Question

What is the molarity of H2SO3 if 250 mL of the acid was used to neutralize 200 mL of 1.5 M solution of Mg(OH)2?

Answer 1

Answer: Molarity of
`H_2SO_3` is 1.2 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
`H_2SO_3`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is
`Mg(OH)_2`

We are given:

`n_1=2\\M_1=?\\V_1=250mL\\n_2=2\\M_2=1.5M\\V_2=200mL`

Putting values in above equation, we get:

`2* M_1* 250=2* 1.5* 200\\\\M_1=1.2M`

Thus molarity of
`H_2SO_3` is 1.2 M if 250 mL of the acid was used to neutralize 200 mL of 1.5 M solution of
`Mg(OH)_2`