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Postby JamieVu_2C » Mon Jan 20, 2020 12:03 am

Calculate the pH of 0.15 M H2SO4(aq) at 25 degrees C.

For polyprotic acids, I thought you only needed to calculate the first Ka value and consider the other deprotonations as insignificant. However, the solutions manual shows that the first ionization is complete but the second one isn't, so you calculate the second Ka value. But if H2SO4 is already a strong acid, why can't you just consider that [H3O+] = .15 M and find the pH from the first ionization only?

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Re: 6E.1

Postby Chem_Mod » Mon Jan 20, 2020 12:10 am

Most of the time you can disregard the deprotonations other than the first for polyprotic acids because the Kas for subsequent deprotonations are extremely small compared to the first deprotonation and can therefore be ignored. However, in the case of H2SO4, the Ka for the second deprotonation is relatively large (1.2 x 10^-2), so the additional amount of H3O+ generated is actually significant enough to affect the pH.

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