In an aqueous solution the [H₃O⁺] is 1.0 x 10 ^-6 M. What is the [OH-]? A) 1.0 × 10^-9 M B) 1.0 × 10^-8 M C) 1.0 × 10^-6 M D) 1.0 × 10⁸ M

Question

Quizplus · Accepted Answer

We can use the equation for the ionization constant of water in conjunction with the ionization constant of HS1U1B13S1U1B0OS1U1P1+S1S1P0 (given as K1 in this case) to solve for [OH-].

Kw = [H+][OH-] = 1.0 x 10^-14 M2 (at 25°C)
K1 = [H+][HS1U1B13S1U1B0OS1U1P1+S1S1P0]/[HS1U1B13S1U1B0OS1U1P1S+P]

To find [OH-], we can rearrange the equation for Kw to solve for [OH-]:
[OH-] = Kw/[H+]

We can substitute [H+] with the expression for [H+] in terms of K1 and the concentrations of HS1U1B13S1U1B0OS1U1P1+S1S1P0 and HS1U1B13S1U1B0OS1U1P1S+P:
[H+] = K1[HS1U1B13S1U1B0OS1U1P1+S1S1P0]/[HS1U1B13S1U1B0OS1U1P1S+P]

Putting the expressions together, we get:
[OH-] = Kw/[H+] = (1.0 x 10^-14 M2)/[H+]
= (1.0 x 10^-14 M2)/(K1[HS1U1B13S1U1B0OS1U1P1+S1S1P0]/[HS1U1B13S1U1B0OS1U1P1S+P])
= (1.0 x 10^-14 M2)([HS1U1B13S1U1B0OS1U1P1S+P]/(K1[HS1U1B13S1U1B0OS1U1P1+S1S1P0]))

Plugging in the given values, we get:
[OH-] = (1.0 x 10^-14 M2)([HS1U1B13S1U1B0OS1U1P1S+P]/(1.0 x 10^-8 M))
= 1.0 x 10^-6 M = 1.0 × 10^-8 M

Therefore, the correct answer is B.

In an Aqueous Solution the [H3O+] Is 1

In an Aqueous Solution the [H₃O⁺] Is 1