Question

1. You need to prepare 1.000 L of 0.50 M phosphate buffer, pH=6.21. Use the Henderson-Hasselbalch...

1. You need to prepare 1.000 L of 0.50 M phosphate buffer, pH=6.21. Use the Henderson-Hasselbalch equation with a value of 6.64 for pK2 to calculate the quantities of K2HPO4 and K2H2PO4 you need to add to the flask. Record the steps of these calculations and use them as a guide for the calculation you will have to make in the laboratory. (Calculate the intermediate values to at least one more significant figure than required.)

What is the [K2HPO4]/[K2H2PO4] ratio you will need? What concentrations of [K2HPO4] and [K2H2PO4] will you need to make the total concentration 0.50 M? (Units required.)

How many mole of K2HPO4 and K2H2PO4 will you need? (unit required.)

What mass of K2HPO4 and K2H2PO4 will you need? (unit required.) (FW of K2HPO4= 174.3 g/mol. FW of K2H2PO4 = 136.1 g/mol.)

Homework Answers

Answer #1

pH = 6.21

pK2 = 6.64

pH = pK2 + log([Base]/[Acid]) (Using HH equation)

6.21 = 6.64 + log ([K2HPO4] / [KH2PO4])

[K2HPO4] / [KH2PO4] = 0.65 ... (1)

Moles of K2HPO4 and KH2PO4 required = 0.5 M * 1L = 0.5 moles

[K2HPO4] + [KH2PO4] = 0.5 ... (2)

Using equations 1 and 2, we get:

[KH2PO4] = 0.30 M

[K2HPO4] = 0.20 M

Moles of KH2PO4 = 0.30 M * 1 L = 0.30 moles

Moles of K2HPO4 = 0.20 M * 1 L = 0.20 moles

Mass of KH2PO4 = 0.30 moles * 136.1 g/mol = 41.23 g

Mass of K2HPO4 = 0.20 moles * 174.3 g/mol = 34.35 g

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