Question

181 mL of an unknown HCl solution was neutralized in a titration with 36.2 mL of 0.250 M NaOH. What is the molarity of the unknown HCl solution

Answer 1

Answer: The molarity of the unknown HCl solution is 0.05 M

Step-by-step explanation:

To calculate the molarity 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
`HCl`

`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=?\\V_1=181mL\\n_2=1\\M_2=0.250M\\V_2=36.2mL`

Putting values in above equation, we get:

`1* M_1* 181=1* 0.250* 36.2\\\\M_1=0.05M`

Thus the molarity of the unknown HCl solution is 0.05 M