid still bothers me. In this case where there is half an equivalent of a strong base with a weak base, wouldn't this result in a buffer solution? And according to HH equation the mol of base equals the mol of acid for pH=pKa in a buffer solution.
Yet these calculations gives .1M of conjugate base and .20M of the acid.

And yeah I would definitely be interested in that RSO. Sounds awesome.

 

[jowen88]

again, to put it in mmol terms
starting:
-3mmol of NaOh (.3M x 10ml)
-6mmol of HOAc (.3 x 20ml

ending
-6mmol - 3mmol (from rxn of 3mmol of the HOAc and 3mmol of NaOH)
=3 mmol of OAc in 30 ml (total) of solution
thus, [HOAc] is also 3mmol in 30ml of solution, so it is also @.1M concentration, both are drawn down to .1M by the dilution

Remember, that initial .3M of HOAc had its concentration halved since half of it is OAc-, but it also had its concentration 2/3'ered by the addition of 10ml of NaOH's volume. I.e. [.3]/2 (conversion to [A-], then (20/30ml)[.15M] for dilution effects leads to .10 all around

and can you shoot me your email addy, mines [email protected], we can meet up sometimes this week and ill finish explaining it if you still seems weird hah.