Q.10.112 - An Introduction to General, Organic, and Biological Chemistry

Question

The daily output of stomach acid (gastric juice) is 1000 ml to 2000ml. Prior to a meal, stomach acid (HCl) typically has a pH of 1.42.

a. What is the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$of stomach acid?

b. One chewable tablet of the antacid Maalox contains 600mg of$CaC{O}_3$ Write the neutralization equation and calculate the milliliters of stomach acid neutralized by 2 tablets of Maalox.

c. The antacid milk of magnesia contains 400 mg per teaspoon. Write the neutralization equation, and calculate the number of milliliters of stomach acid neutralized by 1 tablespoon of milk of magnesia ( 1 tablespoon =3 teaspoons).

Step-by-Step Solution

Answer

Part (a). $3.8\times 10-2 \mathrm{M}$

Part (b). $630 ml$

Part (c). $1100 ml$

Step 1: Given Information (Part a)

The pH has some control over the accessibility of supplements, organic capacities, microbial action, and the way of behaving of synthetic compounds.

Step 2: Explanation Part (a)

We know, pH  =$1.42$

Using,

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-1.42} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.8\times {10}^{-2}\mathrm{M} \end{gathered}$$

Step 3: Explanation Part (b)

Given,

The reaction is,

$2\mathrm{HCl}\left(aq\right)+{\mathrm{CaCO}}_3\left(s\right)\to {\mathrm{CaCl}}_2\left(aq\right)+{\mathrm{CO}}_2\left(g\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)$

One chewable tablet of the antacid Maalox contains $600$mg of $CaC{O}_3$

In two tablets = $12000mg$

We know from the reaction,

$1 \mathrm{mol} {\mathrm{CaCO}}_3=100.09 \mathrm{g} {\mathrm{CaCO}}_3$ and $2 \mathrm{mol} \mathrm{HCl}=1 \mathrm{mol} {\mathrm{CaCO}}_3$ and $1\mathrm{L} \mathrm{HCl}=3.8\times {10}^{-2}\mathrm{mol} \mathrm{HCl}$

Calculating the stomach acid neutralized,

$1.20\times {10}^3\mathrm{mg}{ \mathrm{CaCO}}_3\times \frac{{\displaystyle 1\mathrm{g}}}{{\displaystyle 1000 \mathrm{mg}}}\times \frac{1 \mathrm{mol} {\mathrm{CaCO}}_3}{100.09 \mathrm{g} {\mathrm{CaCO}}_3}\times \frac{2 \mathrm{mol} \mathrm{HCl}}{1 \mathrm{mol} {\mathrm{CaCO}}_3}\times \frac{1\mathrm{L} \mathrm{HCl}}{3.8\times {10}^{-2}\mathrm{mol} \mathrm{HCl}}\times \frac{1000 \mathrm{mL} \mathrm{HCl}}{1\mathrm{L} \mathrm{HCl}}$

$=630 ml$

Step 4: Explanation Part (c)

The chemical reaction is,

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

We know,

$1\mathrm{mol} \mathrm{Mg}(\mathrm{OH}{)}_2=58.32\mathrm{g} \mathrm{Mg}(\mathrm{OH}{)}_2$

From the reaction,

$2 \mathrm{mol} \mathrm{HCl}=1\mathrm{mol} \mathrm{Mg}(\mathrm{OH}{)}_2$ and $1 \mathrm{L} \mathrm{HCl}=3.8\times {10}^{-2}\mathrm{mol} \mathrm{HCl}$

Calculating the stomach acid neutralized by one tablespoon of milk of magnesia,

$1.20\times {10}^3\mathrm{mg} \mathrm{Mg}(\mathrm{OH}{)}_2\times \frac{1\mathrm{g}}{1000\mathrm{mg}}\times \frac{1 \mathrm{mol} \mathrm{Mg} (\mathrm{OH}{)}_2}{58.32 \mathrm{g} \mathrm{Mg}(\mathrm{OH}{)}_2}\times \frac{2 \mathrm{mol} \mathrm{HCl}}{1\mathrm{mol} \mathrm{Mg}(\mathrm{OH}{)}_2}\times \frac{1\mathrm{L} \mathrm{HCl}}{3.8\times {10}^{-2}\mathrm{mol} \mathrm{HCl}}\times \frac{1000 \mathrm{mL} \mathrm{HCl}}{1\mathrm{L} \mathrm{HCl}}$

$=1100 mL$