Calculate the pH of a 0.4310 M aqueous solution of sodium dihydrogen phosphate ,

Question

Calculate the pH of a 0.4310 M aqueous solution of sodium dihydrogen phosphate, NaH2PO4.

Use the Tables link on the toolbar for any equilibrium constants that are required

Acid/Base Ionization Constants at 25 oC

Acid

Formula

Ka1

Ka2

Ka3

Acetic acid

CH3COOH

1.8 × 10-5

Acetylsalicylic acid (aspirin)

HC9H7O4

3.0 × 10-4

Aluminum ion

Al(H2O)43+

1.2 × 10-5

Arsenic acid

H3AsO4

2.5 × 10-4

5.6 × 10-8

3.0 × 10-13

Ascorbic acid

H2C6H6O6

7.9 × 10-5

1.6 × 10-12

Benzoic acid

C6H5COOH

6.3 × 10-5

Carbonic acid

H2CO3

4.2 × 10-7

4.8 × 10-11

Ferric ion

Fe(H2O)63+

4.0 × 10-3

Formic acid

HCOOH

1.8 × 10-4

Hydrocyanic acid

HCN

4.0 × 10-10

Hydrofluoric acid

HF

7.2 × 10-4

Hydrogen peroxide

H2O2

2.4 × 10-12

Hydrosulfuric acid

H2S

1.0 × 10-7

1.0 × 10-19

Hypochlorous acid

HClO

3.5 × 10-8

Nitrous acid

HNO2

4.5 × 10-4

Oxalic acid

H2C2O4

5.9 × 10-2

6.4 × 10-5

Phenol

C6H5OH

1.0 × 10-10

Phosphoric acid

H3PO4

7.5 × 10-3

6.2 × 10-8

3.6 × 10-13

Sulfuric acid

H2SO4

very large

1.2 × 10-2

Sulfurous acid

H2SO3

1.7 × 10-2

6.4 × 10-8

Zinc ion

Zn(H2O)42+

2.5 × 10-10

Base

Formula

Kb

Ammonia

NH3

1.8 × 10-5

Aniline

C6H5NH2

7.4 × 10-10

Caffeine

C8H10N4O2

4.1 × 10-4

Codeine

C18H21O3N

8.9 × 10-7

Diethylamine

(C2H5)2NH

6.9 × 10-4

Dimethylamine

(CH3)2NH

5.9 × 10-4

Ethylamine

C2H5NH2

4.3 × 10-4

Hydroxylamine

NH2OH

9.1 × 10-9

Isoquinoline

C9H7N

2.5 × 10-9

Methylamine

CH3NH2

4.2 × 10-4

Morphine

C17H19O3N

7.4 × 10-7

Piperidine

C5H11N

1.3 × 10-3

Pyridine

C5H5N

1.5 × 10-9

Quinoline

C9H7N

6.3 × 10-10

Triethanolamine

C6H15O3N

5.8 × 10-7

Triethylamine

(C2H5)3N

5.2 × 10-4

Trimethylamine

(CH3)3N

6.3 × 10-5

Urea

N2H4CO

1.5 × 10-14

Acid/Base Ionization Constants at 25 oC

Acid

Formula

Ka1

Ka2

Ka3

Acetic acid

CH3COOH

1.8 × 10-5

Acetylsalicylic acid (aspirin)

HC9H7O4

3.0 × 10-4

Aluminum ion

Al(H2O)43+

1.2 × 10-5

Arsenic acid

H3AsO4

2.5 × 10-4

5.6 × 10-8

3.0 × 10-13

Ascorbic acid

H2C6H6O6

7.9 × 10-5

1.6 × 10-12

Benzoic acid

C6H5COOH

6.3 × 10-5

Carbonic acid

H2CO3

4.2 × 10-7

4.8 × 10-11

Ferric ion

Fe(H2O)63+

4.0 × 10-3

Formic acid

HCOOH

1.8 × 10-4

Hydrocyanic acid

HCN

4.0 × 10-10

Hydrofluoric acid

HF

7.2 × 10-4

Hydrogen peroxide

H2O2

2.4 × 10-12

Hydrosulfuric acid

H2S

1.0 × 10-7

1.0 × 10-19

Hypochlorous acid

HClO

3.5 × 10-8

Nitrous acid

HNO2

4.5 × 10-4

Oxalic acid

H2C2O4

5.9 × 10-2

6.4 × 10-5

Phenol

C6H5OH

1.0 × 10-10

Phosphoric acid

H3PO4

7.5 × 10-3

6.2 × 10-8

3.6 × 10-13

Sulfuric acid

H2SO4

very large

1.2 × 10-2

Sulfurous acid

H2SO3

1.7 × 10-2

6.4 × 10-8

Zinc ion

Zn(H2O)42+

2.5 × 10-10

Explanation / Answer

NaH2PO4 is salt of NaOH and H3PO4.

pKa = - log Ka = - log [7.5 x 10-3]

pKa = 2.12

for such salts

pH = 1/2 [pKw + pKa + log C]

pH = 1/2 [14 + 2.12 + log 0.4310]

pH = 7.88

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