Q. 70 - An Introduction to General, Organic, and Biological Chemistry

Question

How many grams of  ALOH), are required to neutralize 150, mL of stomach acid with a pH of $1.5$ ?

Step-by-Step Solution

Answer

The usual volume of stomach fluid is $20\text{ to }100\mathrm{mL}$, and the pH of the stomach is acidic $\left(1.5\text{ to }3.5\right)$. In some circumstances, these figures are translated to milliequivalents per hour  (mEq/hr), which represents actual acid production.

Step:1 Introduction

One ${\mathrm{H}}^{+}$ usually responds with ${\mathrm{OH}}^{-}$ in a neutralising reactions. As a result, a neutralisation equation may necessitate parameters to balance the ${\mathrm{H}}^4$ from the acid and the ${\mathrm{OH}}^{-}$ from either the base.

$\text{ Number of moles }=\text{ molarity }x\text{ volume }\left(L\right)$

$\text{ Molar mass of }\mathrm{Al}(\mathrm{OH}{)}_3=78.0036\mathrm{g}/\mathrm{mole}$

$\text{ Mass of solute }=\text{ molar mass }x\text{ mol }$

Step:2 Explanation

Given: - stomach acid volume $=150.0\mathrm{mL}=0.150\mathrm{L}$stomach acid pH $=1.5$

Acid reflux neutralisation response The combination $\mathrm{HCl}$ with $\mathrm{Al}(\mathrm{OH}{)}_3$ is as follows:

$\mathrm{HCl}\left(\mathrm{aq}\right)+\mathrm{Al}\left(\mathrm{OH}{)}_3\right(s)?{\mathrm{AlCl}}_3(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(l)$

${\mathrm{H}}^{+}$

Balance On both sides, three molecules of $\mathrm{HCl}$ are required.

$3\mathrm{HCl}\left(aq\right)+\mathrm{Al}\left(\mathrm{OH}{)}_3\right(s)?{\mathrm{AlCl}}_3(aq)+3{\mathrm{H}}_2\mathrm{O}(\circ )$

Because the number of $\mathrm{H}$ on both sides is equivalent and other such elements are equal, the initial response is balanced for neutralisation.

To neutralise one mole of $\mathrm{Al}(\mathrm{OH}{)}_3$, we require three moles of $\mathrm{HCl}$

Step: 3 Calculation of Moles

acid reflux concentration $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-1.5}=0.031622\mathrm{M}$

Contents of the stomach moles $=0.031622\mathrm{M}\times 0.150\mathrm{L}=0.004734\text{ moles }$ 

In order to neutralise stomach acid and $\mathrm{Al}(\mathrm{OH}{)}_3$

caption Molar mass of the startaligned

Moles of  $\mathrm{Al}(\mathrm{OH}{)}_3=\frac{n}{3}=\frac{0.004734}{3}$

                                      $=0.0016$

Moles of  $\mathrm{Al}(\mathrm{OH}{)}_3=0.0016\text{ moles }$

Moles of  $\mathrm{Al}(\mathrm{OH}{)}_3=0.0016\text{ moles }\times 78.0036\mathrm{g}/\mathrm{mol}$

Moles of $\mathrm{Al}(\mathrm{OH}{)}_3=0.125\mathrm{g}$