Question

A student wants to calculate the concentration (molarity) of a 10.0 mL vinegar student uses 40.5 mL sample. The of 0.25 M NaOH. What is the concentration (molarity) of the vinegar? Credit will only be given if you show your work. The chemical equation is:

CH₃COOH+NaOH- NaCH COO+ H₂O

Answer 1

The molarity of the vinegar is calculated by determining the moles of NaOH used in the titration and applying that number to the corresponding amount of acetic acid in vinegar, considering the one-to-one molar ratio of the reaction. The result is a vinegar molarity of 1.0125 M.

Step-by-step explanation:

To calculate the molarity of the vinegar, we can use the provided information about the titration process. The equation given is: CH3COOH + NaOH → NaCH3COO + H2O. This tells us that the acetic acid in the vinegar reacts with sodium hydroxide in a one-to-one molar ratio.

First, calculate the number of moles of NaOH used. Since the concentration of NaOH (0.25 M) and the volume used (40.5 mL) are given, we apply the formula for calculating moles: moles NaOH = M × V. Converting milliliters to liters by dividing by 1000, we get moles NaOH = 0.25 M × 0.0405 L = 0.010125 moles.

Because the reaction ratio is one-to-one, the moles of acetic acid in the vinegar will also be 0.010125. The concentration of acetic acid is then calculated by dividing the moles of acetic acid by the volume of vinegar in liters. Converting 10.0 mL of vinegar to liters gives us 0.01 L, so the concentration (molarity) of acetic acid in the vinegar is 0.010125 moles ÷ 0.01 L = 1.0125 M.