Counting hydrogen ions: How many H+ ions are present in 1 c.c (1 mL) of an aqueous solution with pH = 13 at 25°C?

Introduction / Context:
Translating pH values into actual counts of ions sharpens intuition for how dilute acidic or basic solutions are. This is useful in biochemical buffers, environmental sampling, and micro-scale analytical chemistry where tiny volumes are involved.

Given Data / Assumptions:

• pH = 13 at 25°C, so [H+] = 10^-13 mol L^-1.
• Volume considered: 1 c.c = 1 mL = 1 × 10^-3 L.
• Avogadro’s number NA ≈ 6.023 × 10^23 mol^-1.

Concept / Approach:
The number of hydrogen ions equals moles of H+ in the sample times Avogadro’s number. Compute moles using concentration times volume, then scale by NA to get the particle count. High pH means extremely low [H+], so even in a millilitre, the count is surprisingly modest compared with NA.

Step-by-Step Solution:
Determine [H+]: [H+] = 10^-13 mol L^-1 (from pH definition).Compute moles in 1 mL: n = 10^-13 * 10^-3 = 10^-16 mol.Convert to number of ions: N = n * NA = 10^-16 * 6.023 × 10^23.Evaluate: N = 6.023 × 10^7 ions.Match to the option: 6.023 × 10^7.

Verification / Alternative check:
Using pOH = 1 implies [OH-] = 0.1 mol L^-1; the product [H+][OH-] = 10^-14 holds at 25°C, reaffirming [H+] = 10^-13 and the above count.

Why Other Options Are Wrong:
10^13 and 6.023 × 10^13 are many orders too high; they would correspond to concentrated acidic conditions, not pH 13.6.023 × 10^10 is still three orders too large given the tiny molar amount in 1 mL.

Common Pitfalls:
Confusing millilitres with litres; forgetting to multiply by Avogadro’s number; mixing up [H+] and [OH-] at high pH.

Final Answer:
6.023 × 10^7