Q. 22 - An Introduction to General, Organic, and Biological Chemistry

Question

 In a teaspoon $5 mL$ of a common liquid antacid, there are
$400 mg$ of $Mg{\left(OH\right)}_2$ and $400 mg$ of $Al{\left(OH\right)}_3$. A $0.080 M$$HCl$ solution, which is similar to stomach acid, is used to
neutralize $5.0 mL$ of the liquid antacid.

a. Write the balanced chemical equation for the neutralization
of $HCl$  and $Mg{\left(OH\right)}_2$
b. Write the balanced chemical equation for the neutralization
of $HCl$ and $Al{\left(OH\right)}_3$
c. What is the pH of the $HCl$ solution?
d. How many milliliters of the $HCl$ solution are needed to
neutralize the $Mg{\left(OH\right)}_2$?
e. How many milliliters of the $HCl$ solution are needed to
neutralize the $Al{\left(OH\right)}_3$?

Step-by-Step Solution

Answer

Part(a) The balanced chemical equation is $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right) + 2\mathrm{HCl}\left(\mathrm{aq}\right) \to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) + {\mathrm{MgCl}}_2\left(\mathrm{aq}\right)$

Part(b) The balanced chemical equation is $Al{\left(OH\right)}_3+ 3HCl\Rightarrow AlC{l}_3+3H_{2}O$

Part(c) pH of $HCl$ solution is $1.5-3.5$

Part(d)  $86.25\times {10}^{-3} mL$ of $HCl$ solution is needed to neutralize.

Part(e)  $64.125\times {10}^{-3} mL$ of  $HCl$ solution is needed to neutralize.

Part(a) Step 1 : Given information

We are given that a teaspoon of common liquid antacid is taken and to neutralize it we have taken $HCl$

Part(a) Step 2 : Simplify

As we are using $Mg{\left(OH\right)}_2$ to neutralize $HCl$, therefore,

The balanced chemical equation is $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right) + 2\mathrm{HCl}\left(\mathrm{aq}\right) \to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) + {\mathrm{MgCl}}_2\left(\mathrm{aq}\right)$

Part(b) Step 1 : Given information

We are given that a teaspoon of common liquid antacid is taken and to neutralize it we have taken $HCl$

Part(b) Step 2 : Simplify

As we are using $Al{\left(OH\right)}_3$ to neutralize $HCl$, therefore,

The balanced chemical equation is data-custom-editor="chemistry" $Al{\left(OH\right)}_3+ 3HCl\Rightarrow AlC{l}_3+3H_{2}O$

Part(c) Step 1 : Given information

We are given that a teaspoon of common liquid antacid is taken and to neutralize it we have taken $HCl$

Part(c) Step 2 : Simplify

As we Know $HCl$ is highly acidic. so its pH should be less than $7$

pH of $HCl$ solution is $1.5-3.5$

Part(d) Step 1 : Given information

We are given that a teaspoon of common liquid antacid is taken and to neutralize it we have taken $HCl$

Part(d) Step 2 : Simplify

From the balanced reaction $\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right) + 2\mathrm{HCl}\left(\mathrm{aq}\right) \to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) + {\mathrm{MgCl}}_2\left(\mathrm{aq}\right)$

We can say millimoles of $Mg{\left(OH\right)}_2$ is equal to two times millimoles of $HCl$

molar mass of $Mg{\left(OH\right)}_2=58 g$

moles of $Mg{\left(OH\right)}_2=\frac{400\times 10-3}{58}=6.9\times {10}^{-3}$

Therefore,

$$\begin{gathered} 6.9\times {10}^{-3}=0.080\times V \\ V=86.25\times {10}^{-3} mL \end{gathered}$$

Part(e) Step 1 : Given information

We are given that a teaspoon of common liquid antacid is taken and to neutralize it we have taken $HCl$

Part(e) Step 2 : Simplify

From the balanced reaction is $Al{\left(OH\right)}_3+ 3HCl\Rightarrow AlC{l}_3+3H_{2}O$

We can say millimoles of $Al{\left(OH\right)}_3$ is equal to three times millimoles of $HCl$

molar mass of $Al{\left(OH\right)}_3=78g$

moles of $Al{\left(OH\right)}_3=\frac{400\times {10}^{-3}}{78g}=5.13\times {10}^{-3}$

Therefore, $$\begin{gathered} 5.13\times {10}^{-3}=0.080V \\ V=64.125\times {10}^{-3} mL \end{gathered}$$