Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The mass of $\text{HCl}$ neutralized by a dose of milk of magnesia containing $3.\text{26}\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is to be calculated.

Concept introduction: $\text{HCl}$ is a strong acid and magnesium hydroxide is a strong base. When both react with each other, the products formed are magnesium chloride and water. This is a type of neutralization reaction in which the strong acid that is $\text{HCl}$ react with strong base that is $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ to form salt and water.

To determine: The mass of $\text{HCl}$ neutralized by a dose of milk of magnesia containing $3.\text{26}\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ .

Answer to Problem 4.1P

Solution:

The mass of $\text{HCl}$ neutralized is $4.08\text{g}$ .

Explanation of Solution

Given information:

The mass of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is $3.26\text{g}$ .

The given reaction is,

$\text{Mg}{\left(\text{OH}\right)}_2\left(aq\right)+2\text{HCl}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{MgCl}}_2\left(aq\right)$

The molar mass of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is $58.32\text{g}/\text{mol}$ .
The molar mass of $\text{HCl}$ is $36.4\text{6}\text{g}/\text{mol}$ .
The number of moles of $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is calculated as,

$\begin{array}{c}\text{58}\text{.32}\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}=\text{1}\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\\ \text{1}\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}=\frac{\text{1}}{\text{58}\text{.32}}\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\\ \text{3}\text{.26}\text{g}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}=\frac{\text{1}}{\text{58}\text{.32}}\times \text{3}\text{.26}\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\\ =0.\text{056}\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\end{array}$

According to the given reaction, $\text{1}\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ neutralizes $2\text{mol}\text{HCl}$ .
Therefore, $0.\text{056}\text{mol}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ neutralizes $\begin{array}{c}\text{0}\text{.056}\times 2\text{mol}\text{HCl}\\ =\text{0}\text{.112}\text{mol}\text{HCl}\end{array}$

Mass of $\text{HCl}$ is calculated by the formula,

$\text{Mass}\text{of}\text{HCl}=\text{Number}\text{of}\text{moles}\text{of}\text{HCl}\times \text{Molar}\text{mass}\text{of}\text{HCl}$

Substitute the values of number of moles and molar mass of $\text{HCl}$ .

$\begin{array}{c}\text{Mass}\text{of}\text{HCl}=\text{Number}\text{of}\text{moles}\text{of}\text{HCl}\times \text{Molar}\text{mass}\text{of}\text{HCl}\\ =0.1\text{12}\text{mol}\times 36.46\text{g}/\text{mol}\\ =4.08\text{g}\end{array}$

Conclusion

The mass of $\text{HCl}$ required is $4.08\text{g}$ .