Question

Calculate either [H3O+]or [OH−] for each of the solutions at 25 °C.

Solution A: [OH−]=1.49×10−7 M; [H3O+]=

Solution B: [H3O+]=9.93×10−9 M; [OH−]=

Solution C: [H3O+]=7.73×10−4 M; [OH−]=

Answer 1

To calculate the concentration of either hydronium ions ([H3O+]) or hydroxide ions ([OH-]) for each solution, we can use the fact that in pure water at 25 °C, the product of the concentrations of hydronium and hydroxide ions is constant:

[H3O+] × [OH-] = 1.0 × 10^-14 M^2

Let's calculate the missing concentration for each solution:

Solution A:
Given: [OH-] = 1.49 × 10^-7 M
To find [H3O+]:
[H3O+] × [OH-] = 1.0 × 10^-14 M^2
[H3O+] × (1.49 × 10^-7 M) = 1.0 × 10^-14 M^2
[H3O+] ≈ (1.0 × 10^-14 M^2) / (1.49 × 10^-7 M)
[H3O+] ≈ 6.71 × 10^-8 M

Solution B:
Given: [H3O+] = 9.93 × 10^-9 M
To find [OH-]:
[H3O+] × [OH-] = 1.0 × 10^-14 M^2
(9.93 × 10^-9 M) × [OH-] = 1.0 × 10^-14 M^2
[OH-] ≈ (1.0 × 10^-14 M^2) / (9.93 × 10^-9 M)
[OH-] ≈ 1.01 × 10^-6 M

Solution C:
Given: [H3O+] = 7.73 × 10^-4 M
To find [OH-]:
[H3O+] × [OH-] = 1.0 × 10^-14 M^2
(7.73 × 10^-4 M) × [OH-] = 1.0 × 10^-14 M^2
[OH-] ≈ (1.0 × 10^-14 M^2) / (7.73 × 10^-4 M)
[OH-] ≈ 1.29 × 10^-11 M

So, the calculated concentrations are:
Solution A: [H3O+] ≈ 6.71 × 10^-8 M; [OH-] ≈ 1.49 × 10^-7 M
Solution B: [H3O+] ≈ 9.93 × 10^-9 M; [OH-] ≈ 1.01 × 10^-6 M
Solution C: [H3O+] ≈ 7.73 × 10^-4 M; [OH-] ≈ 1.29 × 10^-11 M