Question #00338

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Answer 1 · 2017-07-31T04:12:32

$0.0286 M$ $0.0286M$

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

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

$NaOH (a q) + HCl (a q) \to NaCl (a q) + H_{2} O (l)$ $"NaOH"(aq) + "HCl"(aq) rarr "NaCl"(aq) + "H"_2"O"(l)$

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

$M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1 = M_2V_2$

where

We have:

Plugging in known values, we have

$(0.0249 M) (28.7 l mL) = (M_{2}) (25.00 l mL)$ $(0.0249M)(28.7color(white)(l)"mL") = (M_2)(25.00color(white)(l)"mL")$

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