Question

A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6 mL of the base is added. The concentration of acetic acid in the sample was ________ M.

Answer 1

Answer: The concentration of $CH_3COOH$ comes out to be 0.16 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:

$n_1=1\\M_1=?M\\V_1=25mL\\n_2=1\\M_2=0.175M\\V_2=26.6mL$

Putting values in above equation, we get:

$1\times M_1\times 25=1\times 0.175\times 26.6\\\\M_1=0.1862M$

Hence, the concentration of $CH_3COOH$ comes out to be 0.1862 M.