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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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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