# Calculating amount of Disodium ethylenediaminetetraacetate dihydrate (EDTA.Na2.2H2O) for preparation of 1 litre of 0.5 M EDTA solution

Molecular weight of EDTA.Na2.2H2O = 372.24 g/mol
Solution volume = 1 L
Desired molarity of solution = 0.5 M
Amount of substance (in grams) = To be calculated

Formula : Place all values in the formula
Amount of substance (in grams) = 0.5 x 372.24 x 1
= 186.12 grams

Conclusion: To prepare 1 litre of 0.5 M solution, you need 186.12 grams of Disodium ethylenediaminetetraacetate dihydrate (EDTA.Na2.2H2O).

# Calculating volume of 0.5M EDTA required for preparation of 100 ml of 1 mM EDTA

Concentrated 0.5 M EDTA stock solution can be diluted to prepare 1 mM EDTA solution. Following procedure can be used to calculate amount of 0.5M EDTA solution required to prepare 100 ml of 1 mM EDTA solution

# Procedure:

Step 1: Convert all concentration unit to either molar or millimolar.

The SI prefix “milli” represents a factor of 10-3. Here we convert stock solution concentration (0.5M EDTA) to millimolar (mM).

1 mM   =     10-3 M = 0.001 M

OR                               1 M      =      1000 mM
0.5 M   =      0.5 x 1000 mM

0.5 M   =      500 mM

Stock solution concentration is 500 mM.

Step 2: Calculate required amount of stock solution (500 mM EDTA) to prepare 1000 ml of 1 mM EDTA

Formula: Cf: Concentration of final solution i.e., 1 mm EDTA
Vf: Volume of the final solution i.e., 100 ml
CS: Conc. of stock solution i.e., 500 mM

VS: Volume of stock solution – To be calculated

Place all values in the formula

1 x 100 = 500 x VS VS = 0.2 ml

Step 2: Calculate the amount of water

You want to prepare 100 ml solution of which 0.2 ml is 0.5M EDTA stock solution. Rest of the amount will be water which is calculated by subtracting 0.2 ml from 100 ml.

Vf     : Volume of the final solution i.e., 100 ml
Vs    : Volume of the stock solution i.e., 0.2 ml

Amount of diluent i.e., water: 100 – 0.2 = 99.8 ml

Conclusion: To prepare 100 ml of 1 mM solution of EDTA from 0.5M EDTA stock solution, you need to mix 0.2 ml of 0.5M EDTA stock solution and 99.8 ml water.

# Overview:

• Ammonium Hydroxide (NH4OH) is a clear colourless liquid. 56.6% (w/w) concentrated Ammonium Hydroxide can be obtained from different suppliers.
• When ammonia dissolves in water, it forms Ammonium Hydroxide. 28% aqueous solution of Ammonia (NH3) is equal to approximately 56.6% Ammonium Hydroxide.
• 56.6% (w/w) Ammonium Hydroxide means that 100 grams of Ammonium Hydroxide contains 56.6 grams of NH4OH.
• The density of 56.6% (w/w) Ammonium Hydroxide is 0.9 g/ml at 25°C that means that weight of the 1 ml of Ammonium Hydroxide is 0.9 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of NH4OH is equal to 35.05 grams of NH4OH (molecular weight = 35.05).

# Calculation procedure:

Known values:
Density of Ammonium Hydroxide :                        0.9 g/ml
Molecular weight of NH4OH:                                  35.05 g/mole

Concentration of Ammonium Hydroxide:            56.6% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Ammonium Hydroxide.

Formula:
Density =  weight / volume    or
Volume =  weight / density    or

Volume of 100 gram of Ammonium Hydroxide : 100/0.9 =  111.11 ml

Note: 56.6% (w/w) Ammonium Hydroxide means that 100 grams of Ammonium Hydroxide contain 56.6 grams of NH4OH.

The volume of 100 grams of Ammonium Hydroxide is 111.11 ml. That means 56.6 grams of NH4OH is present in 111.11 ml of Ammonium Hydroxide.

