You are instructed to create 900. mL of a 0.47 M phosphate buffer with a pH of 6

Question

You are instructed to create 900. mL of a 0.47 M phosphate buffer with a pH of 6.7. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available.

Ka1 = 6.9103

Ka2 = 6.2108

Ka3 = 4.81013

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?

How many moles of base are needed for the buffer?

How many grams of acid are needed for the buffer?

How many grams of base are needed for the buffer?

H3PO4(s) + H2O(l) H3O+(aq) + H2PO4(aq)    

Ka1 = 6.9103

H2PO4(aq) + H2O(l) H3O+(aq) + HPO42(aq)    

Ka2 = 6.2108

HPO42(aq) + H2O(l) H3O+(aq) + PO43(aq)

Ka3 = 4.81013

Explanation / Answer

ANSWER

Use H-H equation

pH = pKa + log[base]/[Acid]

Since there are three Ka Values hence three pKa values. We will choose that pKa value which is close to the rquired pH. Henc we will choose pKa2

pKa = -logKa = - log 6.2 X 10-8 = 7.2

6.7 = 7.2 + log[base]/[Acid]

log[base]/[Acid] = - 0.5

log[Acid] / [base] = 0.5

[Acid] / [base] = 0.301

[Acid] = 0.301 [base]

As per question [Acid] + [base] = 0.47M

0.301 [base] +  [base] = 0.47 M

1.301 [base] = 0.47M

[base] = 0.47 / 1.301 = 0.36M Answer to part A

[Acid] = 0.47 - 0.36 = 0.109M     Answer to part B

(C) No. of moles of acid = Volume X Molarity = 0.9L X 0.109 = 0.098 moles

900mL = 0.9L

(D) No. of moles of base = Volume X Molarity = 0.9 x 0.36 = 0.324 Moles

(E) mass of acid = Molar mass of of acid (H2PO4)  X No. of moles = 96 X 0.109 = 10.46g

(F)Mass of base = Molar mass of of base  X No. of moles

Base is not mentioned in he question. Suppose it is NaOH

Mass of base = 40 X 0.36 = 1.44g

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