Question

Hi there,

I am asked to calculate for the millimoles of acetic acid in a sample for a lab (Determination of the molar mass and ionization constant of a weak acid) this lab was performed with titration of 0.10 M NaOH into a 30.615g of acetic acid.

Measured pH of the acetic acid sol’n	Trial 1
Mass of acetic acid sol’n taken for the tritation	30.615g
Initial buret reading of NaOH tritrant	0.00ml
Final buret reading of NaOH tritrant	12.6ml
Net vol. of NaOh	12.6ml
Millimoles of acetic acid in sample	1.27ml
Molar [acetic acid sol’n]
Calculated molar mass of acetic acid

Chemical Equation: CH3COOH + NaOH -> CH3COONa + H2O

Trial1: millimoles of NaOH to endpoint of tritration:

0.1009M NaOH   12.6ml

0.1009M NaOH/ 1ml * 12.6 ml = 1.27134mmol = 1.27 mmol NaOH

Reasoning:

1mmol HC2H3O2 / 1mmol NaOH = 1.27 mmol, Because the mmol of NaOH is 1.27 mmol NaOH and there is a 1 to 1 ratio between HC2H3O2

Millimoles of acetic acid in sample:

30.615g CH3COOH * 1 mole CH3COOH/ 60.6g * 1000mmol CH3COOH/ 1 moleCh3COOH = 509.7 mmol CH3COOH

I am unsure of what value to use as the millimoles of acetic acid.

Answer 1

ANSWER-

Consider the following reaction-

CH₃COOH + NaOH -> CH₃COONa + H₂O

Trial-1

mmoles of NaOH to endpoint of tritration= volume of NaOH used x Molarity

=12.6 ml x0.10 M

=1.26 mmol NaOH

from the above given equation of titration ,

it is clear that one mole of CH₃COOH require 1 mole of NaOH for nuetralisation

So, no of mmol of CH₃COOH used ( in the titration )= 1.26 mmol of CH₃COOH

actual no of moles acetic acid present in the solution = mass of the substance / molar mass

=30.615/60.05

=0.5098 moles

=509.82 mmoles of CH₃COOH   

after titration,

no of mmoles of CH₃COOH left =(509.82-1.26)

=508.56 mmol CH₃COOH

Note- 508.56 mmol CH₃COOH can be further used fo find out ionisation constant and molarity etc by knowing the volume in which this acetic acid present.