Quantitative Analysis V
Water of Crystallization

Some chemical substances contain water of crystallization. Water of crystallisation is water that is chemically bonded into a crystal structure. Substances which contains water of crystallization are termed hydrated. While those without water of crystallization are called anhydrous substances.

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$$\text{% Water of crystallization} = \frac{\text{ mass of anhydrous}}{\text{molar mass of anhydrous}} = \frac{\text{mass of water lost}}{nH_2O} \tag{1}$$ $$ \text{ % Water of crystallization} = \frac{\text{mass conc. of water}}{\text{mass conc. of hydrated}} × 100 \tag{2}$$ $$\frac{\text{ % of hydrated}}{\text{ % of anhydrous}} = \frac{\text{molar mass of hydrated}}{\text{molar mass of anhydrous}} \tag{3}$$

Example 1: 3.05g of a hydrated Na2CO3.xH2O were dissolved in water and the solution was made up to 500cm³. 25cm³ of this solution required 23.50cm³ of 0.105 mol/dm³ hydrochloric acid for complete neutralization. Determine x and hence write the formula of the hydrated salt.

Solution

$$\text{The equation of the equation} $$

Na2CO3.XH2O + 2HCl ——> 2Nacl + (x+1)H2O + CO2

$$ \text{Find the conc. of B in mol/dm³ }$$ $$ {C_A} = 0.105 mol/dm³ $$ $${V_A} = 23.50 cm³ $$ $${V_B} = 25cm³ $$ $${C_B} = \frac{0.105 × 23.50}{2 × 25} $$ $$ = 0.0494mol/dm³ $$ $$\text{Find the conc. of B in g/dm³} $$ $$ \text{If 3.05g of B is in 500cm³} $$ $$\text{xg will be in 1000cm³} $$ $$ \text{Conc of B in g/dm³} = \frac{3.05 × 1000}{500} $$ $$ = 6.10g/dm³ $$ $$\text{Find the molar mass of B }$$ $$ \text{M.m of B} = \frac{\text{Conc of B in g/dm³}}{\text{Conc. of B in mol/dm³}} $$ $$ = \frac{6.10}{0.494} = 123.5 g/mol $$ $$ \text{ since } {Na_2CO_3.xH_2O} = 123.5 $$ $$ 106 + 18x = 123.5 $$ $$ 18x = 123.5 - 106 $$ $$ x = \frac{17.5}{18} ≈ 1 $$ $$ Formula = {Na_2CO_3.H_2O} $$

Example 2: Some crystals of Washing soda on exposure to the atmosphere effloresced. 5.88g of this partly effloresced Washing soda, Na2CO3.xH2O were then dissolved in 500cm³ of water. 25cm³ of this trioxocarbonate (IV) solution required 30.0cm³ of 0.1M hydrochloric acid for complete neutralization. Calculate x and write the formula of the effloresced salt.

Solution

$$ \text{Equation of reaction} $$

Na2CO3.XH2O + 2HCl ——> 2Nacl + (x+1)H2O + CO2

$$ {C_A} = 0.1M $$ $$ {V_A} = 30cm³ $$ $$ {V_B} = 25cm³ $$ $$ {C_B} = \frac{0.1 × 30}{2 × 25} $$ $$\text{Molar conc. of B} = 0.06mol/dm³ $$ $$\text{mass conc. of B is given by:} $$ $$ {C_B} = \frac{5.88 × 1000}{500} = 11.76g/dm³ $$ $$\text{Molar mass of B is given as} $$ $$ M = \frac{\text{mass conc. of B}}{\text{molar conc. of B}} $$ $$ M = \frac{11.76}{0.06} $$ $$ M = 196 g/mol $$ $$\text{Thus, }{Na_2CO_3.xH_2O} = 196 $$ $$ 106 + 18x = 196 $$ $$ 18x = 196 - 106 $$ $$ 18x = 90 $$ $$ x = \frac{90}{18} = 5 $$ $$\text{Formula} = {Na_2CO_3.5H_2O} $$

Example 3: 25cm³ of a 0.025mol/dm³ solution of a trioxocarbonate(IV) salt required 24.50cm³ of a 0.050mol/dm³ hydrochloric acid for complete neutralization. Given that B contains 7.2g/dm³ of the hydrated trioxocarbonate(IV) salt, calculate the:

  1. Concentration of the anhydrous salt in B in g/dm³[Molar mass of the anhydrous salt in B = 106g/mol]

  2. Percentage of water of crystallization in the hydrated salt

Solution

$$\text{Mass conc. of B} = \text{Molar conc. × M} $$ $$ = 0.025 × 106 = 2.65g/dm³ $$ $$ \text{Conc of water} = 7.2 - 2.65 $$ $$ = 4.55 g/dm³ $$ $$ \text{% of water of crystallization = } $$ $$ \frac{\text{Conc. of water}}{\text{Conc. of hydrated}} × 100 $$ $$ = \frac{4.55}{7.2} × 100 $$ $$ = 63.2 \text{%} $$
Summary