Question

A volume of 50.0 mL of a 0.510 M HNO3 solution is titrated with 0.810 M...

A volume of 50.0 mL of a 0.510 M HNO3 solution is titrated with 0.810 M KOH. Calculate the volume of KOH required to reach the equivalence point.

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Solution:

Balanced chemical equation,

HNO3 (aq)       +     KOH (aq)   -----> KNO3 (aq) + H2O (l)

M1 = 0.510 M        M2 = 0.810 M

V1 = 50.0 mL        V2 = ?

n1 = 1                   n2 = 1

At equivalence point, the number of moles of an acid is exactly equal to number of moles of a base,

(M1V1 / n1) = (M2V2 / n2)

V2 = (M1V1 / n1) * (n2 / M2)

    = (( 0.510 * 50.0) / (1)) * (1 / 0.810)

V2 = 31.5 mL

A:- The volume of 0.810 M KOH required to reach the equivalence point is 31.5 mL

Add a comment
Know the answer?
Add Answer to:
A volume of 50.0 mL of a 0.510 M HNO3 solution is titrated with 0.810 M...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT