Respuesta :
The question refers to a titration of vinegar with standardized NaOH and the following data are provided:
Molarity of NaOH = 0.3849 mol/L
Volume of NaOH used = 18.70 mL
Initial volume of buret = 0.30 mL
Final volume of buret = 19.00 mL
Volume of vinegar = 10.00 mL
First, we need to understand the experiment and data provided. The experiment consists in a buret filled with standardized NaOH that will be added to vinegar. As we add NaOH to vinegar, the H+ ions from vinegar (an acid) are neutralized by the OH- from NaOH (a base), up to the point where there aren't any more H+ ions in solution. When the color of the solution turns pink, it means that all H+ ions were neutralized and the titration is done.
From the initial and final volume of the buret, we know how much, in mL, of NaOH was used to neutralize the acid in solution (18.70 mL) and, together with the molarity of NaOH, we can calculate the number of moles of NaOH used.
[tex]n=\text{molarity}\times volume\text{ }\rightarrow n=0.3849\text{ mol/L }\times(18.70\times10^{-3})L\rightarrow n_{NaOH}=0.007198\text{ mol}[/tex]Now, from the reaction between NaOH and vinegar (acetic acid, chemical formula CH3COOH), we can calculate the amount of acetic acid that reacts with 0.007198 moles of NaOH:
[tex]CH_3COOH\mleft(aq\mright)+NaOH\mleft(aq\mright)\longrightarrow CH_3COONa\mleft(aq\mright)+H2O\mleft(l\mright)[/tex][tex]n_{NaOH}=n_{CH3COOH}\rightarrow n_{CH3OOH}=0.007198\text{ mol}[/tex]From the above calculations, we know that there were 0.007198 moles of CH3COOH in solution and, from the initial volume of CH3COOH (10 mL), we can calculate its molarity:
[tex]\text{molarity = }\frac{\nu mber\text{ of moles (mol)}}{\text{volume (L)}}\rightarrow molarity_{CH3COOH}=\frac{0.007198\text{ mol}}{10\times10^{-3}L}=0.7198\text{ mol/L}[/tex]Therefore, from the titration with NaOH, we have that the molarity of the vinegar solution (acetic acid) is 0.7198 mol/L.
