Chemistry - acids, bases, equilibrium

AI Thread Summary
To compute the pH of a neutral aqueous solution at a temperature where Kw is 3.49e-13, the equation Kw = [H+][OH-] is applied. Assuming [H+] equals [OH-] in a neutral solution, the concentration of hydrogen ions is calculated as [H+] = 3.49e-6. The pH is then determined using the formula pH = -log[H+], resulting in a pH of approximately 5.457. The discussion emphasizes the relationship between Kw and pH, highlighting that Kw varies with temperature. Understanding these concepts is crucial for solving equilibrium problems in chemistry.
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Homework Statement


Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.49e-13 at some temperature, compute the pH of a neutral aqueous solution at that temperature.

Homework Equations


Kw = [H+][OH-]
pH = -log[H+]

The Attempt at a Solution


3.49e-13 = [H+]*10e-7
[H+] = 3.49e-6
pH = 5.457
 
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remember Kw is the ionproduct of water at a given temperature. So {H+}*{OH-}= 3.49e-13 next {H+}*{OH-} can be seen as {H+}e2 so what to do next to gain {H+} from Kw?
 
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