Answer:
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
