Show that the hydroxide-ion concentration in an aqueous solution is 1x10-4 M when the hydronium-ion concentration is 1x10-10 M.
(H^+)(OH^-) = Kw = 1E-14
1 × 10-9
To show the relationship between the hydroxide-ion concentration ([OH-]) and the hydronium-ion concentration ([H3O+]) in an aqueous solution, we can use the concept of the autoionization of water.
In water, there is a self-ionization process in which water molecules act as both acids and bases, resulting in the formation of both hydronium ions (H3O+) and hydroxide ions (OH-). This can be represented by the following equilibrium reaction:
H2O(l) ⇌ H3O+(aq) + OH-(aq)
The equilibrium constant for this reaction is known as the ion product of water, Kw, and it is equal to the product of the concentration of hydronium ions and hydroxide ions:
Kw = [H3O+][OH-]
At 25°C, the value of Kw is approximately 1.0 × 10^-14.
Given that the hydronium-ion concentration is [H3O+] = 1.0 × 10^-10 M, we can use the value of Kw to determine the hydroxide-ion concentration.
Kw = [H3O+][OH-]
1.0 × 10^-14 = (1.0 × 10^-10) [OH-]
To find [OH-], we rearrange the equation:
[OH-] = (1.0 × 10^-14)/(1.0 × 10^-10)
[OH-] = 1.0 × 10^-14-(-10)
[OH-] = 1.0 × 10^-4 M
Thus, the hydroxide-ion concentration ([OH-]) in the aqueous solution is 1.0 × 10^-4 M when the hydronium-ion concentration ([H3O+]) is 1.0 × 10^-10 M.
To show the relation between the hydroxide-ion concentration ([OH-]) and the hydronium-ion concentration ([H3O+]), we can use the fact that the concentration of these ions in any aqueous solution is related by the self-ionization of water, given by the Kw expression:
Kw = [H3O+][OH-]
where Kw is the equilibrium constant for the self-ionization of water.
In neutral water at 25°C, Kw is equal to 1.0 x 10^-14. This means that the product of the hydronium-ion concentration and the hydroxide-ion concentration always equals 1.0 x 10^-14 in any aqueous solution.
Now, we are given the hydronium-ion concentration [H3O+] as 1.0 x 10^-10 M. We can substitute this value into the Kw expression:
Kw = [H3O+][OH-]
1.0 x 10^-14 = (1.0 x 10^-10)[OH-]
Now, let's solve for [OH-], the hydroxide-ion concentration:
[OH-] = (1.0 x 10^-14) / (1.0 x 10^-10)
[OH-] = 1.0 x 10^-14 / 1.0 x 10^-10
[OH-] = 1.0 x 10^-14-(-10)
[OH-] = 1.0 x 10^-4
Thus, the hydroxide-ion concentration in the aqueous solution is 1.0 x 10^-4 M when the hydronium-ion concentration is 1.0 x 10^-10 M.