How would you prepare 100.00 ml of 0.100 M phosphate buffer at pH 7.00 using solid KH,PO, and K2HPO4? MM KH,PO4: 136.08 MM K,HPO: 174.2 Ka - 6.2 x 10 H2PO, = HPO,2- + H*
How would you prepare 100.00 ml of 0.100 M phosphate buffer at pH 7.00 using solid KH,PO, and K2HPO4? MM KH,PO4: 136.08 MM K,HPO: 174.2 Ka - 6.2 x 10 H2PO, = HPO,2- + H*
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 75QAP: A diprotic acid, H2B(MM=126g/moL), is determined to be a hydrate, H2B xH2O. A 10.00-g sample of this...
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How would you prepare 100.00 mL of 0.100 M phosphate buffer at pH 7.00 using solid KH,PO, and K2HPO,?
MM KH;PO4: 136.08
MM KHPO: 174.2
Ka = 6.2 x 108
H,PO,
HPO,2- + H*
RUSH. Please show the solution. Step by
step. Thank you. Will upvote after.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5cb018ed-b69c-48f3-887f-b464f37d9a5a%2F07e0bf64-fd48-4f50-a2ec-ae7b40e48486%2Fk5h7uv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Chemistry
How would you prepare 100.00 mL of 0.100 M phosphate buffer at pH 7.00 using solid KH,PO, and K2HPO,?
MM KH;PO4: 136.08
MM KHPO: 174.2
Ka = 6.2 x 108
H,PO,
HPO,2- + H*
RUSH. Please show the solution. Step by
step. Thank you. Will upvote after.
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Assuming that the buffer has y concentration of KH2PO4.
Hence the concentration of K2HPO4 in buffer = 0.100 - y
Does the 0.100 came from the conversion of 100 mL? Or is it from the given 0.100 M buffer?
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