Essay on Titration Andrew Harvey

Submitted By Green-tea
Words: 1577
Pages: 7

CHEMISTRY - Practical Task Preparation (v3)

By Andrew Harvey

-Acid/Base Titrations can be acid to base or base to acid. The following assumes you add acid to base. This is the best method for the exam. My calcs will not fit together as I used different numbers for different parts that should be the same. This document may not be 100% correct!.- THIS IS BASED ON THE MARKING GUIDELINES AS SHOWN IN THE BLUE TITLES.

The question will be along the lines of:
Using sodium hydroxide NaOH provided at 0.100 M, determine the concentration of acetic acid
(CH3COOH) in vinegar. There is approximately 10% w/v acetic acid in vinegar.

Perform appropriate dilutions using correct calculations and equipment

A volumetric flask (above) can be used to measure out a substance to an exact volume.
��������������� ��� �−1 =
������ �� �����
������ (�)

In our practical task we will be given a base of known concentration and we will need to find the concentration of an acid in a common household substance. The base (NaOH) is our primary standard, we should use 25mL of it in the flask for each titration. We do not need to dilute the base.

However the acid will need to be diluted, this is because the titration works best and is more accurate when we have equal concentration of the acid and the base. So just say we are given NaOH at 0.100 M, we will need to dilute the acid to make it roughly this concentration.

We are told the approximate amount of acetic acid in vinegar to be 10% w/v. This means that there is 10g of acetic acid per 100 mL of the solution (as 10% = 10/100). However concentration is moles per Litre, so firstly we make it per 1 L, 100g per 1L. Now we need to change the 100g into moles, and as � = �
��
= 100
12.01×2+1.008 ×4+16.00×2
= 100
60.052
≈ 1.665���. So we have 1.665 mol per 1 L, so concentration is approximately = 1.665 M. This is not close enough to the concentration of the base (0.100 M), so we need to dilute the acid. As the acid is a bit over 16 times more concentrated than the base, we should add 16 times the amount of water than the acid (per volume). So using a
250mL volumetric flask, we would need 15mL of acid and then fill the rest with water, but we can only measure out multiples of 25mL with the pipette, so the closest we can get with a 250mL
1.665
volumetric flask is
10
= 0.1665 M, this is close enough to the concentration of the base (0.100 M), if it was not we could use a 500mL volumetric flask, and/or different multiples of 25mL of the acid.

So as we use 25mL of acid, and then add 225mL of water, the acid has been diluted by a factor of
10…