[jacky0620]

Procedure:
a.   Measure accurately 10.0 cm3 of glacial ethanoic acid and 10.0 cm3 of propan-1-ol into a clean, dry 100 cm3 distillation flask. Mix thoroughly.
b.   Transfer 1.00 cm3 of the mixture by pipette to a 125-conical flask containing about 25 cm3 distilled water and 2 drops of phenolphthalein indicator solution. Titrate to end point with 0.50 M NaOH. Record this volume. Repeat the titration to obtain an average volume of  0.50 M NaOH used.

Calculation:
Conc. of propan-1-ol before mixing = 13.33 mol/dm3
Conc. of ethanoic acid before mixing = 17.50 mol/dm3
Conc. of propan-1-ol after mixing =13.33/2   = 6.665 (mol/dm3 )
Conc. of ethanoic acid after mixing =  17.5/2 = 8.75 (mol/dm3)


Titration is done, but why that result of titration did not used for calculating Conc. of ethanoic acid after mixing ? And why /2 ?

[enahs]

You had 10mL of one compound, and then you added another of 10mL of another.
Concentration (molarity) is moles per liter (Mol/L). If you double the volume of solution, you decrease the concentration by half.

[jacky0620]

You had 10mL of one compound, and then you added another of 10mL of another.
Concentration (molarity) is moles per liter (Mol/L). If you double the volume of solution, you decrease the concentration by half.

if not 10 & 10, is:
acid:14.3
propan-1-ol:18.8
CAN I use ratio?
But, why titration is done, but its result is not used?

[xiankai]

Conc. of propan-1-ol after mixing =13.33/2   = 6.665 (mol/dm3 )
Conc. of ethanoic acid after mixing =  17.5/2 = 8.75 (mol/dm3)

this is wrong, some of the reagents reacted to form propyl ethanoate. you have to use titration results (amt. of NaOH used) to find the actual conc. of reagents left.

[jacky0620]

This is an example from Internet.
I ask here because I don't know if it is true or not too.

[jacky0620]

{Procedure:
a.   Measure accurately 14.3 cm3 of glacial ethanoic acid and 18.8 cm3 of propan-1-ol into a clean, dry 100 cm3 distillation flask. Mix thoroughly.
b.   Transfer 1.00 cm3 of the mixture by pipette to a 125-conical flask containing about 25 cm3 distilled water and 2 drops of phenolphthalein indicator solution. Titrate to end point with 0.50 M NaOH. Record this volume. Repeat the titration to obtain an average volume of  0.50 M NaOH used.}

CH3COOH+NaOH->CH3COONa+H2O

no of mole of NaOH= 0.5M* 20/1000dm^3
no of mole of CH3COOH= no of mole of NaOH=0.01mol
Molarity of CH3COOH =mole / Volume

But What is the volume of CH3COOH?

1CM^3?But contains many things....OR Others?

HELP AGAIN

[xiankai]

1CM^3?But contains many things....OR Others?

is this not your mixture, from which you extracted 1cm³ of it?

14.3 cm3 of glacial ethanoic acid and 18.8 cm3 of propan-1-ol

also, concentration of CH₃COOH uses same value of volume as concentration of C₃H₇OH, as both are in the mixture together.

[jacky0620]

Mixture :
14.3 cm^3 ethanoic acid and 18.8cm^3 propan-1-ol


In the flask
there are" 1cm^3 of mixture of ethanoic acid and propan-1-ol " , "25cm^3 of water " and indicator

And titrate with NaOH

In calculation, what is the volume of ethanoic acid???

[xiankai]

you are given volume of ethanoic acid. but you must use the volume of the mixture of ethanoic acid and propan-1-ol.

[jacky0620]

CH3COOH+NaOH->CH3COONa+H2O

no of mole of NaOH= 0.5M* 20/1000dm^3
no of mole of CH3COOH= no of mole of NaOH=0.01mol
Molarity of CH3COOH =mole / Volume=0.01/(1/1000)

[xiankai]

so far you're good!