Q. 10.96 - An Introduction to General, Organic, and Biological Chemistry

Question

A 10.0-mL sample of vinegar, which is an aqueous solution of acetic acid,${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2$, requires $16.5 \mathrm{mL}$of a $0.500\mathrm{MNaOH}$ solution to reach the endpoint in a titration. What is the molarity of the acetic acid solution? $(10.7)$

${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\left(aq\right)+\mathrm{NaOH}\left(aq\right)⟶{\mathrm{H}}_2\mathrm{O}\left(f\right)+{\mathrm{NaC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\left(aq\right)$

Step-by-Step Solution

Answer

Molarity ${\mathrm{CH}}_3\mathrm{COOH}=\frac{0.0083{\mathrm{moleCH}}_3\mathrm{COOH}}{0.010 \mathrm{L}}$

$=0.830{\mathrm{MCHCOOH}}_3\mathrm{OOH}$

Step 1 : Given information

That are given numbers are calculated  by: 

Equalities /conversion factors

Step 2: Explanation

Step 1 Given $16.5 \mathrm{ml}$ of$0.500 \mathrm{MNaOH};10.0 {\mathrm{mlCH}}_3\mathrm{COOH}=0.010 {\mathrm{LCH}}_3\mathrm{COOH}$

Need molarity of${\mathrm{CH}}_3\mathrm{COOH}$.

Step 2 Plan

Metric Molarity mole Divide by 

$16.5 \mathrm{mlL}$moles$\mathrm{NaOH}$ mole ${\mathrm{CH}}_3\mathrm{COOHM}$

Factor factor factor liters

Step 3 Equalities/conversion factors

$\begin{array}{r}1L\mathrm{NaOH}=0.500 \mathrm{moleNaOH}\\ \frac{1L}{0.500 \mathrm{mole}}=\frac{0.500 \mathrm{mole}}{1L}\\ 1L\mathrm{NaOH}=1000\mathrm{mlNaOH}\\ \frac{1L}{1000\mathrm{ml}}=\frac{1000\mathrm{ml}}{1L}\end{array}$

Step 3: Given information

That are given equations are: 

Step 4: Explanation

Step 2 of 2

l mole$\mathrm{NaOH}=1 {\mathrm{moleCHCOOH}}_3\mathrm{COOH}$

Step 4 Set up problem

$16.5\mathrm{mLNaOH}\times \frac{1\text{ L SaOH }}{1000\mathrm{mLNaOH}}\times \frac{0.500\mathrm{mole}.\mathrm{AaOH}}{1\mathrm{LNeOH}}\times \frac{1{\mathrm{moleCHCOOH}}_3}{1 \mathrm{moleNaOH}}$

$=0.0083{\mathrm{moleCHCHOH}}_3\mathrm{OOH}$

Molarity ${\mathrm{CH}}_3\mathrm{COOH}=\frac{0.0083{\mathrm{moleCH}}_3\mathrm{COOH}}{0.010 \mathrm{L}}$

$=0.830{\mathrm{MCH}}_3\mathrm{COOH}$