Describe how you would prepare 250 ml of 0.2M phosphate buffer, pH =12.5, given solid Na2HPO4.2H2O...

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Describe how you would prepare 250 ml of 0.2M phosphate buffer, pH =12.5, given solid Na2HPO4.2H2O...

Describe how you would prepare 250 ml of 0.2M phosphate buffer, pH =12.5, given solid Na2HPO4.2H2O and Na3PO4.H2O. pKa= 4.75

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Answer 1

Answer #1

The pKa value is given wrong. The actual value of pKa3 = 12.35

According to Henderson-Hasselbulch equation:

pH = pKa + Log[Na3PO4.H2O/Na2HPO4.2H2O]

12.5 = 12.35 + Log[Na3PO4.H2O/Na2HPO4.2H2O]

i.e. Log[Na3PO4.H2O/Na2HPO4.2H2O] = 0.15

i.e. [Na3PO4.H2O/Na2HPO4.2H2O] = 10^0.15 = 1.4125 ---- Equation 1

[Na2HPO4.2H2O + Na3PO4.H2O] = 250 mL * 0.2 mmol/mL = 50 mmol ---- Equation 2

From equations 1 and 2:

[Na2HPO4.2H2O] = 50/(1.4125+1) = 20.725 mmol

Now, [Na3PO4.H2O] = 50 - 20.725 = 29.275 mmol

The mass of Na2HPO4.2H2O required to prepare the given buffer = 20.725 mmol * 178 g/mol= 3.689 g

The mass of Na3PO4.H2O required to prepare the given buffer = 29.275 mmol * 182 g/mol = 5.328 g

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Answer 2

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