You are instructed to create 400. mL of a 0.31 M phosphate buffer with a pH of 7

Question

You are instructed to create 400. mL of a 0.31 M phosphate buffer with a pH of 7.2. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.)

Ka1 = 6.9103

Ka2 = 6.2108

Ka3 = 4.81013

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

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

How many moles of acid are needed for the buffer?

______mol

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

How many grams of acid are needed for the buffer?

______g

How many grams of base are needed for the buffer?

______g

Also given:

Buffer

pKa

Phosphate (pKa1)

2.15

Citrate (pKa1)

3.06

Formate

3.75

Succinate (pKa1)

4.21

Citrate (pKa2)

4.76

Acetate

4.76

Pyridine

5.23

Citrate (pKa3)

5.40

Succinate (pKa2)

5.64

PIPES (Piperazine-N,N-bis(2-ethanesulphonic acid)

6.76

ACES (N-2-Acetamido-2-aminoethanesulphonic acid)

6.90

Imidazole

6.95

MOPS (3-(N-Morpholino)propanesulphonic acid)

7.20

Phosphate (pKa2)

7.20

TES (2-[Tris(hydroxymethyl)methylamino]ethanesulphonic acid])

7.50

HEPES (N-2-Hydroxyethylpiperazine-N`-2-ethanesulphonic acid

7.55

Tris H2NC(CH2OH)3

8.06

Tricine (N-[Tris(hydroxymethyl)methyl]glycine)

8.15

Glycine

9.78

Phosphate (pKa3)

12.43

Magnesium Oxide (MgO)

13.09

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

You need to use the following equation:

pH = pKa + log([B]/[A])

Now, you want to prepare a buffer with a pH of 7.2, so you need to use the pKa value that it's closer to the pH. In this case, if you calculate the pKa for all the Ka:

pK1 = -log(6.9x10-3) = 2.16

pK2 = -log(6.2x10-8) = 7.21

pK3 = -log(4.8x10-13) = 12.3

So the pK value to use it's the second one so:

7.20 = 7.21 + log[B]/[A]

-0.01 = log[B]/[A]

[B]/[A] = 0.977

We also know that the buffer should have 0.31 M so:

A + B = 0.31 and if B/A = 0.977 then B = 0.977A

A + 0.977A = 0.31

1.977A = 0.31

A = 0.31/1.977

[A] = 0.157 M

[B] = 0.977(0.157)

[B] = 0.153 M

Moles of A = 0.157 x 0.4 = 0.0628 moles of A

Moles of B = 0.153 x 0.4 = 0.0612 moles of B

mass of A = moles x MM

MMA = 3x1 + 31 + 4x16 = 98 g/mol

mass of A = 0.0628 x 98 = 6.15 g of A

MMB (Na2HPO4) = 23x2 + 1 + 31 + 4x16 = 142 g/mol

mass of B = 0.0612 x 142 = 8.69 g of B

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