Chapter 16, Problem 76E

Interpretation Introduction

Interpretation:

The grams of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ that are required to prepare $600.0\text{mL}$ of $0.150\text{N}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution are to be calculated.

Concept introduction:

Solution is a homogeneous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. The normality of a solution is defined as the number of equivalents per liter of the solution. One equivalent of an acid is the quantity that gives $1$ mole of hydrogen ions in a chemical reaction. One equivalent of a base is the quantity that reacts with one mole of hydrogen ions.

Answer to Problem 76E

The grams of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ that are required to prepare $600.0\text{mL}$ of $0.150\text{N}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution are $7.71\text{g}$.

Explanation of Solution

The formula to calculate the normality is given below.

$\text{Normality}=\frac{\text{Equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liters}}$ …(1)

The normality of the solution is $0.150\text{N}$.

The relation between $\text{N}$ and $\text{eq}/\text{L}$ as shown below.

$1\text{N}=1\text{eq}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{N}}{1\text{eq}/\text{L}}\text{and}\frac{1\text{eq}/\text{L}}{1\text{N}}$

The conversion factor to determine $\text{eq}/\text{L}$ from $\text{N}$ is given below.

$\frac{1\text{eq}/\text{L}}{1\text{N}}$

Therefore, $0.150\text{N}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& 0.150\text{N}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\times \frac{1\text{eq}/\text{L}}{1\text{N}}\\ & =& \frac{\text{0}\text{.150}\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{\text{L}}\end{array}$

The volume of the solution is $600.0\text{mL}$.

The relation between $\text{L}$ and $\text{mL}$ is given below.

$\text{1}\text{L}=\text{1000}\text{mL}$

The probable conversion factors are given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{and}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}$

The conversion factor to determine $\text{L}$ from $\text{mL}$ is given below.

$\frac{\text{1}\text{L}}{\text{1000}\text{mL}}$

Therefore, the volume in liters is calculated below.

$\begin{array}{rll}\text{Volume}& =& 600.0\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ & =& \text{0}\text{.600}\text{L}\end{array}$

Substitute the values of normality and volume of solution in equation (1).

$\frac{\text{0}\text{.150}\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{\text{L}}=\frac{\text{Equivalents}\text{of}\text{solute}}{\text{0}\text{.600}\text{L}}$

Rearrange the above equation for the value of equivalent of solute.

$\begin{array}{rll}\text{Equivalents}\text{of}\text{solute}& =& \left(\frac{\text{0}\text{.150}\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{\text{L}}\right)\left(\text{0}\text{.600}\text{L}\right)\\ & =& 0.09\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

Therefore, the equivalents of solute, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $0.09\text{eq}$.

The reaction of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is given below.

$2\text{HCl}+\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\to {\text{BaCl}}_2+2{\text{H}}_{\text{2}}\text{O}$

In this reaction, $1$ mole of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ gives two moles of ${\text{OH}}^{-}$ ions during the reaction. Therefore, the number of equivalents per mole is $2$.

The formula to calculate the equivalent mass is given below.

$\text{Equivalent}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{\text{Number}\text{of}\text{equivalents}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}$ …(2)

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of barium is $137.3\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.008\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 137.3\text{g}{\text{mol}}^{-1}+2\times \left(1.008\text{g}{\text{mol}}^{-1}+16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 137.3\text{g}{\text{mol}}^{-1}+2\times \left(17.008\text{g}{\text{mol}}^{-1}\right)\\ & =& 137.3\text{g}{\text{mol}}^{-1}+34.016\text{g}{\text{mol}}^{-1}\\ & =& 171.3\text{g}{\text{mol}}^{-1}\end{array}$

Substitute the molar mass and number of equivalents of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ in equation (2).

$\begin{array}{rll}\text{Equivalent}\text{mass}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}& =& \frac{171.3\text{g}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{2\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}\\ & =& \frac{85.65\text{g}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{1\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}\end{array}$

The mass of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ can be determined by the formula given below.

$\text{Mass}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}=\text{Equivalents}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\times \text{Equivalent}\text{mass}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ …(3)

The equivalents of solute, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $0.09\text{eq}$.

Substitute the equivalents of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ and the equivalent mass of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ in equation (3).

$\begin{array}{rll}\text{Mass}\text{of}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}& =& 0.09\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\times \frac{85.65\text{g}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}{1\text{eq}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}\\ & =& 7.71\text{g}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

Therefore, the grams of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ that are required to prepare $600.0\text{mL}$ of $0.150\text{N}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution are $7.71\text{g}$.

Conclusion

The grams of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ that are required to prepare $600.0\text{mL}$ of $0.150\text{N}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution are $7.71\text{g}$.