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Zolol [24]
3 years ago
11

When [H+] = 4.0 × 10–9 M in water at 25°C, then

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
5 0

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

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<h3>What is an equilibrium constant?</h3>

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The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.

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