Q. 10.52 - An Introduction to General, Organic, and Biological Chemistry

Question

What is the molarity of an acetic acid solution if $25.0 \mathrm{mL}$  of the ${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2$ solution is titrated with $29.7 \mathrm{mL}$ of a $0.205\mathrm{M}$ $\mathrm{KOH}$ solution$?$

${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\left(aq\right)+\mathrm{KOH}\left(aq\right)⟶{\mathrm{H}}_2\mathrm{O}\left(l\right)+{\mathrm{KC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\left(aq\right)$

Step-by-Step Solution

Answer

The molarity of ${\mathrm{HC}}_2{\mathrm{H}}_3\mathrm{O}\left(aq\right)$ is $0.24354\mathrm{M}$

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: Neutralization reaction

In the neutralization reaction, $H^{+}$ always reacts with $OH¯$. Therefore, the neutralization equation may require a factor to balance the acid $H^{+}$ with the base $OH¯$.

Step 1: Write down the reactants and products. 

Step 2: Balance the  acid $H^{+}$ and the base $OH¯$ and set the coefficients.

Step 3: Balance $H2 O$ with $H^{+}$ and $OH¯$.

Step 4: Write a  balance formula for the reaction. 

Neutralize acids and bases using the same number of $H^{+}$ and $OH¯$.

 

Find the number of moles using the concentration formula 

Use concentration expression to find number of moles.

$\text{ Molarity }=\frac{\text{ Number of moles of solute }}{\text{ Volume of the solution }}$

                                          $=\frac{\mathrm{n}}{\mathrm{v}}$

Number of moles $=$Molarity $\times$ volume of solution $\left(\mathrm{L}\right)$

 

                                     $M_1{V}_1=M_2{V}_2$

Step 3: Number of moles

Write down the balance response and state all the values given and the amount required.

${\mathrm{HC}}_2{\mathrm{H}}_3\mathrm{O}\left(aq\right)+\mathrm{KOH}\left(aq\right)\to {\mathrm{H}}_2\mathrm{O}\left(l\right)+{\mathrm{KC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\left(aq\right)$

Given: - Volume of ${\mathrm{HC}}_2{\mathrm{H}}_3\mathrm{O}\left(aq\right)=25.0 \mathrm{mL}$

                             $=0.025 \mathrm{L}$

Need to find the molar concentration of ${\mathrm{HC}}_2{\mathrm{H}}_3\mathrm{O}\left(aq\right)$

$Molarity of KOH=0.205 M$

$Volume of KOH=29.7 mL$

 

                           $=0.0297 \mathrm{L}$

Find number of moles of $\mathrm{KOH}=$ Molarity $x$ Volume

$\mathrm{n}=0.205\mathrm{M}\times 0.0297 \mathrm{L}$

$n=0.0060885 moles$

At the titration point, $OH¯$ and $H^{+}$are equal. in short,

Molarity of ${\mathrm{HC}}_2{\mathrm{H}}_3\mathrm{O}\left(\mathrm{aq}\right)=\frac{\mathrm{n}}{\mathrm{v}}$

                                         $=\frac{0.0060885\mathrm{moles}}{0.025 \mathrm{L}}$

 

                                           $=0.24354\mathrm{M}$