Question

In: Chemistry

Once the stock solutions are prepared, describe how you would produce your buffers (four). Your explanation...

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

Solutions

Expert Solution

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.

queen_honey_blossom answered 2 years ago
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