Like all equilibrium constants, Kw varies
somewhat with temperature. Given that Kw
is 5.02×10−13 at some temperature, compute
the pH of a neutral aqueous solution at that
temperature.
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To compute the pH of a neutral aqueous solution at a given temperature, we need to know the value of the dissociation constant of water (Kw) at that temperature. The dissociation constant of water is the product of the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in water, and it is defined as Kw = [H+][OH-].
Given that Kw = 5.02×10^(-13) at the given temperature, we can determine the concentration of either H+ or OH- ions in a neutral solution.
In a neutral solution, the concentration of H+ ions is equal to the concentration of OH- ions. Let's assume their concentration is x.
So, [H+] = x and [OH-] = x.
Using the definition of Kw, we have:
Kw = [H+][OH-]
5.02×10^(-13) = x * x
Taking the square root of both sides, we get:
sqrt(5.02×10^(-13)) = sqrt(x^2)
7.08×10^(-7) = x
Now that we know the concentration of H+ ions in a neutral solution is 7.08×10^(-7) mol/L, we can calculate the pH.
The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions. Mathematically, pH = -log[H+].
Therefore, pH = -log(7.08×10^(-7))
pH = -log(7.08) + log(10^(-7))
pH = -(-6.150) + (-7) (taking log values)
pH = 6.150 - 7
pH = -0.85
Therefore, at the given temperature, the pH of a neutral aqueous solution is approximately -0.85.