Titration Essay example:
Introduction
A titration is a laboratory procedure involving the carefully measured and controlled addition of a solution from a buret into a measured volume of a sample solution. By definition, volumetric analysis is quantitative analysis using accurately measured titrated volumes of standard chemical solutions. Titration is an example of volumetric analysis because it involves accurately measured volumes of a sample solution for analysis. The type of titration carried out is acid base titration using HCl and Na2CO3. In this titration, three major steps involved were first measure out an amount of known solution, add indicator and three, to titrate using unknown solution. At the first colour change the titration would stop.
Abstract
The purpose to this lab is to standardize an approximately 1 mol•L-1 solution of hydrochloric acid. There were many steps taken in this titration but the main steps were measuring out the amount of known solution, adding the indicator and finally, titrating the solution until the first colour change. In each trial, the initial reading, final reading and the volume of HCl used was recorded down as quantitative results. The average volume of hydrochloric acid was found to be 12.03mL. The amount of sodium carbonate in the 10.00ml of solution was found to b 0.05 mol. The amount of hydrogen chloride that was dissolved in the average volume of acid is 0.365g. Through these calculations, the concentration of hydrochloric acid was found to be 8.3 mol•L-1.
Purpose
Refer to handout “Standardize an Approximately 1 mol•L-1 Hydrochloric Acid”
Material and Method
Refer to handout “Standardize an Approximately 1 mol•L-1 Hydrochloric Acid”
Results
Quantitative Results
Trial Initial Final Reading (in mL) Volume HCl Used (in mL)
1 0.2 12.4 12.2
2 12.4 23.9 11.5
3 23.9 36.3 12.4
Calculations
Average volume of HCl
12.2mL + 11.5mL + 12.4mL = 12.03mL
3
Balanced chemical equation
2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2CO2(l)
2NaCl(aq) + H20(l) + CO2(g)
Amount of sodium carbonate in 10.0mL of solution
1) Find number of moles of in 10.00mL @ 0.500mol •L-1
nNa2CO3 = 10.00mL x 1L x 0.500mol
1000ml 1L
= 0.005mol
2) Find mass of Na2CO3 using number of moles of Na2CO3 and molar mass
MNa2CO3= 105.99 g/mol
mNa2CO3= 0.005 mol x 105.99g
mol
=0.530g
Hydrogen chloride dissolved in the average volume of acid used:
1) Find out how many moles of HCl reacted using molar ratio and the number of mole of
Na2CO3
nHCl = 0.005 mol Na2CO3 x 2 HCL
1 Na2CO3
= 0.01mol
2) Find mass of HCl using number of moles of HCl and molar mass
MHCl = 36.46 g/mol
mHCl = 0.01mol x 36.46g
mol
= 0.365g
Concentration of HCL
C= nd
V
[HC] = 0.01mol
0.01203L
= 8.3 mol•L-1
Conclusion
The purpose of the lab was to standardize an approximately 1 mol•L-1 solution of hydrochloric acid. To standardize the solution of hydrochloric acid the actual concentration of the solution was needed. In conclusion, to standardize the solution of hydrochloric acid the calculations of the concentration are needed to be done first.
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