Question

Sulfuric acid (H2SO4) reacts with potassium hydroxide (KOH) as follows. H2SO4(aq) + 2 KOH(aq) K2SO4(aq) + 2 H2O(l) Calculate the volume of 0.100 M sulfuric acid required to neutralize 25.0 mL of 0.0821 M KOH.

Answer 1

Answer: The volume of
`H_2SO_4` comes out to be 10.2625 mL.

Step-by-step explanation:

To calculate the concentration 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_4`

`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=2\\M_1=0.100M\\V_1=?mL\\n_2=1\\M_2=0.0821M\\V_2=25mL`

Putting values in above equation, we get:

`2* 0.100* V_1=1* 0.0821* 25\\\\V_1=10.2625mL`

Hence, the volume of
`H_2SO_4` comes out to be 10.2625 mL.