Question

How to calculate the equivalent mass of Barium hydroxide octahydrate?

Answer 1

How to calculate the equivalent mass of Barium hydroxide octahydrate?

**Solution:**

1. $Ba(OH)_2 \times 8H_2O$ is a crystalline hydrate. It is the formula to calculate the equivalent mass of crystalline hydrate:

$$M_{e(\text{substance} \times nH_2O)} = M_{e(\text{substance})} + M_{e(H_2O)} \times n;$$ $$M_{e(Ba(OH)_2 \times nH_2O)} = M_{e(Ba(OH)_2)} + M_{e(H_2O)} \times n;$$

2. $Ba(OH)_2$ is a hydroxide. It is the formula to calculate the equivalent mass of hydroxide:

$$M_{e(\text{hydroxide})} = \frac{M(\text{hydroxide})}{acidity};$$ $$M(Ba(OH)_2) = 171 \text{ g/mol};$$ $$M_{e(Ba(OH)_2)} = \frac{171}{2} = 85,5 \text{ g/mol};$$

3. $H_2O$ is an oxide. It is the formula to calculate the equivalent mass of oxide.

$M_{e(\text{oxide})} = \frac{M(\text{oxide})}{n \times \text{valence}}$; n- is a number of atoms of the oxide-forming element in the oxide molecule:

$$M(H_2O) = 18 \text{ g/mol};$$ $$M_{e(H_2O)} = \frac{18}{2 \times 1} = 9 \text{ g/mol}.$$

4. I am calculating the equivalent mass of $Ba(OH)_2 \times 8H_2O$:

$$M_{e(Ba(OH)_2 \times 8H_2O)} = 85,5 + 8 \times 9 = 157,5 \text{ g/mol}.$$

**Answer:** the equivalent mass of $Ba(OH)_2 \times 8H_2O$ is 157,5 g/mol.

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