Question

Analysis of Acetic Acid Unknown

M Acetic Acid

Average M Unknown

Average % Acetic Acid

Analysis of Acetic Acid Unknown ID of Unknown used Trial 1 Trial 2 Vol. Unknown used Vol. NaOH used Mmol NaOH used M Acetic Acid, calc. Average M Unknown ml mmol 7. mmol Average % Acetic Acid HoAc Audlcnstn 17.4 (assume density - 1 g/mL) Show the calculations used for Sample 1 for the determination of the molarity of unknown Acetic acid sample and for the % Acetic acid in this sample. For the calculation of the % acetic acid in the sample, assume that the density of the sample is 1 g/ml

Answer 1

1) first we calculate molarity of acetic acid in trial 1 :

for that we calculate the moles of acid for trial 1

=7.1 ml NaOH * 1L / 10^3 ml *0.1 mol NaOH / 1L NaOH soln * 1mol unknown acid/ 1mol NaOH

=7.1 *10^-3*0.1

=7.1 * 10^-4 moles acid

Molarity = M1=moles of solute / volume in litre

we have 30 ml unknown soln so divided by 1000 =0.03 L

=7.1 * 10^-4 /0.03

Molarity =0.0236 M unknown acid

Trial 2 :

moles for trial 2

= 9.4ml NaOH * 1L/ 10^3 * 0.1 mol NaOH / 1L NaOH soln * 1mol unknown acid/ 1mol NaOH

= 9.4 * 10^-3 * 0.1

=9.4 * 10^-4

Molarity = 9.1 * 10^-4 / 0.03

=0.0303 M

2) Average molarity = 0.0236 + 0.0303 /2

=0.02695

3) % average aceitic acid = mas of aceitic acid / total mass of solution (acetic acid+ water) * 100

=60.05 g/ mol / 625 g water

=0.096 * 100

=9.608