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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 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*

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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.
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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Please show how you got 0.06173. 

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