The titrate 20.00 milliliters of an unknown hc2h3o2 smaple, use 43.00 of a standard 0.1 m naoh titrant. what is the molarity of the hc2h3o2?
0.215 m
0.1000m
860.0m
0.04651 m

0.1000M

Explanation:

Given parameters:

Volume of acid = 20mL, to liter gives 2 x 10⁻²L

Volume of NaOH = 43mL to liter gives 4.3 x 10⁻²L

Molarity of NaOH = 0.1M

Unknown:

Molarity of acid = ?

Solution;

The acid is easily identified as a carboxylic acid which is called ethanoic acid. With the identification of the acid, we can write the reaction equation as shown below:

                      CH₃COOH + NaOH → CH₃COONa + H₂

Now that we have established the reaction equation, we proceed to solve for molarity of the acid given.

To solve for the molarity of the acid, we solve from the known parameters to the unknown ones. We know the details of the base NaOH used.

We first find the number of moles of NaOH as shown below:

     Number of moles of NaOH = molarity x volume of NaOH

    Number of moles of NaOH = 0.1 x 4.3 x 10⁻² = 0.0043mole

From the reaction equation, we know that:

        1 mole of NaOH reacted with 1 mole of ethanoic acid

      0.0043mole of NaOH will produce 0.0043mole of the acid

Now to find the molarity of the acid, we use the expression below:

  Molarity of acid =

Molarity of acid =   = 0.1M

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