 CHEMISTRY : Quantitative Analysis

Finding the Concentration of NITRIC ACID

 PROJECT CODE: 1.33 SECTION: QUANTITATIVE ANALYSIS PROJECT TITLE: Finding the Concentration of NITRIC ACID [Concentrated solution] RELEASE DATE: 1997 LAST UPDATE: 1 Sep 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)

HNO3 (aq) + NaOH (aq) --- > NaNO3 (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 : Solution N - Analar Conc. HNO3 [70%] (tested in 1997)
###### In 1997 a solution of conc HNO3 from a 2.5L bottle (Analar) marked as 70-71% concentration (labelled in this document as Solution N) provided the following results:
Reading Sol. N Volume (mL) Weight (g) Specific Gravity (m/Vol)
1 25 34.93 1.397
2 50 69.75 1.395
Average Specific Gravity of Solution N is 1.40g/mL

##### Example 2 : Solution N2 - Analar Conc. HNO3 [70%] (Solution N retested in 2009)
###### In 2009 (12 years Later) the same solution N was tested and labelled here as Solution N2 . The following results were obtained:
Reading Sol. N2 Volume (mL) Weight (g) Specific Gravity (m/Vol)
1 25 35.0 1.400
2 25 34.94 1.398
Average Specific Gravity of Solution N is 1.40g/mL

## Molarity [M]

#### Calculations

##### Example 1 : Solution N - Analar Conc. HNO3 [70%] (tested in 1997)
###### The calculation of the Molarity of HNO3 is given using an example of Sol. N with the following data: Molarity of NaOH used [M-NaOH] = 0.8M Dilution factor of HNO3 [Dil] = 20 (5mL in 100mL)) Volume of Acid used in Titration [V-HNO3] = 20mL Volume of NaOH required to neutralise HNO3 [V-NaOH] = 19.5mL (average of 3 runs) The reaction equation of this Acid-Alkali neutralisation is

NaOH (aq) + HNO3 (aq) --- > NaNO3 (aq) + H2O (l)

###### From the equatoion above: 1 mole NaOH : 1 mole HNO3 0.0156 moles NaOH :0.0156 moles HNO3 0.0156 moles HNO3 are present in 20mL volume ? moles of HNO3 present in 1000mL are (0.0156 x 1000) / 20 = 0.78M HNO3 However, there was a dilution factor of 20 and so the molarity of the concentrated HNO3 is 0.78 x 20 = 15.6M The following formula can be used:

 Molarity HNO3 = [M-NaOH] x [V-NaOH] x Dil [V-HNO3]

## Percentage Concentration [%]

###### The percentage of Acid present should theoretically be worked out by the ratio of the weight of actual HNO3 molecules present in 1000mL and the actual weight of 1000mL Concentrated HNO3 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]