When [H+] = 4.0 × 10–9 M in water at 25°C, then Kw =1.0×10–14.
b. pOH = 7.
c. [H3O+] = [OH–].
d. pH= 7
e. A-D are all correct

a. Kw = 1.0 × 10⁻¹⁴

Explanation:

a. Let's consider the self-ionization of water.

2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻

The ion-product of water (Kw) at 25 °C is:

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴

c. Considering [H⁺] = [H₃O⁺] = 4.0 × 10⁻⁹ M, the concentration of OH⁻ is:

[OH⁻] = 1.0 × 10⁻¹⁴/[H₃O⁺] = 2.5 × 10⁻⁶ M

b. We can calculate the pOH using the following expression.

pOH = -log [OH⁻] = -log 2.5 × 10⁻⁶ = 5.6

d. We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 4.0 × 10⁻⁹ = 8.4