You are instructed to create 700. mL of a 0.40 M phosphate buffer with a pH of 7

Question

You are instructed to create 700. mL of a 0.40 M phosphate buffer with a pH of 7.3. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s)H2O) H30 (aq) H2PO4 (aq) H2PO4-(aq) + H20(,)--H3O+(aq) + HPO42-(aq) HP042-(aq) + H20(,)--H3O+(aq) + PO43-(aq) Kal = 6.9x10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13 4 (a Which of the available chemicals will you use for the acid component of your buffer? H3PO4 NaH2PO4 Na2HPO4 Na3PO4 Which of the available chemicals will you use for the base component of your buffer? H3PO4 NaH2PO4 Na2HPO4 Na3PO4 What is the molarity needed for the acid component of the buffer? What is the molarity needed for the base component of the buffer? How many moles of acid are needed for the buffer? mol

Explanation / Answer

1) we will use the combinaiton of acid and salt where the pKa values approximate the desired pH

so here we will use

Acid: NaH2PO4

Salt : Na2HPO4

pKa = 7.2

we will use the Henderson-Hasselbach equation in order to find out the ratio of acid and conjugate base that is required.

pH = pKa + log[HPO4--]/[H2PO4-]

7.3 = 7.20 + log[HPO4--]/[H2PO4-]

[HPO4--]/[H2PO4-] = 1.25

[HPO4--] =1.25[H2PO4-] ......................1

we want to create 0.4 M

so [HPO4--] + [H2PO4-] = 0.4 .....................2

Put 1 in 2

1.25[H2PO4-] + [H2PO4-] = 0.4

2.25 [H2PO4-] = 0.4

[H2PO4-] = 0.178 M (molarity of acid component)

[H2PO4-] = 0.4 -0.178 = 0.222 M

Moles of acid = molarity x volume = 0.178 X 0.7 = 0.125 moles of acid

Moles of base = moalrity X volume = 0.222 X 0.7 = 0.155 moles of base

Mass of acid = moles X molecular weight of acid = 0.125 X 120 = 15 grams

Mass of base = moles X mol wt = 0.155 x 142 = 22.01 grams

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