You are instructed to create 200. mL of a 0.63 M phosphate buffer with a pH of 6

Question

You are instructed to create 200. mL of a 0.63 M phosphate buffer with a pH of 6.0. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.)

H3PO4(s) + H2O(l) equilibrium reaction arrow H3O+(aq) + H2PO4(aq) Ka1 = 6.9103

H2PO4(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + HPO42(aq) Ka2 = 6.2108

HPO42(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + PO43(aq) Ka3 = 4.81013

What is the molarity needed for the acid component of the buffer? 0.593

What is the molarity needed for the base component of the buffer? 0.0371

How many moles of acid are needed for the buffer?

How many moles of base are needed for the buffer? 0.00741

How many grams of acid are needed for the buffer?

How many grams of base are needed for the buffer?

Explanation / Answer

To create 200.0 mL of a 0.63 M phosphate buffer, total moles of buffer = 200 ml x 0.63M = 126 m moles

pH of buffer = 6.0 ,so use K2

pKa = -log (Ka)

PKa = - log (6.2 x 10^8) = 7.2

We can calculate (base)/(acid) by the Henderson-Hasselbalch equation

pH = pKa + log ( base)/( acid)

6 = 7.2 + log (base)/(acid)

log (base)/(acid) = 6 - 7.2 = -1.2

(base)/ (acid) = 10^ -1.2

(base)/ (acid) = 0.0630

Acid + base = 126 m moles

Moles of acid = 118.532 m moles of acid = 0.1185 moles

Moles of base = 7.468 m moles = 0.007468 moles

i)The molarity needed for the acid component of the buffer = moles of acid/volume of buffer in L

The molarity of acid = 0.1185 moles/0.200L = 0.593M

ii) The molarity needed for the base component of the buffer = moles of base/volume of buffer in L

The molarity of base = 0.0075 moles/0.200L = 0.037M

iii) Moles of acid are needed for the buffer    = 0.119 moles

iv) Moles of base are needed for the buffer = 0.007 moles

v) Grams of acid are needed for the buffer -

Molar mass of Monosodium phosphate = 119.98 g/mol

Grams of acid = Moles of acid x Molar mass of Monosodium phosphate

0.119 moles x 119.98 g/mol = 14.277 g = 14.3 g

vi Grams of base are needed for the buffer

Sodium hydrogen phosphate Molar mass= 141.96 g/mol

Grams of base = Moles of base x Molar mass of Sodium hydrogen phosphate

Grams of base = 0.0075moles x 141.96 g/mol = 1.06 g

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