Question

A 25.0 mL of a solution of acetic acid of unknown concentration is titrated with 0.09926 M NaOH and the equivalence point volume was determined by graphical means to be 22.4 mL. What is the concentration of the acetic acid?

Answer 1

Answer: The concentration of acetic acid is 0.0889 M

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $CH_3COOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=25.00mL\\n_2=1\\M_2=0.09926M\\V_2=22.4mL$

Putting values in above equation, we get:

$1\times M_1\times 25.00=1\times 0.09926\times 22.4\\\\M_1=\frac{1\times 0.09926\times 22.4}{1\times 25.00}=0.0889M$

Hence, the concentration of acetic acid is 0.0889 M