What is the pH of pure water at 40.0°C if the K_w at this temperature is 2.92 × 10^-14? A) 6.767 B) 0.465 C) 7.000 D) 7.233 E) 8.446

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Quizplus · Accepted Answer

The pH of pure water is calculated using the formula pH = -log[H⁺]. At 40.0°C, Kw = [H⁺][OH⁻] = 2.92 × 10⁻¹⁴. Since [H⁺] = [OH⁻] in pure water, [H⁺] = √(2.92 × 10⁻¹⁴) = 1.71 × 10⁻⁷. Therefore, pH = -log(1.71 × 10⁻⁷) ≈ 6.767.

What Is the PH of Pure Water at 40