Question

2.

Calculate [H3O+]total, [OH-]total, [H3O+]water, [OH-]water, the pH and the pOH for a 3.2 x

10-2 M aqueous solution of perchloric acid, HClO4.

Compare your value for [H3O+]water in the perchloric acid solution with the value for [H3O+]water in pure water. Are they the same or different? If they are different, explain why they are different.

3. Calculate [H3O+]total, [OH-]total, [H3O+]water, [OH-]water, [Ca2+]total, the pH and the pOH for a 0.00025 M aqueous solution of calcium hydroxide, Ca(OH)2.

Answer 1

3.2 x10-2 M aqueous solution of perchloric acid, HClO4.

[H₃O⁺]_total = 3.2 x 10^-2M [H₃O⁺]_HClO4 + 10^-7M [H₃O⁺]_H2O (contribution from water)

= 3.2 x 10^-2M [H₃O⁺] ([H₃O⁺]_H2O is negligible compared to [H₃O⁺]_HClO4)

[H₃O⁺]_water [OH^-]_water = 10^-14

[OH^-]_water = 10^-7M

[H₃O⁺]_total [OH^-]_total= 10^-14

[OH^-]_total = 10^-14/[H₃O⁺]_total

= 10^-14/3.2 x 10^-2

= 0.3125 x 10^-12

p^H = -log [H₃O⁺]

   = - log 3.2 x 10^-2

= 1.495

p^OH = 14 - p^H

= 14 - 1.495

= 12.505

[H₃O⁺] is same in both HClO₄ and water

0.00025 M aqueous solution of calcium hydroxide, Ca(OH)₂.

[OH^-]_Ca(OH)2 = 2 x 0.00025 M

=0.00050 M

= 5 x 10^-4 M

  [OH^-]_water = 10^-7 M

[OH^-]_total =  [OH^-]_Ca(OH)2 +  [OH^-]_water

= 5 x 10^-4 + 10^-7 (negligible compared with [OH^-]_Ca(OH)2)

= 5 x 10^-4

[H₃O⁺]_Ca(OH)2 = 0

[H₃O⁺]_water = 10^-7

[H₃O⁺]_total[OH^-]_total = 10^-14

[H₃O⁺]_total = 10^-14/5 x 10^-4

   ^{= 0.2 x} 10^-10

[Ca₂⁺]_total = 2.5 x 10^-4

p^OH = -log[OH^-]_total

= -log(5 x 10^-4)

= 3.3

p^H = 14 - p^OH

= 14-3.3

=10.7