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:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.05M\\V_2=25.0mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.5\times V_1=1\times 0.05\times 25.0\\\\V_1=2.5mL[/tex]
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
