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

Step-by-step 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* M_1* 10.0=1* 1.0* 13.5\\\\M_1=1.35M`

Thus molarity of
`HNO_3` is 1.35 M