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You are instructed to create 400. mL of a 0.39 M phosphate buffer with a pH of 6.2. You have phosphoric acid and the sodium s

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Answer #1

Which of the available chemicals will you use for the acid component of your buffer? : NaH2PO4

Which of the available chemicals will you use for the base component of your buffer? : Na2HPO4

What is the molarity needed for the acid component of the buffer? : 0.355 M

What is the molarity needed for the base component of the buffer? : 0.0349 M

How many moles of acid are needed for the buffer? : 0.142 mol

How many moles of acid are needed for the buffer? : 0.0140 mol

How many grams of acid are needed for the buffer? : 17.0 g

How many grams of base are needed for the buffer? : 1.98 g

Explanation

Total phosphate concentration = 0.39 M

[NaH2PO4] + [Na2HPO4] = 0.39 M ...(1)

Ka2 = 6.2 x 10-8

pKa2 = -log(Ka2)

pKa2 = -log(6.2 x 10-8)

pKa2 = 7.21

According to Henderson - Hasselbalch equation,

pH = pKa + log([conjugate base] / [weak acid])

pH = pKa2 + log([Na2HPO4] / [NaH2PO4])

6.2 = 7.21 + log([Na2HPO4] / [NaH2PO4])

log([Na2HPO4] / [NaH2PO4]) = 6.2 - 7.21

log([Na2HPO4] / [NaH2PO4]) = -1.01

[Na2HPO4] / [NaH2PO4] = 10-1.01

[Na2HPO4] / [NaH2PO4] = 0.0983   ...(2)

Solving equations (1) and (2)

[NaH2PO4] = 0.355

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