Open Question
What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 × 10-14?
17. Acid and Base Equilibrium
Auto-Ionization
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What is the ion product constant (Kw) of pure water at 50.0°C, if the pH is 6.630?
A
5.48 x 10^-14
B
1.00 x 10^-14
C
7.00 x 10^-14
D
2.00 x 10^-14
Verified step by step guidance1
Understand that the ion product constant of water (Kw) is the product of the concentrations of hydrogen ions [H⁺] and hydroxide ions [OH⁻] in water: Kw = [H⁺][OH⁻].
Recognize that the pH is related to the concentration of hydrogen ions by the formula: pH = -log[H⁺]. Given the pH of 6.630, calculate [H⁺] using the formula: [H⁺] = 10^(-pH).
Calculate the concentration of hydrogen ions [H⁺] using the given pH value: [H⁺] = 10^(-6.630).
Since pure water is neutral, the concentration of hydroxide ions [OH⁻] is equal to the concentration of hydrogen ions [H⁺]. Therefore, [OH⁻] = [H⁺].
Substitute the calculated [H⁺] value into the expression for Kw: Kw = [H⁺][OH⁻] = [H⁺]^2. Calculate Kw using this expression.
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