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The value of the water dissociation constant Kw varies with temperature. Its value is normally given as 1.00 x 10-14 mol2 dm-6 at room temperature but 1.00 x 10-13 mol2 dm-6 at 60 °C. What is the pH of pure water at 60 °C?
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Correct Answer: D. Less that 7.0 but the water is neutral.
It is not necessary to calculate anything to solve this problem but we will go through the steps.
The dissociation of water (note carefully that the ratio of the products is 1:1; also keep in mind that this is the basis of the neutrality of pure water: acid units = base units):
H2O + H2O <-> H3O+ + OH–
The expression for the self-ionisation of water or the water dissociation constant:
Kw = [H3O+][OH–] = [H+][OH–]
At room temperature, keeping in mind that the dissociation of the ions is 1:1, we get: 10-14 = [H+][OH–] = [H+]2
And so [H+] = 10-7 and then -log[H+] = 7 = pH
Doing the identical math but using Kw = 10-13 yields a pH of 6.5 BUT because the ratio of the ions is still 1:1 (if not, you would not have done the calculation) thus it must be neutral.
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