Question

Titration problem?

The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion:
CH3COOH(aq)+NaOH(aq)→H2O(l)+NaC2H3O2(aq)

If 3.45 mL of vinegar needs 43.0 mL of 0.150 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 2.00 qt sample of this vinegar?

Answer 1

WE gots a concentration of under `2*mol*L^-1`....

Explanation:

Well, we use the equation.....

`NaOH(aq) + H_3C-CO_2H(aq) rarr H_3C-CO_2^(-)Na^(+) + H_2O(l)`

And thus there is 1:1 equivalence.

`"Moles of NaOH"=43.0xx10^-3*Lxx0.150*mol*L^-1=6.45xx10^-3*mol`

And thus `"concentration of vinegar"=(6.45xx10^-3*mol)/(3.45xx10^-3*L)` `=1.87*mol^-1`.

I will let you do the second part of this question as I have no idea what you mean by a `"quart"`.