Question

At what volume of added NaOH solution would you expect to see an end point Answer in units of mL (full question is in photo :) )

Answer 1

endpointVinegar is also called as acetic acid, and we can write it in the following way CH3COOH. When we talk about titration, it refers to the amount of base needed to neutralize the acid. So there is a chemical reaction involved. The reaction between vinegar and NaOH base is as follows:

$CH_3CO_{}OH+Na_{}OH\rightarrow CH_3CO_{}O_{}Na+H_2O$

We have a 1 to 1 ratio, i.e. to neutralize x moles of vinegar we need the same moles of NaOH. Let's calculate how many moles of acid are believed to be in the solution. We will use the molarity = 0.78M = 0.78 mol/L.

$\begin{gathered} \text{Moles of vinegar = Volume}\times Molarity \\ \text{Moles of vinegar = }0.02061L\times0.78\frac{mol}{L} \\ \text{Moles of vinegar = }0.016\text{mol} \end{gathered}$

Then we need 0.016 moles of NaOH, the volume required will be:

$\begin{gathered} \text{Volume of NAOH=0.016 mol}\times\frac{1L}{0.933\text{mol}}\times\frac{1000mL}{1L} \\ \text{Volume of NAOH=}17.23mL \end{gathered}$

So, the volume that would be needed of NaOH to see an end point is 17.23 mL