Chapter 15, Problem 75A

(a)

Interpretation Introduction

Interpretation: The equation for the formation of octahydrate of barium hydroxide from the anhydrous salt needs to be written.

Concept Introduction: A hydrate is defined as a compound in which water is present in the form of ${\text{H}}_{\text{2}}\text{O}$ molecules with a definite water content by weight. They are generally crystalline solids.

Explanation of Solution

The anhydrous salt is $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ . Here, it can form hydrate after a reaction with water molecules. Since the given salt forms an octahydrate, thus, it reacts with 8 water molecules. The reaction is represented as follows:

  $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}+8{\text{H}}_{\text{2}}\text{O}\to \text{Ba}{\left(\text{OH}\right)}_{\text{2}}.8{\text{H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation: The percent by mass of water in barium hydroxide octahydrate needs to be determined.

Concept Introduction: The percentage of water by mass can be simply calculated by taking the ratio of the molar mass of water and that of barium hydroxide octahydrate.

Explanation of Solution

The formula of barium hydroxide octahydrate is $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}·8{\text{H}}_{\text{2}}\text{O}$ . Here, the molar mass of water that is ${\text{H}}_{\text{2}}\text{O}$ is 18 g/mol and molar mass of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}·8{\text{H}}_{\text{2}}\text{O}$ can be calculated as follows:

  $\begin{array}{c}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}·8{\text{H}}_{\text{2}}\text{O}=137.327+\left(2\left(16+1\right)\right)+8\left(18\right)\\ =\left(137.327+34+144\right)\text{ g/mol}\\ =315.327\text{ g/mol}\end{array}$

The mass percent of water in hydrate can be calculated as follows:

  $\begin{array}{c}\%m=\frac{8\left(18\right)}{315.327}\times 100\%\\ =45.67\%\end{array}$

Therefore, the mass percent of water in barium hydroxide octahydrate is $45.67\%$ .