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)

INTRODUCTION:
This procedure is useful to find the concentration level of a concentrated solution of Nitric Acid which perhaps is old and is not as strong as its original concentration marked on the commercial bottle. Srong acids easily loose their concentration by decompostion and absorption of moisture from air. The procedure is hence aimed for concentrated solution more than 1Molar. The calculation giives the following results:
 Specific Gravity of Acid (g/mL)
 Molar Concentration (M)
 Percentage Concentration (%)
A number of examples are given, all based on true solution that I have in my home Lab.
PRINCIPLE:
Specific Gravity
Specific gravity is easily found by measuring the weight of a fixed volume of acid and calculated it by the formula Mass / Volume.
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)
From the NaOH titre volume, the number of moles of NaOH are calculated. From the reaction above it can be noted that the ratio [ Acid : Alkali ] = 1 : 1, and thus the moles of HNO3 are half that of NaOH. The moles of acid present in the volume titrated are converted to Molarity of the Acid after taking into account the dilution.
Percentage Acid [%]
Once the specific gravity and Molarity are determined, the percentage is calculated from the fraction of the weight of theoretical Acid molecules in 1000g solution (Molarity x RMM) divided by the actual weight of 1000g acid (specific gravity x 1000).
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 7071% 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]
Procedure
1) Preparation of NaOH Standard Titration Solution
Prepare 0.5M NaOH by dissolving 5g NaOH into 250ml Water in a volumetric flask. Other molarites can be used but ideally in the range of 0.4M to 1.2M
Note the molarity used [MNaOH].
1M NaOH = 40g NaOH in 1000 mL water
0.5M NaOH = Xg NaOH in 1000 mL water
= > X = 0.5 x 40 / 1 = 20g
0.5M NaOH are made by 20g in 1000mL water
0.5M NaOH are made by X g in 250mL water
=> X = (20 x 250) / 1000 = 5g in 250mL
Short cut : to prepare 0.X M NaOH in 250mL water simply measure X g NaOH !
2) Preparation of Conc. HNO3 solution
Transfer 5mL of Concentrated HNO3 using a volumetric pipette to a 100mL volumetric flask and gently add water to the mark to make a 1:20 dilution (5:100)
Note the dilution factor [Dil].
3) Titration
Transfer 20mL of the HNO3 dilution to three 100mL flasks.
Note the volume of acid used [VHNO3].
Add few drops of indicator such as Bromothymol blue which has the following pH colour properties
 pH < 5 > pale yellow
 pH 6.0 > yellow
 pH 7.0 > green/torquise
 pH 7.2 > blue


Pour the NaOH Standard solution in a measuring burette
Titrate the acid solution with the NaOH to achieve neutralisation that is when the indicator colour turns torquise or just blue. Measure the volume of NaOH used by the burette for the three flasks and average the results out Note the volume of NaOH used [VNaOH].
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 [MNaOH] = 0.8M
 Dilution factor of HNO3 [Dil] = 20 (5mL in 100mL))
 Volume of Acid used in Titration [VHNO3] = 20mL
 Volume of NaOH required to neutralise HNO3 [VNaOH] = 19.5mL (average of 3 runs)
The reaction equation of this AcidAlkali neutralisation is
NaOH (aq) + HNO3 (aq)  > NaNO3 (aq) + H2O (l)
in 1000mL NaOH there are 0.8 moles
in 19.5mL NaOH, moles present (19.5 x 0.8) / 1000 ( = 0.0156 moles NaOH)
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 = 
[MNaOH] x [VNaOH] x Dil 



[VHNO3] 
Short Cut : Since the dilution factor [Dil] and the Volume of Acid used [VHNO3] are both 20, they cancel each other out and so the Molarity of HNO3 in this case is [MNaOH] x [VolNaOH] = (0.8 x 19.5 ) = 15.6M !!
Example 2 : Solution N2  Analar Conc. HNO3 [70%] (Solution N retested in 2009)
 Molarity of NaOH used [MNaOH] = 0.5M
 Dilution factor of HNO3 [Dil] = 20 (5mL in 100mL))
 Volume of Acid used in Titration [VHNO3] = 20mL
 Volume of NaOH required to neutralise HNO3 [VNaOH] = 30.4mL (average of 3 runs)
Molarity HNO3 = [MNaOH] x [VNaOH] x Dil x [VHNO3]
= (0.5 x 30.4) x 20 / 20 = 15.2M
Example 3 :Solution O  General Purpose Conc. HNO3
 Molarity of NaOH used [MNaOH] = 0.5M
 Dilution factor of HNO3 [Dil] = 20 (5mL in 100mL))
 Volume of Acid used in Titration [VHNO3] = 20mL
 Volume of NaOH required to neutralise HNO3 [VNaOH] = 25.8mL (average of 3 runs)
Molarity HNO3 = [MNaOH] x [VNaOH] x Dil x [VHNO3]
= (0.5 x 25.8) x 20 / 20 = 12.9M
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] 
Example 1 : Solution N  Analar Conc. HNO3 [70%] (tested in 1997)
1 mole HNO3 in 1000 mL weigh 63g (RMM of HNO3 = 1 + 14 + 48)
However Sol. N was found to have 15.58 moles in 1000ml (15.58M)
Hence the weight of HNO3 molecules in 1000mL is 15.58 x 63 = 981.54g
The actual weight of 1000mL of Sol. N = its Specific Gravity (g/mL) x 1000 = 1.4 x 1000 = 1400g
% HNO3 = Weight of HNO3 molecules in 1000mL x 100 / Weight of 1000mL Acid
= 981.54 x 100 / 1400 = 70.1%
Example 2 : Solution N2  G.P.R Conc. HNO3 [70%] (= Solution N tested in 2009)
 Specifig gravity = 1.40
 Molarity of HNO3 = 15.2M
% HNO3 = Weight of HNO3 molecules in 1000mL x 100 / Weight of 1000mL Acid
= (63 x 15.2) x 100 / (1.4 x 1000) = 68.4%
CONCLUSIONS
Summary of Results:
Sample Conc. HNO3 Solution
 Test Date
 Specific Gravity (g/mL)
 Molarity (M)
 Percentage w/w (%)

Solution N  Analar Conc. HNO3 [70%] (tested in 1997) 
? 1997 
1.4 
15.8 
70.1 
Solution N2  G.P.R Conc. HNO3 [70%] (= Solution N tested in 2009) 
Aug 2009 
1.4 
15.2 
68.4 
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