Question

How many ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH

Answer 1

Answer: 127.5ml

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 $HNO_3$

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

We are given:

$n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.75M\\V_2=85mL$

Putting values in above equation, we get:

$1\times 0.5\times V_1=1\times 0.75\times 85\\\\V_1=127.5mL$

Thus 127.5 ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH