Question

Construct log(C) vs. pH diagrams for the following two acid-base systems:

0.20 M NH4Cl, 0.10 M H2CO3

List the pH & concentration of ALL species in the following solutions:

(a)

0.10 M HNaCO3

(b) 0.10 M Na2CO3

(c) 0.10 M H2CO3

(d) 0.20 M NH4Cl

(d) 0.20 M NH3

(e) 0.10 (NH4)2CO3

Answer 1

a) 0.1 M HNaCO3:

Log(C) = Log(0.1) = -1

Formula: pH = 7 + (1/2)*(pKa + LogC)

i.e. pH = 7 + (1/2)*(6.35 - 1)

i.e. pH = 9.675

b) 0.10 M Na2CO3

Log(C) = Log(0.1) = -1

Formula: pH = 7 + (1/2)*(pKa + LogC)

i.e. pH = 7 + (1/2)*(10.33 - 1)

i.e. pH = 11.665

(c) 0.10 M H2CO3

Log(C) = Log(0.1) = -1

Formula: pH = (1/2) (pKa - LogC)

i.e. pH = (1/2) (6.35 + 1)

i.e. pH = 3.675

(d) 0.2 M NH4Cl:

Log(C) = Log(0.2) = -0.699

Formula: pH = 7 - (1/2)*(pKb + LogC)

i.e. pH = 7 - (1/2)*(4.74 - 0.699)

i.e. pH = 4.98

(e) 0.20 M NH3

Log(C) = Log(0.2) = -0.699

Formula: pH = 14 - {(1/2) (pKb - LogC)}

i.e. pH = 14 - {(1/2) (4.74 + 0.699)}

i.e. pH = 11.28