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Chemistry 67 Online
OpenStudy (anonymous):

Assume that your initial aqueous solution contains only 0.100 mol of acetic acid and has a volume of 100 mL. To this acetic acid solution you will add a 1.0 mol/L NaOh solution. Let's see what happens after adding different amounts of NaOH. Suppose you have added a total of 23.0 mL of NaOH solution since the beginning of the titration. How many mol of acetate ions are present in the reaction mixture at this point?

OpenStudy (anonymous):

CH3COOH (aq) + NaOH(aq) --> CH3COONa (aq) + H2O(l)

OpenStudy (abb0t):

I think you can use what is known as an "ice" table?

OpenStudy (abb0t):

Are you familiar with that?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I got CH3COOH= 1.00mol/L and NaOH=1.0mol/L But i'm sure what to do with the volumes...

OpenStudy (abb0t):

ok giv me a moment to write it out 4 u

OpenStudy (anonymous):

okay, thank you

OpenStudy (abb0t):

Ok. I won't give you the answer, but I will show you how to start it. Hopefully this will help guide you:

OpenStudy (abb0t):

Does that sort of make sense to you? I did it fairly quick, but I think i covered most of the work

OpenStudy (abb0t):

The change in the product is same as the change in the reactant, because this is a 1:1 ratio, does that make sense?

OpenStudy (abb0t):

You gain, that amount of moles.

OpenStudy (abb0t):

i realize you aren't looking for concentration, so ignore the find V\(\sf _T\) in the end.

OpenStudy (anonymous):

Is the answer 0.023?

OpenStudy (abb0t):

Yes.

OpenStudy (anonymous):

Thanks @abb0t :)

OpenStudy (anonymous):

One quick question: Suppose that you have added a total of 100 mL of the 1.0 mol/L NaOH solution since the beginning of the titration. What is the concentration, in mol/L, of acetate ions in the reaction mixture at this point? Would it be 0.023/.1L = 0.23mol/L @abb0t

OpenStudy (anonymous):

@abb0t

OpenStudy (abb0t):

No, because you added 23 mL of base to the initial solution, remember?

OpenStudy (abb0t):

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