Question

How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125...

How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125 M potassium hydrogen phthalate to give a buffer of pH 5.80? The Ka's for phthalic acid are 1.12 x 10-3 and 3.90 x 10-6.

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

Ka2   = 3.9*10^-6

PKa2   = -logKa2

           = -log(3.9*10^-6)

           = 5.41

no of moles of KHP   = molarity * volume in L

                                 = 0.125*0.1 = 0.0125moles

dipotassium phthalate (K2P)

H2P(aq) -----------> H^+ + HP^-          Ka1

HP^- ----------> H^+   + P^2-                   Ka2

PH   = PKa + log[P^2-]/[HP^-]

5.80   = 5.41 + log[P^2-]/0.0125

log[P^2-]/0.0125     = 5.80-5.41

log[P^2-]/0.0125   = 0.39

[P^2-]/0.0125   =   2.4547

[P^2-]    = 2.4547*0.0125

[P^2-]       = 0.0307 moles

no of moels of K2P( dipotassium phthalate)   = 0.0307moles

mass of K2P   = no of moles * gram molar mass

                       = 0.0307*242.3   = 7.44g

dipotassium phthalate   = 7.44g

Add a comment
Know the answer?
Add Answer to:
How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125...
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