Question

Pre-Lab Calculations 1. Use the Henderson-Hasselbalch equation to calculate the mass of solid sadium acetate required to mix with 50.0 mL of 0.10 M acetic acid to prepare a pH 4 buffer. Ka for acetic acid is 18 x 10 Use the Henderson-Hasselbalch equation to calculate the mass of solid ammonium chloride required to mix with 50.0 mL of 0.10 M ammonia to prepare a pH 10 buffer. The Kb for ammonia is 1.8 x 10 2. Purpose The purpose of this experiment is to explore the concept of equilibria in the context of acids and bases, to understand the nature of acid-base equilibrium, buffer solutions, and pH. Safety Precautions As with all experiments, you will be working in a potentially hazardous environment and must obey all lab safety rules. Hydrochloric acid, acetic acid, and sodium hydroxide are corrosive and will cause burns if spilled on skin or inhaled. Thi s experiment doe s not generate any hazardous waste. The solutions will be collected for pretreatment by your instructor. Materials To perform this experiment, you will need the following: Laboratory tape, la 0.1 M bels, or markers acid (HCI) Several small beakers beakers ah
0 0
Add a comment Improve this question Transcribed image text
Answer #1

POH: 14-PH = 14-10, 4 4 : 4.74 + log [4H23 [0-107 4-4.74- .. log Co 1o 0-74 Lo.10] 0.74 I0 N H 0.018197 4 0.018197 mol NH4 4Solu tion Henderson Hasselbach Eguation; base AcidJ Pkalog Kalog Cx 0) 474 8X10 [b a seJ 4 = 4.74 + log LbaseJ o.10] 4-4.74 0

Add a comment
Know the answer?
Add Answer to:
Use the Henderson-Hasselbalch equation to calculate the mass of solid sodium acetate required to mix with...
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
ADVERTISEMENT