CHEMISTRY : Quantitative Analysis

Finding the Concentration of HYDROCHLORIC ACID

INTRODUCTION:

This procedure is useful to find the concentration level of a concentrated solution of Hydrochloric 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:

1. Specific Gravity of Acid (g/mL)
2. Molar Concentration (M)
3. 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)

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

From the NaOH titre volume, the number of moles of NaOH are calculated, which from the reaction [ Acid : Alkali ] = 1 : 1, it is equal to the moles of HCl . These moles are converted to Molarity of the Acid 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 : Conc Hydrochloric acid Sol. A in 1996

In 1996 a solution of conc HCl (labelled as Sol A) provided the following results:

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:

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

Molarity [M]

Procedure

1) Preparation of NaOH Standard Titration Solution

Prepare 0.3M NaOH by dissolving 3g NaOH into 250ml Water in a volumetric flask. Other molarites can be used but ideally in the range of 0.2M to 0.6M

Note the molarity used [M-NaOH].

1M NaOH = 40g NaOH in 1000 mL water
0.3M NaOH = Xg NaOH in 1000 mL water
= > X = 0.3 x 40 / 1 = 12g

0.3M NaOH are made by 12g in 1000mL water
0.3M NaOH are made by X g in 250mL water
=> X = (12 x 250) / 1000 = 0.3g in 250mL

Short cut : to prepare 0.X M NaOH in 250mL water simply measure X g NaOH !

2) Preparation of Conc. HCl solution

Transfer 5mL of Concentrated HCl 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 HCL dilution to three 100mL flasks.

Note the volume of acid used [V-HCl].

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 [V-NaOH].

Calculations

The calculation of the Molarity of HCl is given using an example of Sol A with the following data:

1. Molarity of NaOH used [M-NaOH] = 0.3M
2. Dilution factor of HCl [Dil] = 20 (5mL in 100mL))
3. Volume of Acid used in Titration [V-HCl] = 20mL
4. 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)

in 1000mL NaOH there are 0.3 moles
in 37.7mL NaOH, moles present (37.3 x 0.3) / 1000 ( = 0.0119 moles NaOH)

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]

Short Cut : Since the dilution factor [Dil] and the Volume of Acid used [V-HCl] are both 20, they cancel each other out and so the Molarity of HCl in this case is [M-NaOH] x [Vol-NaOH] = 0.3 x 37.7 !!

Example 2: Solution A2 (Same solution A but tested 12 years later):

1. Molarity of NaOH used [M-NaOH] = 0.2M
2. Dilution factor of HCl [Dil] = 20 (5mL in 100mL))
3. Volume of Acid used in Titration [V-HCl] = 20mL
4. Volume of NaOH required to neutralise HCl [V-NaOH] = 34.7mL (average of 3 runs)

Molarity HCl = [M-NaOH] x [V-NaOH] x Dil / [V-HCl] = (0.2 x 34.7) x 20 / 20 = 6.94M

Example 3 : New Commercial Concentrated Hydrochloric Acid (2009) - Solution B

1. Molarity of NaOH used [M-NaOH] = 0.5M
2. Dilution factor of HCl [Dil] = 20 (5mL in 100mL))
3. Volume of Acid used in Titration [V-HCl] = 20mL
4. Volume of NaOH required to neutralise HCl [V-NaOH] = 23.6mL (average of 3 runs)

Molarity HCl = [M-NaOH] x [V-NaOH] x Dil / [V-HCl] = (0.5 x 23.6) x 20 / 20 = 11.8M

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]

Example 1 : Concentrated Hydrochloric Acid Solution A

1 mole HCl in 1000 mL weigh 36.5g (RMM of HCL = 1 + 35.5)
However Sol. A was found to have 11.26 moles in 1000ml (11.26M)
Hence the weight of HCl molecules in 1000mL is 11.26 x 36.5 = 410.8g

The actual weight of 1000mL of Sol. A = its Specific Gravity (g/mL) x 1000 = 1.16 x 1000 = 1160g

% HCl = Weight of HCl molecules in 1000mL x 100 / Weight of 1000mL Acid
       = 410.8 x 100 / 1160 = 35.4%

Example 2: Solution A2 (Same solution A but tested 12 years later):

1. Specifig gravity = 1.11
2. Molarity of HCl = 6.94M

% HCl = Weight of HCl molecules in 1000mL x 100 / Weight of 1000mL Acid
       = (36.5 x 6.94) x 100 / (1.11 x 1000) = 22.8%

Example 3 : New Commercial Concentrated Hydrochloric Acid (2009) - Solution B

1. Specifig gravity = 1.17
2. Molarity of HCl = 11.8M

% HCl = Weight of HCl molecules in 1000mL x 100 / Weight of 1000mL Acid
       = (36.5 x 11.8) x 100 / (1.17 x 1000) = 36.8%

CONCLUSIONS

Summary of Results:

Sample Conc. HCl Solution	Test Date	Specific Gravity (g/mL)	Molarity (M)	Percentage w/w (%)
Solution A Conc. HCl	1997	1.16	11.26	35.4
Solution A Conc. HCl (Tested 12 years later)	Aug 2009	1.11	6.94	22.8
Solution B (New stock 2009)	Aug 2009	1.17	11.8	36.8

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