Question

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

Short 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-by-step solution

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