Question

Calculate the following quantities for each of the following solutions. Show your calculations on a separate sheet of paper.

	pH	pOH	[H+]	[OH-]
0.500 M HCL
0.300 M KOH
0.180 M NH4Cl

note: Kb NH3 = 1.8 * 10^-5

Answer 1

HCl is strong acid

pH = -log [H+]

= -log[0.5]

= 0.3

pH + pOH = 14

pOH = 14-0.3

= 13.7

[OH-] = 10^-13.7 = 2 x 10^-14

[H+] = 0.5

part B

since it is strong base

pOH = log[0.3]

pOH = 0.52

pH = 14-0.52 = 13.48

[H+] = 10^-pH = 3.3 x 10^-14

[OH] = 0.3

c)

0.18 M NH4Cl releases 0.18M NH4+

NH4+ in water --> H+ & NH3  

K = Kw / Kb

1 x 10^-14 / 1.8 x 10^-5

=5.56 x 10^-10  

K = [H+] [NH3] / [NH4+]

5.56 x 10^-10 = x2 / 0.18-x

x2 + 5.56 x 10^-10 x - 1 x 10^-10

x = [H+] = 9.110^-6 molar

pH = 5.04

pOH = 14 - 5.04

= 8.95

[OH] = 10^-pOH = 10 ^{- 8.95} = 1.12 x 10^-9