Titration

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Titration Lab

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 NaCO. 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.

AbstractBuy Titration term paper

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 1.0mL. 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 0.65g. Through these calculations, the concentration of hydrochloric acid was found to be 8. 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. 1.4 1.

1.4 . 11.5

. 6. 1.4

Calculations

Average volume of HCl

1.mL + 11.5mL + 1.4mL = 1.0mL

Balanced chemical equation

HCl(aq) + NaCO(aq)  NaCl(aq) + HCO(l)

 NaCl(aq) + H0(l) + CO(g)

Amount of sodium carbonate in 10.0mL of solution

1) Find number of moles of in 10.00mL @ 0.500mol •L-1

nNaCO = 10.00mL x 1L x 0.500mol

1000ml 1L

= 0.005mol

) Find mass of NaCO using number of moles of NaCO and molar mass

MNaCO= 105. g/mol

mNaCO= 0.005 mol x 105.g

mol

=0.50g

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

NaCO

nHCl = 0.005 mol NaCO x HCL

1 NaCO

= 0.01mol

) Find mass of HCl using number of moles of HCl and molar mass

MHCl = 6.46 g/mol

mHCl = 0.01mol x 6.46g

mol

= 0.65g

Concentration of HCL

C= nd

V

[HC] = 0.01mol

0.010L

= 8. 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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