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? $(10.7)$

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

Short 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-by-step solution

Given information 

That are given numbers are calculated by:

Equalities /conversion factors

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

localid="1654585928067" $16.5\mathrm{mlL}$moles$\mathrm{NaOH}$mole ${\mathrm{CH}}_3\mathrm{COOHM}$

Factor factor factor liters

Step 3 Equalities/conversion factors

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

Given information

That are given equations are:

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

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