Answer:
Kw = 1×10⁻¹⁴
Explanation:
The product between the [H₃O⁺] . [OH⁻] talks about the
autoionization of water.
2H₂O ⇄ H₃O⁺ . OH⁻ Kw
This is an equilibrium of 2 molecules of water that produces 1 mol of hydronium and 1 mol of hydroxides.
In order to determine pH and pOH, parameters to indicate acidic or basic solutions we know:
- log [H₃O⁺] = pH
- log [OH⁻] = pOH
When [H₃O⁺] or [OH⁻] = 1×10⁻⁷ , pH or pOH = 7 (Neutral solutions)
So if we see the equation above, we have an equilibrium.
In order to determine Kw, we need [H₃O⁺] . [OH⁻]
As [H₃O⁺] and [OH⁻] = 1×10⁻⁷ , Kw = 1×10⁻¹⁴
The product of [H₃O⁺] and [OH⁻] is 1 × 10⁻¹⁴. The correct option is the first option - 1 x 10-14
The ionization constant of water, Kw, is the product of [H₃O⁺] and [OH⁻]
That is,
Kw = [H₃O⁺][OH⁻]
In pure water, [H₃O⁺] = [OH⁻] = 1 × 10⁻⁷
∴ Kw = [H₃O⁺][OH⁻] = 1 × 10⁻⁷ × 1 × 10⁻⁷
Kw = [H₃O⁺][OH⁻] = 1 × 10⁻¹⁴
Hence, the product of [H₃O⁺] and [OH⁻] is 1 × 10⁻¹⁴. The correct option is the first option - 1 x 10-14
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