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}$solution$?$

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

A conditionthat 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.

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 role="math" localid="1652628313664" $OH¯$.

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 $H2O$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.

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

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

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

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 ${\mathrm{HC}}_2{\mathrm{H}}_3\mathrm{O}\left(aq\right)$

$MolarityofKOH=0.205M$

localid="1652629388721" $VolumeofKOH=29.7mL$

$=0.0297\mathrm{L}$

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

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

$n=0.0060885moles$

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

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

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

$=0.24354\mathrm{M}$