Chapter 16, Problem 73E

Interpretation Introduction

Interpretation:

The mass of ${\text{NaHSO}}_{\text{4}}$ that must be dissolved in $7.50\times {10}^2\text{mL}$ of solution to produce $\text{0}\text{.200}\text{N}{\text{NaHSO}}_{\text{4}}$ is 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 73E

The mass of ${\text{NaHSO}}_{\text{4}}$ that must be dissolved in $7.50\times {10}^2\text{mL}$ of solution to produce $\text{0}\text{.200}\text{N}{\text{NaHSO}}_{\text{4}}$ is $18.0\text{g}$.

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 $\text{0}\text{.200}\text{N}$.

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, $\text{0}\text{.200}\text{N}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& \text{0}\text{.200}\text{N}{\text{NaHSO}}_{\text{4}}\times \frac{1\text{eq}/\text{L}}{1\text{N}}\\ & =& \frac{\text{0}\text{.200}\text{eq}{\text{NaHSO}}_{\text{4}}}{\text{L}}\end{array}$

The volume of the solution is $7.50\times {10}^2\text{mL}$.

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.

$\begin{array}{rll}\text{Volume}& =& 7.50\times {10}^2\text{mL}\times \frac{\text{1}\text{L}}{\text{1000}\text{mL}}\\ & =& \text{0}\text{.750}\text{L}\end{array}$

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

$\frac{\text{0}\text{.200}\text{eq}{\text{NaHSO}}_{\text{4}}}{\text{L}}=\frac{\text{Equivalents}\text{of}\text{solute}}{\text{0}\text{.750}\text{L}}$

Rearrange the above expression for the value of equivalents of solute in the above expression.

$\begin{array}{rll}\text{Equivalents}\text{of}\text{solute}& =& \frac{\text{0}\text{.200}\text{eq}{\text{NaHSO}}_{\text{4}}}{\text{L}}\times \text{0}\text{.750}\text{L}\\ & =& 0.15\text{eq}{\text{NaHSO}}_{\text{4}}\end{array}$

Therefore, the equivalents of solute, ${\text{NaHSO}}_{\text{4}}$ is $0.15\text{eq}$.

The ionization reaction of ${\text{NaHSO}}_{\text{4}}$ is given below.

${\text{HSO}}_{\text{4}}{}^{-}\to {\text{H}}^{\text{+}}+{\text{SO}}_{\text{4}}{}^{2-}$

In this reaction, $1$ mole of ${\text{NaHSO}}_{\text{4}}$ gives $1$ mole of ${\text{H}}^{\text{+}}$ ions during the reaction. Therefore, the number of equivalents per mole is $1$.

The formula to calculate the equivalent mass is given below.

$\text{Equivalent}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}{\text{NaHSO}}_{\text{4}}}{\text{Number}\text{of}\text{equivalents}\text{of}{\text{NaHSO}}_{\text{4}}}$ …(2)

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

The molar mass of sodium is $22.99\text{g}{\text{mol}}^{-1}$.

The molar mass of sulfur is $32.06\text{g}{\text{mol}}^{-1}$.

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

Therefore, the molar mass of ${\text{NaHSO}}_{\text{4}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 22.99\text{g}{\text{mol}}^{-1}+1.008\text{g}{\text{mol}}^{-1}+32.06\text{g}{\text{mol}}^{-1}+\left(4\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 22.99\text{g}{\text{mol}}^{-1}+1.008\text{g}{\text{mol}}^{-1}+32.06\text{g}{\text{mol}}^{-1}+64.00\text{g}{\text{mol}}^{-1}\\ & =& 120.06\text{g}{\text{mol}}^{-1}\end{array}$

Substitute the molar mass and number of equivalents of ${\text{NaHSO}}_{\text{4}}$ in equation (2).

$\text{Equivalent}\text{mass}\text{of}{\text{NaHSO}}_{\text{4}}=\frac{120.06\text{g}{\text{NaHSO}}_{\text{4}}}{\text{1}\text{eq}{\text{NaHSO}}_{\text{4}}}$

The mass of ${\text{NaHSO}}_{\text{4}}$ can be determined by the formula given below.

$\text{Mass}\text{of}{\text{NaHSO}}_{\text{4}}=\text{Equivalents}\text{of}{\text{NaHSO}}_{\text{4}}\times \text{Equivalent}\text{mass}\text{of}{\text{NaHSO}}_{\text{4}}$ …(3)

The equivalents of solute, ${\text{NaHSO}}_{\text{4}}$ is $0.15\text{eq}$.

Substitute the equivalents of ${\text{NaHSO}}_{\text{4}}$ and the equivalent mass of ${\text{NaHSO}}_{\text{4}}$ in equation (3).

$\begin{array}{rll}\text{Mass}\text{of}{\text{NaHSO}}_{\text{4}}& =& 0.15\text{eq}{\text{NaHSO}}_{\text{4}}\times \frac{120.06\text{g}{\text{NaHSO}}_{\text{4}}}{\text{1}\text{eq}{\text{NaHSO}}_{\text{4}}}\\ & =& 18.0\text{g}{\text{NaHSO}}_{\text{4}}\end{array}$

Therefore, the mass of ${\text{NaHSO}}_{\text{4}}$ that must be dissolved in $7.50\times {10}^2\text{mL}$ of solution to produce $\text{0}\text{.200}\text{N}{\text{NaHSO}}_{\text{4}}$ is $18.0\text{g}$.

Conclusion

The mass of ${\text{NaHSO}}_{\text{4}}$ that must be dissolved in $7.50\times {10}^2\text{mL}$ of solution to produce $\text{0}\text{.200}\text{N}{\text{NaHSO}}_{\text{4}}$ is $18.0\text{g}$.