The given question is incomplete. The complete question is:
How many milliliters of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M HBr to the equivalence point?
Answer: 21.2 ml
Explanation:
To calculate the volume of base, 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]HBr[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.0720M\\V_1=25.0mL\\n_2=1\\M_2=0.0850M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0720\times 25.0=1\times 0.0850\times V_2\\\\V_2=21.2mL[/tex]
Thus 21.2 ml of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M HBr to the equivalence point
