Chapter 4, Problem 4.71P

Interpretation Introduction

Interpretation:

The molarity of the acid solution if $25.98\text{ mL}$ of $0.1180M\text{ KOH}$ solution reacts with $52.50\text{ mL}$ of ${\text{CH}}_3\text{COOH}$ solution is to be calculated.

Concept introduction:

Strong acids and strong bases are the substance that dissociates completely into its ions when dissolved in the solution. They dissociate completely in water to release ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions.

Weak acids and weak bases are the substance that does not dissociate completely into its ions when dissolved in the solution. They dissociate partially in water to release ${\text{H}}^{+}$ ions and ${\text{OH}}^{-}$ ions.

Acetic acid $\left({\text{CH}}_{\text{3}}\text{COOH}\right)$ is a weak acid and potassium hydroxide $\left(\text{KOH}\right)$ is a strong base. Potassium hydroxide $\left(\text{KOH}\right)$ dissociates completely into ions and the acetic acid dissociates to some extent into ions. They both react to form potassium acetate and a water molecule.

The molecular equation for the acid-base reaction of acetic acid and potassium hydroxide is:

${\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+\text{KOH}\left(aq\right)\to {\text{CH}}_{\text{3}}\text{COOK}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$