Respuesta :
Answer:
Approximately [tex]4.1 \times 10^{-13}[/tex].
Explanation:
At a given temperature, the self-ionization constant of water [tex]K_{\rm w}[/tex] is the product of the concentration of [tex]{\rm H_{3}O}^{+}[/tex] and [tex]{\rm OH}^{-}[/tex]:
[tex]K_{\rm w} = [{\rm H_{3}O}^{+}]\, [{\rm OH^{-}}][/tex].
Because this solution is neutral, the concentration of [tex]{\rm H_{3}O}^{+}[/tex] would be equal to that of [tex]{\rm OH}^{-}[/tex]. Since it is given that [tex][{\rm OH}^{-}] = 6.4 \times 10^{-7}\; {\rm M}[/tex]:
[tex][{\rm H_{3}O}^{+}] = [{\rm OH}^{-}] = 6.4 \times 10^{-7}\; {\rm M}[/tex].
Therefore, the [tex]K_{\rm w}[/tex] at this temperature would be:
[tex]\begin{aligned} K_{\rm w} &= [{\rm H_{3}O}^{+}]\, [{\rm OH^{-}}] \\ &= (6.4 \times 10^{-7})\, (6.4 \times 10^{-7}) \\ &\approx 4.1 \times 10^{-13}\end{aligned}[/tex].
At a particular temperature where a neutral solution of water has a concentration of OH⁻ ions of 6.4 × 10⁻⁷ M, the ion product constant, Kw, is 4.096 × 10⁻¹⁴.
The concentration of OH⁻ ions in a neutral solution of water is equal to the concentration of H⁺ ions, which is also 6.4 × 10⁻⁷ M. At this particular temperature, the ion product constant, Kw, can be calculated by multiplying the concentration of H⁺ and OH⁻ ions: Kw = [H⁺][OH⁻] = (6.4 × 10⁻⁷ M)(6.4 × 10⁻⁷ M) = 4.096 × 10⁻¹⁴.
The ion product constant, Kw, is a measure of the strength of water as an acid or base at a particular temperature. It is a constant value at a given temperature and represents the tendency of water to dissociate into its constituent ions. A higher value of Kw indicates a stronger acid or base, while a lower value indicates a weaker acid or base.
In summary, at a particular temperature where a neutral solution of water has a concentration of OH⁻ ions of 6.4 × 10⁻⁷ M, the ion product constant, Kw, is 4.096 × 10⁻¹⁴.
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