What is the [OH–] in an aqueous solution at 25 °C in which [H3O+] = 1.9 10–9 M?

Question

What is the [OH–] in an aqueous solution at 25 °C in which [H3O+] = 1.9  10–9 M?

BlackTom AI · Accepted Answer

First, identify the given data: at 25 °C, the ion-product of water, Kw, is 1.0 × 10^-14. The hydronium concentration is [H3O+] = 1.9 × 10^-9 M.

Next, use the relationship for water autoionization: [H3O+][OH-] = Kw. Solving for [OH-], we get [OH-] = Kw / [H3O+].

Plug in the values: [OH-] = (1.0 × 10^-14) / (1.9 × 10^-9).

Perform the division: 1.0 / 1.9 ≈ 0.5263, and 10^-14 / 10^-9 = 10^-5, so [OH-] ≈ 0.5263 × 10^-5 M.

Expressing with proper significant figures, 0.5263 × 10^-5 M ≈ 5.26 × 10^-6 M. Rounding to two significant figures consistent with the given data, [OH-] ≈ 5.3 × 10^-6 M.

Interpretation: because [H3O+] is greater than the very small neutral-water value (1.0 × 10^-7 M), the solution is acidic enough that [OH-] is suppressed below 1.0 × 10^-7 M, yet the product with [H3O+] still equals Kw. The calculation demonstrates how altering one ion concentration directly dictates the other through Kw.

What is the [OH–] in an aqueous solution at 25 °C in which [H3O+] = 1.9 10–9 M?Single choice

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