Question

a 14.4 ml sample of vinegar, containing acetic acid, was titrated using 0.482 m naoh solution. the titration required 22.91 ml of the base. what was the molar concentration of acetic acid in the vinegar? attempts: 2 of 5 used save for later submit answer your answer is incorrect. 0.767 m

Explanation:

Step1: Write the balanced chemical equation

$CH_3COOH + NaOH
ightarrow CH_3COONa + H_2O$
The mole - ratio of $CH_3COOH$ to $NaOH$ is 1:1.

Step2: Calculate the moles of $NaOH$ used

Use the formula $n = M\times V$, where $M$ is the molarity and $V$ is the volume in liters.
$V_{NaOH}=22.91\ mL = 0.02291\ L$ and $M_{NaOH}=0.482\ M$
$n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.482\ mol/L\times0.02291\ L = 0.01104262\ mol$

Step3: Determine the moles of $CH_3COOH$

Since the mole - ratio of $CH_3COOH$ to $NaOH$ is 1:1, $n_{CH_3COOH}=n_{NaOH}= 0.01104262\ mol$

Step4: Calculate the molarity of $CH_3COOH$ in vinegar

$V_{vinegar}=14.4\ mL = 0.0144\ L$
Use the formula $M=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume in liters.
$M_{CH_3COOH}=\frac{n_{CH_3COOH}}{V_{vinegar}}=\frac{0.01104262\ mol}{0.0144\ L}\approx0.767\ M$

Answer:

$0.767\ M$