Once the stock solutions are prepared, describe how you would produce your buffe
ID: 953491 • Letter: O
Question
Once the stock solutions are prepared, describe how you would produce your buffers (four). Your explanation should be detailed and quantitative.
For my stock solution:
KH2PO4 = (136.08 g/mol)(1M)(100.00ml/1000ml)= 13.608 gm
K2HPO4 = (174.17 g/mol)(1M)(100.00ml/1000ml)= 17.4178 gm
KH2PO4= (136.08 g/mol)(1M)(250.00ml/1000ml)= 34.02 gm
K2HPO4= (174.17 g/mol)(1M)(250.00ml/1000ml)= 43.54 gm
KH2PO4= (136.08 g/mol)(1M)(25.00ml/1000ml) = 3.402 gm
K2HPO4= (174.17 g/mol)(1M)(25.00ml/1000ml) = 4.35425 gm
KH2PO4= (136.08 g/mol)(1M)(10.00ml/1000ml)= 1.3608 gm
K2HPO4= (174.17 g/mol)(1M)(10.00ml/1000ml)= 1.74178 gm
KH2PO4= (136.08 g/mol)(1M)(5.00ml/1000ml)= 0.6804 gm
K2HPO4= (174.17 g/mol)(1M)(5.00ml/1000ml)= 0.87085 gm
Explanation / Answer
For my stock solution:
KH2PO4 = (136.08 g/mol)(1M)(100.00ml/1000ml)= 13.608 gm
This information means that you prepare 100 mL of 1 M solution. Fill about half with water an 100 mL volumetric flask, dissolve the calculated mass of substance, fill with water to mark.
K2HPO4 = (174.17 g/mol)(1M)(100.00ml/1000ml)= 17.4178 gm
This information means that you prepare 100 mL of 1 M solution. Fill about half with water an 100 mL volumetric flask, dissolve the calculated mass of substance, fill with water to mark.
All other/following solutions are 1M, only the final volumes (and the volumetric flask used) are different: 250 mL, 25 mL, 10 mL, 5 mL.
KH2PO4= (136.08 g/mol)(1M)(250.00ml/1000ml)= 34.02 gm
K2HPO4= (174.17 g/mol)(1M)(250.00ml/1000ml)= 43.54 gm
KH2PO4= (136.08 g/mol)(1M)(25.00ml/1000ml) = 3.402 gm
K2HPO4= (174.17 g/mol)(1M)(25.00ml/1000ml) = 4.35425 gm
KH2PO4= (136.08 g/mol)(1M)(10.00ml/1000ml)= 1.3608 gm
K2HPO4= (174.17 g/mol)(1M)(10.00ml/1000ml)= 1.74178 gm
KH2PO4= (136.08 g/mol)(1M)(5.00ml/1000ml)= 0.6804 gm
K2HPO4= (174.17 g/mol)(1M)(5.00ml/1000ml)= 0.87085 gm
You may prepare 0.5 M buffers simply by mixing the two solutions from the same pair.
All have the same ratio HPO42-/ H2PO4-, thus the same pH = pKa2 = 7.21
In fact , I think some information is missing here.
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