# Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 1.9×10−9M

Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 2.1×10−2M

Calculate [H3O+] in the following aqueous solution at 25 ∘C: [OH−]= 6.1×10−12M

## To calculate the concentration of H3O+ in an aqueous solution, you can use the equation for the ion product of water (Kw):

Kw = [H3O+][OH-]

At 25 °C, Kw is equal to 1.0 x 10^-14.

Let's calculate [H3O+] for each given value of [OH-].

For [OH-] = 1.9 x 10^-9 M:
Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+](1.9 x 10^-9)
[H3O+] = (1.0 x 10^-14) / (1.9 x 10^-9)
[H3O+] ≈ 5.26 x 10^-6 M

For [OH-] = 2.1 x 10^-2 M:
Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+](2.1 x 10^-2)
[H3O+] = (1.0 x 10^-14) / (2.1 x 10^-2)
[H3O+] ≈ 4.76 x 10^-13 M

For [OH-] = 6.1 x 10^-12 M:
Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+](6.1 x 10^-12)
[H3O+] = (1.0 x 10^-14) / (6.1 x 10^-12)
[H3O+] ≈ 1.64 x 10^-3 M

Thus, the concentrations of [H3O+] in the given aqueous solutions are approximately:
1) [H3O+] ≈ 5.26 x 10^-6 M
2) [H3O+] ≈ 4.76 x 10^-13 M
3) [H3O+] ≈ 1.64 x 10^-3 M

## To calculate [H3O+] in an aqueous solution given the concentration of [OH-], you need to use the equation for the autoionization of water:

H2O ⇌ H+ + OH-

At 25 ∘C, the concentration of water is assumed to be constant, so we can use the following equation, known as the water autoprotolysis constant (Kw):

Kw = [H+][OH-] = 1.0 × 10^-14

To find [H3O+], it is important to remember that the concentration of [H3O+] is equal to the concentration of [H+]. Therefore, we can rewrite the equation as:

[H+] = [OH-] = x

For the given examples, let's go through the calculations:

1) [OH-] = 1.9 × 10^-9 M

Using the equation Kw = [H+][OH-], we substitute the given values and solve for [H+]:

(1.9 × 10^-9)(x) = 1.0 × 10^-14

Solving this equation gives us:

[H+] = [OH-] = x = 5.26 × 10^-6 M

Therefore, [H3O+] = 5.26 × 10^-6 M

2) [OH-] = 2.1 × 10^-2 M

Using the equation Kw = [H+][OH-], we substitute the given values and solve for [H+]:

(2.1 × 10^-2)(x) = 1.0 × 10^-14

Solving this equation gives us:

[H+] = [OH-] = x = 4.76 × 10^-13 M

Therefore, [H3O+] = 4.76 × 10^-13 M

3) [OH-] = 6.1 × 10^-12 M

Using the equation Kw = [H+][OH-], we substitute the given values and solve for [H+]:

(6.1 × 10^-12)(x) = 1.0 × 10^-14

Solving this equation gives us:

[H+] = [OH-] = x = 1.64 × 10^-3 M

Therefore, [H3O+] = 1.64 × 10^-3 M

Remember to always check your units and significant figures throughout the calculations, as they may affect the final answer.

## (H3O^+)(OH^-) = Kw = 1E-14

Substitute and solve.