Question

1. Calculate the masses of the conjugate acid and base needed to prepare each of t listed in the table on the left. (pKa of H,PO 6.70) Assume the availability of the phosphate salf listed in the table on the right. Fill out the worksheet below. Show all work on separate pages for full credit. кн-РО,/К, НРО, mol NaH,PO/Na2HPO Molar mass (g Salt (0.25 M, 100 mL) (0.25 M, 100 mL) 136.09 кн, РО, K2HPO NaH,PO H2O Na,HPO 7 H2O pH 6.50 (Solution 1) pH 6.70 (Solution 2) pH 6.90 (Solution 3) 174.2 137.99 pH 6.50 (Solution 4) pH 6.70 (Solution 5) pH 6.90 (Solution 6) 268.09 рH .90 pH 6.50 pH 6.70 Moles of H2PO Moles of HPO, Solution 3 Solution 1 Solution 2 Grams of KH2PO Grams of K HPO Solution 6 Solution 5 Solution 4 Grams of NaH2 PO, H2O Grams of Na, HPO 7H2O

Answer 1

For any buffer, according to Henderson Hasselbalch equation:

pH = pka + log(HPO₄^2-)/(H₂PO₄^-)

Given, pka of H₂PO₄^- = 6.70

1 mole of HPO₄^-2 = 1 mole of K₂HPO₄ = 1 mole of Na₂HPO₄.7H₂O

1 mole of H₂PO₄^- = 1 mole of KH₂PO₄ = 1 mole of NaH₂PO₄.H₂O

pH = 6.50

Substituting in Henderson Hasselbalch equation,

6.50 = 6.70 + log [HPO₄^2-]/[H₂PO₄^-]

[HPO₄^2-]/[H₂PO₄^-] = 10^6.50-6.70 = 0.631

Assuming, [HPO₄^2-] = x M

[H₂PO₄^-] = 0.25 - x

So, x / (0.25 - x) = 0.63

On solving, x = 0.097 M

[HPO₄^2-] = 0.097 M

Volume of buffer = 100 ml or 0.100 L

Moles of HPO₄^2- = 0.097 M x 0.100 L = 0.0097 moles

[H₂PO₄^-] = 0.25 M - 0.097 M = 0.153 M

Moles of H₂PO₄^- = 0.153 M x 0.100 L = 0.0153 moles

Solution 1

Moles of KH₂PO₄ required = 0.0153 mol

Grams of KH₂PO₄ = 0.0153 mol*136.09 g/mol = 2.082 g

Moles of K₂HPO₄ required = 0.0097 mol

Grams of K₂HPO₄ = 0.0097 mol * 174.2 g/mol = 1.690 g

Solution 4

Moles of NaH₂PO₄.H₂O required = 0.0153 mol

Grams of NaH₂PO₄.H₂O = 0.0153 mol*137.99 g/mol = 2.111 g

Moles of Na₂HPO₄.7H₂O required = 0.0097 mol

Grams of Na₂HPO₄.7H₂O = 0.0097 mol * 268.09 g/mol = 2.600 g

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pH = 6.70

At pH = pka

[H₂PO₄^-] = [HPO₄^2-] = 0.25/2 = 0.125 M

Moles of H₂PO₄^- = 0.125 M * 0.100 L = 0.0125 mol

Moles of HPO₄^2- = 0.0125 mol

Solution 2

Moles of KH₂PO₄ required = 0.0125 mol

Grams of KH₂PO₄ = 0.0125 mol*136.09 g/mol = 1.701 g

Moles of K₂HPO₄ required = 0.0125 mol

Grams of K₂HPO₄ = 0.0125 mol * 174.2 g/mol = 2.178 g

Solution 5

Moles of NaH₂PO₄.H₂O required = 0.0125 mol

Grams of NaH₂PO₄.H₂O = 0.0125 mol*137.99 g/mol = 1.725 g

Moles of Na₂HPO₄.7H₂O required = 0.0125 mol

Grams of Na₂HPO₄.7H₂O = 0.0125 mol * 268.09 g/mol = 3.351 g

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pH = 6.90

Substituting in Henderson Hasselbalch equation,

6.50 = 6.70 + log [HPO₄^2-]/[H₂PO₄^-]

[HPO₄^2-]/[H₂PO₄^-] = 10^6.90-6.70 = 1.58

Assuming, [HPO₄^2-] = x M

[H₂PO₄^-] = 0.25 - x

x/0.25 - x = 1.58

x = 0.153 M

[HPO₄^2-] = 0.153 M

Volume of buffer = 100 ml or 0.100 L

Moles of HPO₄^2- = 0.153 M x 0.100 L = 0.0153 moles

[H₂PO₄^-] = 0.25 M - 0.153 M = 0.097 M

Moles of H₂PO₄^- = 0.097 M x 0.100 L = 0.0097 moles

Solution 5

Moles of KH₂PO₄ required = 0.0097 mol

Grams of KH₂PO₄ = 0.0097 mol*136.09 g/mol = 1.320 g

Moles of K₂HPO₄ required = 0.0153 mol

Grams of K₂HPO₄ = 0.0153 mol * 174.2 g/mol = 2.665 g

Solution 6

Moles of NaH₂PO₄.H₂O required = 0.0097 mol

Grams of NaH₂PO₄.H₂O = 0.0097 mol*137.99 g/mol = 1.338 g

Moles of Na₂HPO₄.7H₂O required = 0.0153 mol

Grams of Na₂HPO₄.7H₂O = 0.0153 mol * 268.09 g/mol = 4.102 g