 CHEMISTRY : Quantitative Analysis

Finding the Concentration of HYDROCHLORIC ACID

 PROJECT CODE: 1.31 SECTION: QUANTITATIVE ANALYSIS PROJECT TITLE: Finding the Concentration of HYDROCHLORIC ACID [Concentrated solution] RELEASE DATE: 1996 LAST UPDATE: 30 August 2009 VERSION HISTORY: 1.0 - First release 1.1 - Revision of text and formatting. 1.2 - Added new examples (retested samples in Aug 2009)

## PRINCIPLE:

##### Molarity [M]
###### The Principle of this test is to dilute the acid and find the titrate volume of known molarity of NaOH required to neutralize the Acid dilution. The reaction is follows:

ACID (aq) + ALKALI (s) --- > SALT (aq) + WATER (l)

HCl (aq) + NaOH (aq) --- > NaCl (aq) + H2O (l)

## SPECIFIC GRAVITY

#### Procedure

###### Tare an empty dry 100mL beaker over a balance Transfer carefully 10ml (or 25ml) of Acid solution using a volumetric pipette into the beaker. Note the volume used and weight read out. Repeat this for 2 more times to have 3 weight readings and average out the result. One can tare the weight after the previous reading and simply add over the next aliquot volume of acid and take the new weight.

 Specific Gravity (g / mL) = Mass (g) Volume (mL)

#### Calculations

##### Example 1 : Conc Hydrochloric acid Sol. A in 1996
###### In 1996 a solution of conc HCl (labelled as Sol A) provided the following results:
Reading Sol A Volume (mL) Weight (g) Specific Gravity (m/Vol)
1 25 28.95 1.16
2 25 29.05 1.16
3 10 11.55 1.15
Average Specific Gravity of Solution A is 1.16g/mL

##### Example 2 : Conc Hydrochloric acid Solution A2 in 2009
###### In 2009 (12 years Later) the same solution was tested and labelled here as Solution A2 . The following results were obtained:
Reading Sol. A2 Volume (mL) Weight (g) Specific Gravity (m/Vol)
1 25 27.68 1.11
2 25 27.50 1.10
3 10 11.12 1.11
Average Specific Gravity of Solution A is 1.11g/mL

##### Example 3 : Conc Hydrochloric acid Sol. B (New Stock 2009)
###### Brand new Conc HCl solution just bought on the day was measured and gave the following results
Reading Sol. B Volume (mL) Weight (g) Specific Gravity (m/Vol)
1 10 11.72 1.17
2 10 11.69 1.17
Average Specific Gravity of Solution A is 1.17g/mL

## Molarity [M]

#### Calculations

###### The calculation of the Molarity of HCl is given using an example of Sol A with the following data: Molarity of NaOH used [M-NaOH] = 0.3M Dilution factor of HCl [Dil] = 20 (5mL in 100mL)) Volume of Acid used in Titration [V-HCl] = 20mL Volume of NaOH required to neutralise HCl [V-NaOH] = 37.7mL (average of 3 runs) The reaction equation of this Acid-Alkali neutralisation is

NaOH (aq) + HCl (aq) --- > NaCl (aq) + H2O (l)

###### From the equatoion above: 1 mole NaOH : 1 mole HCl 0.0119 moles NaOH : 0.0119 moles HCl 0.0119 moles HCl are present in 20mL volume ? moles of HCl present in 1000mL are (0.0119 x 1000) / 20 = 0.566M HCl However, there was a dilution factor of 20 and so the molarity of the concentrated HCl is 0.566 x 20 = 11.3M The following formula can be used:

 Molarity HCl = [M-NaOH] x [V-NaOH] x Dil [V-HCl]

## Percentage Concentration [%]

###### The percentage of Acid present should theoretically be worked out by the ratio of the weight of actual HCL molecules present in 1000mL and the actual weight of 1000mL Concentrated HCl using the specific gravity index.

 % Acid Conc. = [ Weight of Acid molecules in 1000mL ] x 100 Weight of 1000mL Acid
 % Acid Conc. = = [ RMM of Acid x Molarity] x 100 [Specific Gravity x 1000mL]