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.

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:

$2\times 0.100\times V_1=1\times 0.0821\times 25\\\\V_1=10.2625mL$

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