Molarity of 56.6% Ammonium Hydroxide (28% Aqueous Solution of Ammonia) [NH4OH (NH3 + H2O)]

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

Calculation procedure:

Known values 
Density of 56.6% Ammonium hydroxide solution
0.9 g/ml
Molecular weight of NH4OH35.05 g/mole
Concentration of Ammonium hydroxide solution
56.6% (% by mass, wt/wt)

Step 1: Calculate the volume of 100 g of Ammonium Hydroxide.

Formula:

Density =  weight / volume
or
Volume =  weight / density

The volume of 100 g of Ammonium hydroxide solution: 100/0.9 =  111.11 ml

Note: A 56.6% (w/w) Ammonium hydroxide solution means that 100 g of Ammonium hydroxide solution contains 56.6 g of NH4OH.

The volume of 100 g of Ammonium hydroxide solution is 111.11 ml. It means that 56.6 g of NH4OH is present in 111.11 ml of Ammonium hydroxide solution.

Step 2: Calculate how many grams of NH4OH is present in 1000 ml of Ammonium Hydroxide.
111.11 ml of Ammonium hydroxide solution contain = 56.6 g of NH4OH
1 ml of Ammonium hydroxide solution will contain = 56.6/111.11 g of NH4OH
1000 ml of Ammonium hydroxide solution will contain = 1000 x 56.6/111.11 = 509.405 g of NH4OH

1000 ml of Ammonium hydroxide will contain 509.405 g of NH4OH

Step 3: Calculate the number of moles of NH4OH present in 509.405 g of NH4OH.
35.05 g of NH4OH is equal to 1 mole.
1 g of NH4OH will be equal to 1/35.05 moles.
509.405 g of NH4OH will be equal to = 509.405 x 1/35.05 = 14.534 moles

Therefore, we can say that 1 liter 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.

REFERENCES

Calculator – Calculate Molarity of Ammonium Hydroxide (NH4OH) Solution
Use calculator to calculate the molarity of Ammonium hydroxide solution when concentration is given in % by mass (w/w)

Ammonium hydroxide (NH4OH) Molecular weight: 35.05 g/mol


Concentration of Ammonium Hydroxide in % (wt/wt) : % (wt/wt)

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Density of Ammonium Hydroxide: g/ml


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Molarity of Ammonium Hydroxide: 14.534 M


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