## Step 2: Calculate how many grams of NH4OH is present in 1000 ml of Ammonium Hydroxide.

111.11 ml of Ammonium Hydroxide contain       = 56.6 grams of NH4OH
1 ml of Ammonium Hydroxide will contain           = 56.6/111.11 grams of NH4OH

1000 ml of Ammonium Hydroxide will contain     = 1000 x 56.6/111.11 = 509.405 grams of NH4OH

1000 ml of Ammonium Hydroxide will contain 509.405 grams of NH4OH

## Step 3: Calculate the number of moles of NH4OH present in 509.405 grams of NH4OH.

35.05 grams of NH4OH is equal to 1 mole.
1 gram of NH4OH will be equal to 1/35.05 moles.

509.405 grams will be equal to = 509.405 x 1/35.05 = 14.534 moles

Therefore, we can say that 1 litre of Ammonium Hydroxide contains 14.534 moles or in other words molarity of 56.6% (w/w) Ammonium Hydroxide is equal to 14.534 M.

# Overview:

• Potassium Hydroxide (KOH) is a clear colourless liquid. A 45% (w/w) concentrated Potassium Hydroxide can be obtained from different suppliers.
• 45% (w/w) Potassium Hydroxide means that 100 grams of Potassium Hydroxide contains 45 grams of KOH.
• The density of 45% (w/w) Potassium Hydroxide is 1.456 g/ml at 25°C that means that weight of the 1 ml of Potassium Hydroxide is 1.456 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of KOH is equal to 56.11 grams of KOH (molecular weight = 56.11).

# Calculation procedure:

Known values:
Density of Potassium Hydroxide :                             1.456 g/ml
Molecular weight of KOH:                                         56.11 g/mole

Concentration of Potassium Hydroxide:            45% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Potassium Hydroxide.

Formula:
Density =  weight / volume    or
Volume =  weight / density    or

Volume of 100 gram of Potassium Hydroxide : 100/1.456 =  68.6813 ml

Note: 45% (w/w) Potassium Hydroxide means that 100 grams of Potassium Hydroxide contain 45 grams of KOH.

The volume of 100 grams of Potassium Hydroxide is 68.6813 ml. That means 45 grams of KOH is present in 68.6813 ml of Potassium Hydroxide.

## Step 2: Calculate how many grams of KOH is present in 1000 ml of Potassium Hydroxide.

68.6813 ml of Potassium Hydroxide contain       = 45 grams of KOH
1 ml of Potassium Hydroxide will contain           = 45/68.6813 grams of KOH

1000 ml of Potassium Hydroxide will contain     = 1000 x 45/68.6813 = 655.2 grams of KOH

1000 ml of Potassium Hydroxide will contain 655.2 grams of KOH

## Step 3: Calculate number of moles of KOH present in 655.2 grams of KOH.

56.11 grams of KOH is equal to 1 mole.
1 gram of KOH will be equal to 1/56.11 moles.

655.2 grams will be equal to = 655.2 x 1/56.11 = 11.677 moles

Therefore, we can say that 1 litre of Potassium Hydroxide contains 11.677 moles or in other words molarity of 45% (w/w) Potassium Hydroxide is equal to 11.677 M.

# Overview:

• Sulfuric acid (H2SO4) is a clear colourless liquid. A 95% (w/w) concentrated Sulfuric acid can be obtained from different suppliers.
• 95% (w/w) Sulfuric acid means that 100 grams of Sulfuric acid contains 95 grams of H2SO4.
• The density of 95% (w/w) Sulfuric acid is 1.84 g/ml at 25°C that means that weight of the 1 ml of Sulfuric acid is 1.84 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of H2SO4 is equal to 98.08 grams of H2SO4 (molecular weight = 98.08).

# Calculation procedure:

Known values:
Density of Sulfuric acid :                             1.84 g/ml
Molecular weight of H2SO4:                       98.08 g/mole

Concentration of Sulfuric acid:            95% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Sulfuric acid.

Formula:
Density =  weight / volume    or

Volume =  weight / density    or

Volume of 100 gram of Sulfuric acid : 100/1.84 =  54.3478 ml

Note: 95% (w/w) Sulfuric acid means that 100 grams of Sulfuric acid contain 95 grams of H2SO4.

