If a solution has pH = 3 at 25°C, what is its hydroxide ion concentration [OH-] in mol L^-1?

Introduction / Context:
Relating pH and pOH is a staple skill in aqueous chemistry. At 25°C, water self ionization is quantified by Kw = [H+][OH-] = 1.0 * 10^-14. Given pH, you can determine hydrogen ion concentration and then hydroxide ion concentration using Kw. This type of conversion is common in environmental monitoring, bioprocess pH control, and analytical calculations.

Given Data / Assumptions:

• pH = 3.00 at 25°C.
• Kw at 25°C equals 1.0 * 10^-14.
• Activities are approximated by concentrations for dilute solutions.

Concept / Approach:
Compute [H+] from pH, then use Kw to compute [OH-]. Specifically, [H+] = 10^-pH = 10^-3 mol L^-1. Then [OH-] = Kw / [H+] = 10^-14 / 10^-3 = 10^-11 mol L^-1. Alternatively, use pH + pOH = 14, so pOH = 11 and [OH-] = 10^-11 mol L^-1. Both approaches are equivalent at 25°C.

Step-by-Step Solution:
Calculate hydrogen ion concentration: [H+] = 10^-3 mol L^-1.Apply the water ion product: Kw = [H+][OH-] = 10^-14.Solve for hydroxide: [OH-] = 10^-14 / 10^-3 = 10^-11 mol L^-1.Match to the correct option: 10^-11.

Verification / Alternative check:
Using pOH: pOH = 14 - 3 = 11, hence [OH-] = 10^-11 mol L^-1, confirming the result.

Why Other Options Are Wrong:
10^-10: corresponds to pOH = 10 and pH = 4, not pH 3.10^-3: this is [H+], not [OH-], for pH 3.10^-4: would imply pOH = 4 and pH = 10, a basic solution.

Common Pitfalls:
Forgetting that pH + pOH = 14 applies at 25°C; mixing up [H+] and [OH-]; ignoring that the scale is logarithmic, so a change of 1 pH unit is a tenfold change in [H+].

Final Answer:
10^-11