If [H+] = 1.0 x 10^-3 M at 25°C, what is [OH-]?

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Multiple Choice

If [H+] = 1.0 x 10^-3 M at 25°C, what is [OH-]?

Explanation:
The key idea is that in water at 25°C the ion-product Kw equals [H+][OH−] and Kw = 1.0 × 10^-14. To find [OH−], use [OH−] = Kw / [H+]. Plugging in [H+] = 1.0 × 10^-3 M gives [OH−] = (1.0 × 10^-14) / (1.0 × 10^-3) = 1.0 × 10^-11 M. This also corresponds to pH = 3 and pOH = 11, which sum to 14 as expected at this temperature. The other values do not satisfy the Kw relationship for this [H+], so they aren’t correct.

The key idea is that in water at 25°C the ion-product Kw equals [H+][OH−] and Kw = 1.0 × 10^-14. To find [OH−], use [OH−] = Kw / [H+]. Plugging in [H+] = 1.0 × 10^-3 M gives [OH−] = (1.0 × 10^-14) / (1.0 × 10^-3) = 1.0 × 10^-11 M. This also corresponds to pH = 3 and pOH = 11, which sum to 14 as expected at this temperature. The other values do not satisfy the Kw relationship for this [H+], so they aren’t correct.

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