The volume of 100 grams of Sulfuric acid is 54.3478 ml. That means 95 grams of H2SO4 is present in 54.3478 ml of Sulfuric acid.

## Step 2: Calculate how many grams of H2SO4 is present in 1000 ml of Sulfuric acid.

54.3478 ml of Sulfuric acid contain       = 95 grams of H2SO4
1 ml of Sulfuric acid will contain           = 95/54.3478 grams of H2SO4

1000 ml of Sulfuric acid will contain     = 1000 x 95/54.3478 = 1748 grams of H2SO4

1000 ml of Sulfuric acid will contain 1748 grams of H2SO4

## Step 3: Calculate number of moles of H2SO4 present in 1748 grams of H2SO4.

98.08 grams of H2SO4 is equal to 1 mole.
1 gram of H2SO4 will be equal to 1/98.08 moles.
1748 grams will be equal to = 1748 x 1/98.08 = 17.822 moles

Therefore, we can say that 1 litre of Sulfuric acid contains 17.822 moles or in other words molarity of 95% (w/w) Sulfuric acid is equal to 17.822 M.

# Overview:

• Phosphoric acid (H3PO4) is a clear colourless liquid. A 86% (w/w) concentrated Phosphoric acid can be obtained from different suppliers.
• 85% (w/w) Phosphoric acid means that 100 grams of Phosphoric acid contains 85 grams of H3PO4.
• The density of 85% (w/w) Phosphoric acid is 1.685 g/ml at 25°C that means that weight of the 1 ml of Phosphoric acid is 1.685 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of H3PO4 is equal to 98 grams of H3PO4 (molecular weight = 98).

# Calculation procedure:

Known values:
Density of Phosphoric acid :                             1.685 g/ml
Molecular weight of H3PO4:                           98 g/mole

Concentration of Phosphoric acid:                85% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Phosphoric acid.

Formula:
Density =  weight / volume    or
Volume =  weight / density    or

Volume of 100 gram of Phosphoric acid : 100/1.685 =  59.347 ml

Note: 85% (w/w) Phosphoric acid means that 100 grams of Phosphoric acid contain 85 grams of H3PO4.

The volume of 100 grams of Phosphoric acid is 59.347 ml. That means 85 grams of H3PO4 are present in 59.347 ml of Phosphoric acid.

## Step 2: Calculate how many grams of H3PO4 is present in 1000 ml of Phosphoric acid.

59.347 ml of Phosphoric acid contain        = 85 grams of H3PO4
1 ml of Phosphoric acid will contain        = 85/59.347 grams of H3PO4

1000 ml of Phosphoric acid will contain  = 1000 x 85/59.347 = 1432.254 grams of H3PO4

1000 ml of Phosphoric acid will contain 1432.254 grams of H3PO4

## Step 3: Calculate number of moles of H3PO4 present in 1432.254 grams of H3PO4.

98 grams of H3PO4 is equal to 1 mole.
1 gram of H3PO4 will be equal to 1/98 moles.

1432.254 grams will be equal to = 1432.254 x 1/98 = 14.615 moles

Therefore, we can say that 1 litre of Phosphoric acid contains 14.615 moles or in other words molarity of 85% (w/w) Phosphoric acid is equal to 14.615 M.

# Overview:

• Nitric acid is a clear colourless liquid. A 70% (w/w) concentrated Nitric acid can be obtained from different suppliers.
• 70% (w/w) Nitric acid means that 100 grams of Nitric acid contains 70 grams of HNO3.
• The density of 70% (w/w)Nitric acid is 1.413 g/ml at 25°C that means that weight of the 1 ml of Nitric acid is 1.413 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of HNO3 is equal to 63.01 grams of HNO3 (molecular weight = 63.01).

# Calculation procedure:

Known values:
Density of Nitric acid :                                1.413 g/ml
Molecular weight of HNO3:                       63.01 g/mole

Concentration of Nitric acid:                     70% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Nitric acid.

