Chapter 16, Problem 13PE

Interpretation Introduction

Interpretation:

The number of grams of phosphoric acid that are needed to prepare $2.50\times {10}^2\text{mL}$ of $0.20\text{N}{\text{H}}_3{\text{PO}}_{\text{4}}$, which is used in the reaction ${\text{H}}_3{\text{PO}}_{\text{4}}+\text{2NaOH}\to {\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}+2{\text{H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

The concentration of a solution is measured in term of normality. Normality is defined the number of equivalents of solute present in one liter of solution. The number of equivalent depends on the acidity or basicity of the substance. The number of equivalents is the product of normality and volume of solution.

Answer to Problem 13PE

The number of grams of phosphoric acid $2.45\text{g}$.

Explanation of Solution

The given chemical reaction is shown below.

${\text{H}}_3{\text{PO}}_{\text{4}}+\text{2NaOH}\to {\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}+2{\text{H}}_{\text{2}}\text{O}$

Since, two hydrogen ions that react with two hydroxide ion and results in the formation of two moles of water. So, there is two $\text{eq}/\text{mol}$ of ${\text{H}}_3{\text{PO}}_{\text{4}}$.

The number of grams is calculated by the formula as given below.

$\text{Number}\text{of}\text{grams}=\text{N}\times \text{L}\times \text{equivalent}\text{mass}$…(1)

Where,

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

• $\text{L}$ is the volume of solution in liter.

It is given that the normality of ${\text{H}}_3{\text{PO}}_{\text{4}}$ is $\text{0}\text{.20}\text{N}$.

The volume of solution is $2.50\times {10}^2\text{mL}$.

The conversion of $\text{mL}$ to $\text{L}$ is done as,

$1\text{mL}={10}^{-3}\text{L}$

Therefore, the conversion of $2.50\times {10}^2\text{mL}$ into $\text{L}$ is done below.

$\begin{array}{rll}2.50\times {10}^2\text{mL}& =& 2.50\times {10}^2\times {10}^{-3}\text{L}\\ & =& \text{0}\text{.250}\text{L}\end{array}$

Normality can be expressed in terms of $\text{eq}/\text{L}$.

The equivalent mass is calculated by dividing the molar mass by equivalents per mole.

Therefore,

$\text{Equivalent mass of}{\text{H}}_3{\text{PO}}_{\text{4}}=\frac{\text{Molar}\text{mass}\text{of}{\text{H}}_3{\text{PO}}_{\text{4}}}{\text{equivalents}\text{per}\text{mole}\text{of}{\text{H}}_3{\text{PO}}_{\text{4}}}$…(2)

The molar mass of ${\text{H}}_3{\text{PO}}_{\text{4}}$ is calculated by using the formula as given below.

$\text{Molar}\text{mass}\text{of}{\text{H}}_3{\text{PO}}_{\text{4}}=\left(\begin{array}{c}\text{3×Molar}\text{mass}\text{of}\text{hydrogen}+\\ \text{Molar}\text{mass}\text{of}\text{phosphorous}\text{+}\\ 4\times \text{Molar}\text{mass}\text{of}\text{oxygen}\end{array}\right)$…(3)

The molar mass of hydrogen is $1\text{g}/\text{mol}$.

The molar mass of phosphorous is $31\text{g}/\text{mol}$.

The molar mass of oxygen is $16\text{g}/\text{mol}$.

Substitute the values of molar masses of hydrogen, phosphorous and oxygen in equation (3).

$\begin{array}{rll}\text{Molar}\text{mass}\text{of}{\text{H}}_3{\text{PO}}_{\text{4}}& =& \left(\text{3×}1\text{g}/\text{mol}+31\text{g}/\text{mol}+4\times 16\text{g}/\text{mol}\right)\\ & =& 98\text{g}/\text{mol}\end{array}$

Substitute the values of molar mass and equivalents per mole of ${\text{H}}_3{\text{PO}}_{\text{4}}$ in equation (2).

$\begin{array}{rll}\text{Equivalent mass of}{\text{H}}_3{\text{PO}}_{\text{4}}& =& \frac{\text{98}\text{g}/\text{mol}}{2\text{eq}/\text{mol}}\\ & =& 49\text{g}/\text{eq}\end{array}$

Substitute the values of normality, volume, and equivalent mass of ${\text{H}}_3{\text{PO}}_{\text{4}}$ in equation (1).

$\begin{array}{rll}\text{Number}\text{of}\text{grams}& =& \text{0}\text{.20}\frac{\text{eq}}{\text{L}}\times \text{0}\text{.250}\text{L}\times 49\frac{\text{g}}{\text{eq}}\\ & =& 2.45\text{g}\end{array}$

Therefore, the number of grams of phosphoric acid that are needed to prepare $2.50\times {10}^2\text{mL}$ of $0.20\text{N}{\text{H}}_3{\text{PO}}_{\text{4}}$, which is used in the reaction is $2.45\text{g}$.

Conclusion

The number of grams of phosphoric acid $2.45\text{g}$.