"Calculate the pH of 0.15 M H2SO4(aq) at 25 °C."
I am a little confused on this question. I used the Ka2 value of 1.2x10^-2 to solve for x, but I ended up with the wrong answer. Is this the correct Ka value to use here? This is the equation I used
H2SO4+H2O=HSO4+H3O
x^2/0.15-x= 1.2x10^-2
textbook 6E.1
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Re: textbook 6E.1
Since sulfuric acid is diprotic, you would also have to do a second calculation for HSO4- -> SO42- + H+. This would give you two H+ concentrations that you can sum up to find the total concentration of H+ and thus the pH. Hope this helps!
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Re: textbook 6E.1
Hi Lily,
That is the correct value of Ka2 to use, but it needs to be matched up with the equilibrium equation for the second deprotonation of sulfuric acid. Because H2SO4 is a strong base that dissociates completely for its first deprotonation, you would already have 0.15 M H3O+ in solution. This value should be taken into account when you make your ICE table for the dissociation of HSO4-.
That is the correct value of Ka2 to use, but it needs to be matched up with the equilibrium equation for the second deprotonation of sulfuric acid. Because H2SO4 is a strong base that dissociates completely for its first deprotonation, you would already have 0.15 M H3O+ in solution. This value should be taken into account when you make your ICE table for the dissociation of HSO4-.
Re: textbook 6E.1
Because H2SO4 is a strong acid you can treat the first calculation like any other strong acid where the pH is simply -log of the concentration. There's already 0.15 M of H in solution after the first deprotonation, so you just have to set up an ice table calculation for the 2nd deprotonation to find the H that releases into solution and sum that amount with 0.15. Then you can take -log of that total to find the pH。
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Re: textbook 6E.1
You have the correct Ka2 value to use but the ICE table should be set up for the equation HSO4- (aq) + H2O (l) --> SO42- (aq) + H3O+ (aq) with the initial concentrations being 0.15 M for HSO4- and H3O+ and 0 M for SO42-
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