Formula:
Density =  weight / volume    or

Volume =  weight / density    or

Volume of 100 gram of Nitric acid : 100/1.413 =  70.771 ml

Note: 70% (w/w) Nitric acid means that 100 grams of Nitric acid contain 70 grams of HNO3.

The volume of 100 grams of Nitric acid is 70.771 ml. That means 70 grams of HNO3 is present in 70.771 ml of Nitric acid.

## Step 2: Calculate how many grams of HNO3 is present in 1000 ml of Nitric acid.

70.771 ml of Nitric acid contain        = 70 grams of HNO3
1 ml of Nitric acid will contain           = 70/70.771 grams of HNO3

1000 ml of Nitric acid will contain    = 1000 x 70/70.771 = 989.106 grams of HNO3

1000 ml of Nitric acid will contain 989.106 grams of HNO3

## Step 3: Calculate number of moles of HNO3 present in 989,106 grams of HNO3.

63.01 grams of HNO3 is equal to 1 mole.
1 gram of HNO3 will be equal to 1/63.01 moles.

989.106 grams will be equal to = 989.106 x 1/63.01 = 15.6976 moles

Therefore, we can say that 1 litre of Nitric acid contains 15.6976 moles or in other words molarity of 70% (w/w) Nitric acid is equal to 15.6976 M.

# Overview:

• Sodium hydroxide is a clear colourless liquid. A 50% (w/w) concentrated Sodium hydroxide can be obtained from different suppliers.
• 50% (w/w) Sodium hydroxide means that 100 grams of solution contains 50 grams of NaOH.
• The density of 50% (w/w) Sodium hydroxide is 1.515 g/ml at 25°C that means that weight of the 1 ml of Sodium hydroxide solution is 1.22 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of Sodium hydroxide is equal to 40.00 grams of Sodium hydroxide (NaOH, molecular weight = 40.00).

# Calculation:

Known values:
Density of Sodium Hydroxide :                                1.515 g/ml
Molecular weight of Sodium Hydroxide:                40.00 g/mole

Concentration of Sodium Hydroxide:                      50% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Sodium Hydroxide solution.

Formula:

Density =  weight / volume    or

Volume =  weight / density

Volume of 100 gram of Sodium hydroxide solution: 100/1.515 =  66.0066 ml

Note: 50% (w/w) Sodium Hydroxide means that 100 grams of solution contains 50 grams of Sodium Hydroxide.

The volume of 100 grams of Sodium hydroxide solution is 66.0066 ml. That means 50 grams of NaOH is present in 66.0066 ml of  Sodium hydroxide solution.

## Step 2: Calculate how many grams of NaOH is present in 1000 ml of Sodium hydroxide solution.

66.0066 ml of Sodium Hydroxide contains   = 50 grams of NaOH
1 ml of Sodium Hydroxide will contain          = 50/66.0066 grams of NaOH

1000 ml of Sodium Hydroxide will contain    = 1000 x 50/66.0066 = 757.50 grams of NaOH

1000 ml ofSodium hydroxide solution will contains 757.50 grams of NaOH.

## Step 3: Calculate number of moles of NaOH present in 757.50 grams of NaOH.

40 grams of NaOH is equal to 1 mole.
1 gram of NaOH will be equal to 1/40 moles.

757.50 grams of NaOH will be equal to = 757.50 x 1/40 = 18.9375 moles

Therefore, we can say that 1 litre ofSodium hydroxide solution contains 18.9375 moles or in other words molarity of 50% (w/w) Sodium Hydroxide is equal to 18.9375 M.

# Overview:

• Formic acid is supplied as clear colourless liquid. A 95% (w/w) concentrated formic acid can be obtained from different suppliers. (see suppliers)
• 95% (w/w) formic acid means that 100 grams of Formic contains 95 grams of HCOOH.
• The density of 95% (w/w) formic acid is 1.22 g/ml at 25°C that means that weight of the 1 ml of formic acid is 1.22 grams at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of formic acid is equal to 46.03 grams of HCOOH (molecular weight = 46.03).

