Question

Classify the following each of the 0.10 M solutions as either acidic, basic, or neutral. Show equations and determine/calculate the pH of each solution. You may need K_a or K_b of original acid or base and K_w value to perform each substance.

a) NaCl

b) NH₄Cl

c) KNO₂

d) NaF

e) KNO₃

f) NH₄NO₃

g) KC₂H₃O₂

Answer 1

a) NaCl is the salt of a strong acid and strong base. Hence the solution is neutral and the pH = 7.

b) NH₄Cl is a salt of a strong acid and weak base.

K_b for NH₄OH = 1.81 X 10^-5

pH of the solution is given by equation; pH = 1/2 pK_w - 1/2 log c -1/2 pK_b

pK_w = - log K_w = - log (1 X 10^-14) = 14

pK_b = - log K_b = - log (1.81 X 10^-5) = 4.74

log c = log (10^-1) = -1

Hence, pH = 1/2 (14) - 1/2 (-1) - 1/2 (4.74)

pH = 5.13

The pH shows that the solution is acidic.

c) KNO₂ is the salt of a weak acid, HNO₂, and a strong base KOH.

K_a for KNO₂ = 4.0 x 10^-4​

pOH of the solution is given by equation; pOH = 1/2 pK_w - 1/2 log c -1/2 pK_a

pK_w = - log K_w = - log (1 X 10^-14) = 14

pK_a = - log K_b = - log (4.0 x 10^-4​​) = 3.4

log c = log (10^-1) = -1

Hence, pOH = 1/2 (14) - 1/2 (-1) - 1/2 (3.4) = 5.8

   pH = 14 - pOH = 14 - 5.8 = 8.2

The pH shows that the solution is basic.

d) NaFis the salt of a weak acid, HF, and a strong base NaOH.

K_a for NaF = 6.6 x 10^-4​

pOH of the solution is given by equation; pOH = 1/2 pK_w - 1/2 log c -1/2 pK_a

pK_w = - log K_w = - log (1 X 10^-14) = 14

pK_a = - log K_b = - log (6.6 x 10^-4​​) = 3.18

log c = log (10^-1) = -1

Hence, pOH = 1/2 (14) - 1/2 (-1) - 1/2 (3.18) = 5.91

   pH = 14 - pOH = 14 - 5.91 = 8.09

The pH shows that the solution is basic.