Question

If 72.1 mL of 0.543 M H2SO4 completely titrates 39.0 mL of KOH solution, what is the molarity of the KOH solution? a. 0.317 M b. 0.502 M c. 1.00 M d. 2.01 M

Answer 1

Answer: The correct answer is Option  d.

Explanation:

To calculate the molarity of base, 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 $H_2SO_4$

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

We are given:

$n_1=2\\M_1=0.543M\\V_1=72.1mL\\n_2=1\\M_2=?M\\V_2=39.0mL$

Putting values in above equation, we get:

$2\times 0.543\times 72.1=1\times M_2\times 39.0\\\\M_2=2.01M$

Hence, the correct answer is Option d.