Calculate [H3o+] in each aqueous solution at 25*C, and classify each solution as acidic or basic.
a. [OH-]= 1.1 x 10^-9M
b. [OH-]= 2.9 x 10^-2M
c. [OH-]= 6.9 x 10^-12M
To calculate the [H3O+] in each solution, we will use the equation for the auto-ionization of water:
H2O ⇌ H+ + OH-
The equilibrium constant for this reaction is represented by Kw, which is equal to the concentration of H+ ions multiplied by the concentration of OH- ions. At 25°C, Kw is equal to 1.0 x 10^-14.
Using the equation Kw = [H3O+][OH-], we can rearrange it to solve for [H3O+]:
[H3O+] = Kw / [OH-]
Now, let's calculate the [H3O+] for each solution:
a. [OH-] = 1.1 x 10^-9 M
[H3O+] = 1.0 x 10^-14 / 1.1 x 10^-9 ≈ 9.1 x 10^-6 M
Since the [H3O+] in solution a is greater than 1.0 x 10^-7 M, the solution is acidic.
b. [OH-] = 2.9 x 10^-2 M
[H3O+] = 1.0 x 10^-14 / 2.9 x 10^-2 ≈ 3.4 x 10^-13 M
Since the [H3O+] in solution b is less than 1.0 x 10^-7 M, the solution is basic.
c. [OH-] = 6.9 x 10^-12 M
[H3O+] = 1.0 x 10^-14 / 6.9 x 10^-12 ≈ 1.4 x 10^-3 M
Since the [H3O+] in solution c is greater than 1.0 x 10^-7 M, the solution is acidic.
To summarize:
a. [H3O+] = 9.1 x 10^-6 M, acidic solution
b. [H3O+] = 3.4 x 10^-13 M, basic solution
c. [H3O+] = 1.4 x 10^-3 M, acidic solution