Examples for the maintenance of Kw
Kw is always equal to 1x10-14 in pure water, therefore the concentration of H+
and OH- ions are both equal to 1x10-7
Ex. 1 Determine the concentration of OH- when 1x10-2 M of H+ ions are added to
pure water.
Total [H+] = [H+] water + [H+] added = 1 x10-7M + 1x10-2M = 1x10-2M
Note: the total [H+] is 1x10-2M because this value is much, much larger than
1x10-7M.
The [OH-] can be calculated from the relationship [OH-] = Kw ÷ [H+]. Plug in the
numbers to get:
[OH-] = 1x10-14 ÷ 1x10-2 = 1x10-12 answer
Initially at equilibrium, both the [H+] and [OH-] = 1x10-7 M. At the new equilibrium
(after adding the extra [H+] ions), the [H+] was increased to 1x10-2 M and the [OH-]
decreased to 1x10-12 M to keep the Kw at 1x1014.
Ex.2 Determine the [H+] concentration when 2.0 grams of NaOH(s) are dissolved in
1.0 liters of pure water.
Since we are adding more [OH-], the [H+] will go down.
a. You have to calculate how many moles of NaOH you have.
2.0 grams of NaOH ÷ 40 grams/mole = 5.0 x10-2 moles of NaOH
The molar concentration of NaOH is 5.0 x10-2 moles ÷ 1.0 liters
or 5.0 x10-2 M of [NaOH]
b. Calculate the total [OH-]. This is equal to:
Total [OH-] = [OH-] from water + [OH-] added
= 1x10-7 M + 5.0 x 10-2 M
= 5.0 x 10-2 M
c. The [H+] can be calculated by the relationship:
[H+] = Kw ÷ [OH-]
[H+] = 1.0x10-14 ÷ 5.0 x10-2
= 2.0x10-13