Chapter 9.6, Problem 9.16WE

Interpretation Introduction

Interpretation:

The required volume of $\text{0}\text{.203 M NaOH}$ to neutralize the $\text{25}\text{.0 mL of a 0}{\text{.188 M H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution should be calculated.

Concept introduction:

Neutralization reaction:

• The reaction of acid and base to gives salt is known as neutralization reaction.
• The equal volume and concentrations of acid is equally react with equal volume and concentrated base.
• In end point the ionic concentration is zero.

Volumetric principle:

• The relationship between initial and final concentrations and volumes of solutions in titration process are given in the volumetric equation and it is,

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

Answer to Problem 9.16WE

The required volume of $\text{0}\text{.203 M NaOH}$ to neutralize the $\text{25}\text{.0 mL of a 0}{\text{.188 M H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $\text{23}\text{.15}\text{mL}$

Explanation of Solution

The required volume of $\text{0}\text{.203 M NaOH}$ to neutralize the $\text{25}\text{.0 mL of a 0}{\text{.188 M H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution

The balanced equation is,

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+2NaOH}}_{\text{(aq)}}\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Given,

Molarity of $NaOH=0.203\text{ M}$

Molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}=\text{0}\text{.188 M}$

Volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}=25.0\text{ mL}$

The required volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ to neutralize the given $\text{NaOH}$ solution is,

$\begin{array}{l}\text{NaOH}\text{volume}\text{=}\frac{\text{0}\text{.188 M×25}\text{.0}\text{mL}}{0.203\text{M}}\\ \text{=}\text{23}\text{.15}\text{mL}\end{array}$

• ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is the diprotic acid so it required double the amount of $\text{NaOH}$ solution.
• The given molarities and volume of and are plugged in the above equation to give the required volume of $\text{NaOH}$ to neutralize the given ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.
• The required volume of $\text{0}\text{.203 M NaOH}$ to neutralize the $\text{25}\text{.0 mL of a 0}{\text{.188 M H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $\text{23}\text{.15}\text{mL}$

Conclusion

The required volume of $\text{0}\text{.203 M NaOH}$ to neutralize the $\text{25}\text{.0 mL of a 0}{\text{.188 M H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution was calculated.