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2 years ago

How much of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

Chemistry

1 answer:

Ivan2 years ago

6 0

Answer: 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

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:

$n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.05M\\V_2=25.0mL$

Putting values in above equation, we get:

$1\times 0.5\times V_1=1\times 0.05\times 25.0\\\\V_1=2.5mL$

Thus 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

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