I conduted the following experiment. How do you determine the concentration of v
ID: 916859 • Letter: I
Question
I conduted the following experiment. How do you determine the concentration of vinegar?
Data Table 1. NaOH Titration Volume.
Initial NaOH Volume (mL)
Final NaOH Volume (mL)
Total volume of NaOH used (mL)
Trial 1
9
4
5
Trial 2
5.2
1.2
4.2
Trial 3
6.4
1
5.4
Average Volume of NaOH Used (mL) : 4.867 mL
Data Table 2. Concentration of CH3COOH in Vinegar.
Average volume of NaOH used (mL)
Concentration CH3COOH in vinegar (mol/L)
% CH3COOH in vinegar
4.867
Show your calculations here:
Questions
The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer’s statement?
Initial NaOH Volume (mL)
Final NaOH Volume (mL)
Total volume of NaOH used (mL)
Trial 1
9
4
5
Trial 2
5.2
1.2
4.2
Trial 3
6.4
1
5.4
Average Volume of NaOH Used (mL) : 4.867 mL
Explanation / Answer
Given
Average volume of NaOH = 4.867 mL
[NaOH]= 0.5 M
Volume of acetic acid = 5.0 mL
Lets write reaction between NaOH and CH3COOH
CH3COOH + NaOH -- > CH3COONa + H2O
Mol ratio is 1:1
So we can simply use M1V1 = M2V2
Let M1= Molarity of acid
V1=volume of acid
M2 = molarity of base
V1 = volume of base
M1 = M2V2/V1
= 0.5 M x 4.867 mL / 5.0 mL
= 0.487 M
[CH3COOH]= 0.487 M
Percent error=[ (Accepted value – experimental value )/ accepted value ] x 100
Here lets find Molarity of 5 % acetic acid
Lets assume volume of solution = 1000 mL = 1.000 L
Percent of acetic acid is 5 so we can find mass of it
Mass of acetic acid in 1000 mL
= 1000 mL x 5 g / 100 mL
= 50 g
Moles of acetic acid = 50.0 g / molar mass of acetic acid
= 5.0 g x 1 mol / 60.05 g
=0.832 mol
Molarity =0.832 mol / 1 L = 0.832 M
Percent error = ( 0.832 M – 0.487 M ) / 0.832 M ] x 100
= 41.5 %
Percent error = 41.5 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.