Question

20.0 ml sample of HCl, ist te of Hel is titrated with 0.0300M KOH solution. The volume of potassium droxide required is 27.50mL. What is the molarity of HCI solution? +KOH → KCI+H2O

Answer 1

Answer: The molarity of HCl is 0.041 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 KOH

We are given:

`n_1=1\\M_1=?M\\V_1=20.0mL\\n_2=1\\M_2=0.0300M\\V_2=27.50mL`

Putting values in above equation, we get:

`1* M_1* 20.0=1* 0.0300* 27.50\\\\M_1=0.041M`

Hence, the molarity of HCl is 0.041 M.