Chapter 16, Problem 47E

Interpretation Introduction

Interpretation:

The concentration of $\left[{\text{OH}}^{-}\right]$ and equilibrium constant of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is to be calculated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.