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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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The separation produced is called coupling constant J and is measured in Hz. If the split is produced by two equal protons (equal J) a triple signal known as triplet is produced and if produced by three equal protons, the signal is quadruple and is known as quadruplet. The magnitude of the coupling is varied, depending on the relative disposition of the coupled protons (elevations that separate them, arrangement, spatial arrangement)

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