Chapter 4, Problem 4.91QP

Calculate the volume in milliliters of a 1.420 M NaOH solution required to titrate the following solutions.

(a)    25.00 mL of a 2.430 M HCl solution

(b)    25.00 mL of a 4.500 M H₂SO₄ solution

(c)    25.00 mL of a 1.500 M H₃PO₄ solution

Interpretation Introduction

Interpretation:

The volume of $NaOH$ solution is needed to titrate the given solution has to be determined.

Concept introduction:

Volumetric principle:

• In the neutralization process, the volume and concentration of initial components are equal to the volume and concentration of the final components.
• In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×V}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×V}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{V}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{Final}\text{concentration}\\ {\text{V}}_{\text{d}}\text{=}\text{Final}\text{volume}\end{array}$

Answer to Problem 4.91QP

The volume of $NaOH$ solution is needed to titrate is $\text{42}\text{.78mL}$

Explanation of Solution

Equation for the reaction is

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

Initial concentration of the $\text{HCl}$ solution ${\text{M}}_{\text{c}}\text{ = 2}\text{.430 M}$

Initial volume of the $\text{HCl}$  solution ${\text{V}}_{\text{c}}\text{= 25 mL}$

Final concentration of the $\text{NaOH}$ solution ${\text{M}}_{\text{d}}\text{ = 1}\text{.420M}$

The final volume of the $\text{NaOH}$ solution can be calculated as,

${\text{M}}_{\text{c}}{\text{×V}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×V}}_{\text{d}}$

$\begin{array}{l}\text{(2}\text{.430M)×(25mL)}\text{=}\text{(1}{\text{.420M)×V}}_{\text{d}}\\ \\ {\text{V}}_{\text{d}}\text{=}\frac{\text{(2}\text{.430M)×(25mL)}}{\text{(1}\text{.420M)}}\text{=42}\text{.78mL}\end{array}$

Interpretation Introduction

Interpretation:

The volume of $NaOH$ solution is needed to titrate the given solution has to be determined.

Concept introduction:

Volumetric principle:

• In the neutralization process, the volume and concentration of initial components are equal to the volume and concentration of the final components.
• In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×V}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×V}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{V}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{Final}\text{concentration}\\ {\text{V}}_{\text{d}}\text{=}\text{Final}\text{volume}\end{array}$

Answer to Problem 4.91QP

The volume of $NaOH$ solution is needed to titrate is $\text{158}\text{.4mL}$

Explanation of Solution

Equation for the reaction is

$\text{2NaOH}\text{+}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}2{\text{H}}_{\text{2}}\text{O}$

Initial concentration of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution ${\text{M}}_{\text{c}}\text{ = 4}\text{.500 M}$

Initial volume of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$  solution ${\text{V}}_{\text{c}}\text{= 25 mL}$

Final concentration of the $\text{NaOH}$ solution ${\text{M}}_{\text{d}}\text{ = 1}\text{.420M}$

The final volume of the $\text{NaOH}$ solution can be calculated as,

${\text{M}}_{\text{c}}{\text{×V}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×V}}_{\text{d}}$

$\begin{array}{l}\text{(4}\text{.500 M)×(25mL)}\text{=}\text{(1}{\text{.420M)×V}}_{\text{d}}\\ \\ {\text{V}}_{\text{d}}\text{=}\frac{\text{(4}\text{.500 M)×(25mL)}}{\text{(1}\text{.420M)}}\text{=79}\text{.22mL}\end{array}$

Since two moles of $\text{NaOH}$ are required to titrate with one mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$,

$\text{79}\text{.22mL}\times \text{2=158}\text{.4mL}$

Interpretation Introduction

Interpretation:

The volume of $NaOH$ solution is needed to titrate the given solution has to be determined.

Concept introduction:

Volumetric principle:

• In the neutralization process, the volume and concentration of initial components are equal to the volume and concentration of the final components.
• In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×V}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×V}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{V}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{Final}\text{concentration}\\ {\text{V}}_{\text{d}}\text{=}\text{Final}\text{volume}\end{array}$

Answer to Problem 4.91QP

The volume of $NaOH$ solution is needed to titrate is $\text{79}\text{.22mL}$

Explanation of Solution

Equation for the reaction is

$\text{3NaOH}\text{+}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{Na}}_{\text{3}}{\text{(PO}}_{\text{4}}\text{)}\text{+}{\text{3H}}_{\text{2}}\text{O}$

Initial concentration of the ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution ${\text{M}}_{\text{c}}\text{ = 1}\text{.500 M}$

Initial volume of the ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$  solution ${\text{V}}_{\text{c}}\text{= 25 mL}$

Final concentration of the $\text{NaOH}$ solution ${\text{M}}_{\text{d}}\text{ = 1}\text{.420M}$

The final volume of the $\text{NaOH}$ solution can be calculated as,

${\text{M}}_{\text{c}}{\text{×V}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×V}}_{\text{d}}$

$\begin{array}{l}\text{(1}\text{.500 M)×(25mL)}\text{=}\text{(1}{\text{.420M)×V}}_{\text{d}}\\ \\ {\text{V}}_{\text{d}}\text{=}\frac{\text{(1}\text{.500 M)×(25mL)}}{\text{(1}\text{.420M)}}\text{=26}\text{.40mL}\end{array}$

Since 3 moles of $\text{NaOH}$ are required to titrate with one mole of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$,

$\text{26}\text{.40mL}\times 3\text{=79}\text{.22mL}$