Question

g Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution according to the equation H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2 If you require 30.01 mL of 0.1798 M N a O H solution to titrate 10.0 mL of H C 2 H 3 O 2 solution, what is the molar concentration of acetic acid in the vinegar? Type answer:

Answer 1

Answer: The molar concentration of acetic acid in the vinegar is 0.539 M.

Explanation:

The formula used is:

$M_1V_1=M_2V_2$

where,

$M_1$ and $V_1$ are the concentration and volume of base.

$M_2$ and $V_2$ are the concentration and volume of an acid.

Given:

Molar concentration of NaOH = 0.1798 M

Volume of NaOH = 30.01 mL

Volume of acetic acid = 10.0 mL

Now putting all the given values in the above formula, we get:

$M_1V_1=M_2V_2\\\\0.1798M\times 30.01mL=M_2\times 10.0mL\\\\M_2=0.539M$

Thu, the molar concentration of acetic acid in the vinegar is 0.539 M.