Chapter 19, Problem 136A

Interpretation Introduction

Interpretation : The equilibrium concentration of barium ions is to be calculated.

Concept Introduction : A chemical reaction is considered to be in a state of chemical equilibrium if both the products and the reactants are in the same concentration that does not change over time anymore. The forward reaction rate and the backward reaction rate are equal in this state.

Answer to Problem 136A

The equilibrium concentration of barium ion is $2.0\times {10}^{-8}\text{M}$ .

Explanation of Solution

The saturated solution of ${\text{BaCO}}_3$ is $\text{1}\text{.0}\text{L}$ .

The moles of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{0}\text{.25}\text{mol}$ .

The solubility product of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{5}\text{.1}\times {\text{10}}^{-9}$ .

The dissociation of ${\text{BaCO}}_3$ at equilibrium is given as

  ${\text{BaCO}}_3\left(s\right)\rightleftharpoons {\text{Ba}}^{2+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

The expression for solubility product is given as

  ${\text{K}}_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]$

Let $\left[{\text{Ba}}^{2+}\right]$ be $x$ .

Let $\left[{\text{CO}}_3^{2-}\right]$ be $x+0.25$ .

Since the solubility product is very small, let $\left[{\text{CO}}_3^{2-}\right]$ be $0.25$ .

To calculate the concentration of barium ions, substitute the values in solubility product equation.

  $\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ \text{5}\text{.1}\times {\text{10}}^{-9}=x\times 0.25\\ x=\frac{\text{5}\text{.1}\times {\text{10}}^{-9}}{0.25}\\ =2.0\times {10}^{-8}\text{M}\end{array}$