In pure water at 25°C, what is the relation between pH and pOH?

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Multiple Choice

In pure water at 25°C, what is the relation between pH and pOH?

Explanation:
The relationship is set by the ion-product of water and how pH and pOH are defined. Water autoionizes as H2O ⇌ H+ + OH−, with the product [H+][OH−] equal to Kw. By definition, pH = −log[H+], and pOH = −log[OH−]. Taking logs of Kw = [H+][OH−] gives log[H+] + log[OH−] = log Kw, which rearranges to pH + pOH = pKw. At 25°C, pKw equals 14, so pH + pOH = 14. In pure water at this temperature, [H+] = [OH−] = 1×10^−7, yielding pH = pOH = 7, and 7 + 7 = 14. If temperature changes, Kw changes as well, so the sum becomes pKw (not necessarily 14).

The relationship is set by the ion-product of water and how pH and pOH are defined. Water autoionizes as H2O ⇌ H+ + OH−, with the product [H+][OH−] equal to Kw. By definition, pH = −log[H+], and pOH = −log[OH−]. Taking logs of Kw = [H+][OH−] gives log[H+] + log[OH−] = log Kw, which rearranges to pH + pOH = pKw.

At 25°C, pKw equals 14, so pH + pOH = 14. In pure water at this temperature, [H+] = [OH−] = 1×10^−7, yielding pH = pOH = 7, and 7 + 7 = 14. If temperature changes, Kw changes as well, so the sum becomes pKw (not necessarily 14).

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