Q.10.53 - An Introduction to General, Organic, and Biological Chemistry

Question

If $32.8 \mathrm{mL}$ of a $0.162\mathrm{M} \mathrm{NaOH}$ 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?

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

Step-by-Step Solution

Answer

Molarity of  ${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)$ is $0.1\mathrm{M}$

Step 1: Given information

A state in which acidic and basic ions precisely neutralise each other in solution. In a solution, the concentrations of acidic and basic ions are neutralised by each other.

Step 2-Explanation (part a)

Consider the following example of a reaction:

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

Consider the following information:

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

 $=0.025 \mathrm{L}$

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

$Molarity of NaOH \left(aq\right)=0.162 M$

$Volume of NaOH(a q)=32.8 mL$

 $=0.0328 \mathrm{L}$

Step 3- Explanation (part b)

Find moles of $\mathrm{NaOH}\left(aq\right)=$Molarity  $\times$Volume

$n=0.162\mathrm{M}\times 0.0328 \mathrm{L}$

$n=0.0053136 moles$

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

So,

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

$=\frac{0.0053136}{2}$

 $=0.0026$

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

$=\frac{0.0026\mathrm{moles}}{0.025 \mathrm{L}}$

$=0.104\mathrm{M}$

$=0.1\mathrm{M}$