Q. 10.54 - An Introduction to General, Organic, and Biological Chemistry

Question

If $38.2 \mathrm{mL}$ of a $0.163\mathrm{M} \mathrm{KOH}$ solution is required to titrate $25.0 \mathrm{mL}$ of a solution of ${\mathrm{H}}_2{\mathrm{SO}}_4$, what is the molarity of the ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution$?$

${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{NaOH}\left(aq\right)⟶2{\mathrm{H}}_2\mathrm{O}\left(l\right)+{\mathrm{Na}}_2{\mathrm{SO}}_4\left(aq\right)$

Step-by-Step Solution

Answer

The molar concentration of $H_2{SO}_4\left(aq\right)$ is $0.12M.$ 

Step 1: Given information

A condition that exists when acidic and basic ions  neutralize each other accurately in solution. A state in which the concentrations of acidic ions and basic ions in a solution are neutralized with each other.

Step 2- Explanation (part a)

Imagine the reaction given  as follows: 

${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{NaOH}\left(aq\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(l\right)+{\mathrm{Na}}_2{\mathrm{SO}}_4\left(aq\right)$

Think of the given data as follows: 

$\text{ Volume of }{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)=25.0 \mathrm{mL}$

                     $=0.025 \mathrm{L}$

Molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)$ is calculate.

Molarity of  $\mathrm{KOH}\left(aq\right)=0.163\mathrm{M}$

Volume of   $\mathrm{KOH}\left(aq\right)=38.2 \mathrm{mL}$

                       

                    $=0.0382 \mathrm{L}$

Step 2- Explanation (part b)

 Find the number of moles of volume 

$\text{ Find number moles of }\mathrm{KOH}\left(aq\right)=\text{ Molarity }x\text{ Volume }$

$n=0.163\mathrm{MX}0.0382 \mathrm{L}$

$n=0.0062266 moles$

At the titration point, $2$moles of $KOH\left(aq\right)$ are required to titrate with $1$mole of $H_2 {SO}_4$.

So,

$\text{ Number of moles of }{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\mathrm{n}}{2}$

                                      $=\frac{0.0062266}{2}$

                                           $=0.0031$

Molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)=\frac{\mathrm{n}}{\mathrm{v}}$

                 

                                     $=\frac{0.0031\mathrm{moles}}{0.025 \mathrm{L}}$

                                    $=0.124\mathrm{M}$

                                   $=0.12\mathrm{M}$