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?
CH3COOH (aq) + NaOH(aq) --> CH3COONa (aq) + H2O(l)
I think you can use what is known as an "ice" table?
Are you familiar with that?
yes
I got CH3COOH= 1.00mol/L and NaOH=1.0mol/L But i'm sure what to do with the volumes...
ok giv me a moment to write it out 4 u
okay, thank you
Ok. I won't give you the answer, but I will show you how to start it. Hopefully this will help guide you:
Does that sort of make sense to you? I did it fairly quick, but I think i covered most of the work
The change in the product is same as the change in the reactant, because this is a 1:1 ratio, does that make sense?
You gain, that amount of moles.
i realize you aren't looking for concentration, so ignore the find V\(\sf _T\) in the end.
Is the answer 0.023?
Yes.
Thanks @abb0t :)
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
@abb0t
No, because you added 23 mL of base to the initial solution, remember?
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