# Question 00338

Jul 31, 2017

$0.0286 M$

#### Explanation:

We're asked to find the molarity of the $\text{HCl}$ in the gastric juice with some given titration data.

First, let's write the balanced chemical equation for this neutralization reaction:

$\text{NaOH"(aq) + "HCl"(aq) rarr "NaCl"(aq) + "H"_2"O} \left(l\right)$

• Here's a hint we can use only in the special case of one-to-one molar ratios:

${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

where

• ${M}_{1}$ and ${M}_{2}$ are the molarities of the two solutions

• ${V}_{1}$ and ${V}_{2}$ are the volumes of the two solutions

We have:

• ${M}_{1} = 0.0249 M$ ($\text{NaOH}$)

• ${V}_{1} = 28.7$ $\text{mL}$

• M_2 = ? ($\text{HCl}$)

• ${V}_{2} = 25.00$ $\text{mL}$

Plugging in known values, we have

$\left(0.0249 M\right) \left(28.7 \textcolor{w h i t e}{l} \text{mL") = (M_2)(25.00color(white)(l)"mL}\right)$

M_2 = color(red)(ul(0.0286M#