Question

Formic acid (HCO2H, Ka = 1.8 × 10-4) is the principal component in the venom of stinging ants. What is the molarity of a formic acid solution if 25.00 mL of the formic acid solution requires 39.80 mL of 0.0567 M NaOH to reach the equivalence point?

Answer 1

Answer : The molarity of a formic acid solution is, 0.0903 M

Explanation :

To calculate the molarity of formic 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 formic acid which is $CH_3COOH$

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

As we are given:

$n_1=1\\M_1=?\\V_1=25.00ml\\n_2=1\\M_2=0.0567M\\V_2=39.80ml$

Now put all the given values in above equation, we get:

$1\times M_1\times 25.0ml=1\times 0.0567M\times 39.80ml\\\\M_1=0.0903M$

Hence, the molarity of a formic acid solution is, 0.0903 M