Chapter U4, Problem 15STP

Interpretation Introduction

Interpretation:

The molarity of $\text{NaOH}$ needs to be determined in order to neutralize $\text{34}\text{.1 mL of NaOH}$ by $\text{15}\text{.0 mL of 0}\text{.50 M HCl}$.

Concept introduction:

A laboratory procedure said to be titration is used in which an acid sample is neutralized with a known amount of base and by using indicator in order to determine the end-point (generally determined by color change).

Using dilution law:

${\text{Molarity}}_{\text{acid}}\times {\text{Volume}}_{\text{acid}}\text{=}{\text{Molarity}}_{\text{base}}\times {\text{Volume}}_{\text{base}}$

Answer to Problem 15STP

The correct option is (a), molarity $\text{0}\text{.22 M NaOH}$.

Explanation of Solution

The complete balanced neutralization reaction that occurs between $\text{NaOH}$ and $\text{HCl}$ is:

$\text{NaOH + HCl }\to {\text{ NaCl + H}}_{\text{2}}\text{O}$

It can be observed from the balanced reaction that the mole ratio of $\text{NaOH:HCl}$ is $\text{1:1}$ so, the molarity of $\text{NaOH}$ in order to neutralize $\text{34}\text{.1 mL of NaOH}$ by $\text{15}\text{.0 mL of 0}\text{.50 M HCl}$ is calculated as:

${\text{Molarity}}_{\text{HCl}}{\text{×Volume}}_{\text{HCl}}{\text{ = Molarity}}_{\text{NaOH}}{\text{×Volume}}_{\text{NaOH}}$

Substituting the values:

$\begin{array}{l}\text{15}\text{.0 mL}\times \text{ 0}{\text{.50 M = Molarity}}_{\text{NaOH}}\text{×34}\text{.1 mL}\\ {\text{Molarity}}_{\text{NaOH}}\text{ = }\frac{\text{15}\text{.0 mL}\times \text{ 0}\text{.50 M}}{\text{34}\text{.1 mL}}\\ {\text{Molarity}}_{\text{NaOH}}\text{ = 0}\text{.2199 M }\simeq \text{ 0}\text{.22 M}\end{array}$

Since, the molarity of $\text{NaOH}$ is $\text{0}\text{.22 M}$ so, the correct option is (a).