# Calculation procedure:

Known values:
Density of Formic acid :                                1.22 g/ml
Molecular weight of HCOOH:                      46.03 g/mole

Concentration of Formic acid:                    95% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of formic acid.

Formula:
Density =  weight / volume    or

Volume =  weight / density

Volume of 100 gram of formic acid: 100/1.22 = 81.967 ml

Note: 95% (w/w) Formic acid means that 100 grams of solution contains 95 grams of HCOOH.

The volume of 100 grams of formic acid is 81.967 ml. That means 95 grams of HCOOH is present in 81.967 ml of  formic acid.

## Step 2: Calculate how many grams of HCOOH is present in 1000 ml of Formic acid.

81.967 ml of Formic acid contains        =     95 grams of HCOOH
1 ml of Formic acid will contain           =     95/81.967 grams of HCOOH

1000 ml of Formic acid will contain    =     1000 x 95/81.967 =     1159.003 grams of HCOOH

1000 ml of formic acid will contain 1159.003 grams of HCOOH.

## Step 3: Calculate the number of moles of HCOOH present in 1159.003 grams of HCOOH.

46.03 grams of HCOOH is equal to 1 mole.
1 gram of HCOOH will be equal to 1/46.03 moles.

444.01776 grams will be equal to = 1159.003 x 1/46.03 = 25.179 moles

Therefore, we can say that 1 litre of formic acid contains 25.179 moles of HCOOH or in other words molarity of 95% (w/w) Formic acid is equal to 25.179 M.

# Overview:

• Hydrochloric acid is a clear colourless aqueous solution of hydrogen chloride (HCl) that can be supplied as a 37% (w/w) hydrogen chloride solution (see suppliers).
• 37% (w/w) hydrochloric acid means that 100 grams of hydrochloric acid contains 37 grams of hydrogen chloride.
• The density of 37% (w/w) hydrochloric acid is 1.2 g/ml at 25°C that means that weight of the 1 ml of hydrochloric acid is 1.2 gram at 25°C.
• Molarity refers to number of moles of the solute present in 1 litre of solution.
• In simple words, 1 mole is equal to atomic weight of the substance. For example, 1 mole of hydrogen chloride is equal to 36.46 grams of hydrogen chloride (molecular weight = 36.46).

# Calculation procedure:

Known values:
Density of Hydrochloric acid :                                1.2 g/ml
Molecular weight of Hydrogen chloride:             36.46 g/mole

Concentration of Hydrochloric acid:                     37% (% by mass, wt/wt)

## Step 1: Calculate the volume of 100 grams of Hydrochloric acid.

Formula:
Density =  weight / volume    or

Volume =  weight / density

Volume of 100 gram of Hydrochloric acid   =   100/1.2   =   83.33 ml

Note: 37% (w/w) Hydrochloric acid means that 100 grams of Hydrochloric acid contains 37 grams of Hydrogen chloride.

The volume of 100 grams of Hydrochloric acid is 83.33 ml. That means 37 grams of Hydrogen chloride is present in 83.33 ml of Hydrochloric acid.

## Step 2: Calculate how many grams of Hydrogen chloride is present in 1000 ml of Hydrochloric acid.

83.33 ml of Hydrochloric acid contains       =  37 grams of Hydrogen chloride.
1 ml of Hydrochloric acid will contain          =  37/83.33 grams of Hydrogen chloride.

1000 ml of Hydrochloric acid will contain    =  1000 x 37/83.33   =    444.01776 grams of acetic acid

1000 ml of Hydrochloric acid will contain 444.01776 grams of Hydrogen chloride.

## Step 3: Calculate the number of moles of Hydrogen chloride present in 444.01776 grams of Hydrogen chloride.

36.46 grams of Hydrogen chloride            =     1 mole
1 gram of Hydrogen chloride                     =     1/36.46 moles

444.01776 grams of Hydrogen chloride   =     444.01776 x 1/36.46    =   12.178 moles

Therefore, we can say that 1 liter of Hydrochloric acid contains 12.178 moles or in other words molarity of 37% (w/w) Hydrochloric acid is equal to 12.178.