Question

What volume of 0.250 mol/L sulfuric acid, H2SO4(aq) is needed to react completely with 37.2 mL of 0.650 mol/L potassium hydroxide, KOH(aq)?

Answer 1

Answer: d. 48.4 ml

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 $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=0.250mol/L\\V_1=?mL\\n_2=1\\M_2=0.650mol/L\\V_2=37.2mL$

Putting values in above equation, we get:

$2\times 0.250\times V_1=1\times 0.650\times 37.2\\\\V_1=48.4ml$

Thus volume of 0.250 mol/L sulfuric acid needed is 48.4 ml