Question

A 10.0 mL sample of HNO3 was exactly neutralized by 13.5 mL of 1.0 M KOH. What is the molarity of the HNO3? Use the titrations formula. Show all work.

Answer 1

Answer: Thus molarity of $HNO_3$ is 1.35 M

Explanation:

To calculate the volume 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 $HNO_3$

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

We are given:

$n_1=1\\M_1=?M\\V_1=10.0mL\\n_2=1\\M_2=1.0M\\V_2=13.5mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 1.0\times 13.5\\\\M_1=1.35M$

Thus molarity of $HNO_3$ is 1.35 M