textbook 6E.1

[lily_oneal_2B]

"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

[Alex Yeghikian 1C]

Since sulfuric acid is diprotic, you would also have to do a second calculation for HSO₄^- -> SO₄^2- + 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!

[Brian Diehl 2B]

Hi Lily,

That is the correct value of K_a2 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^-.

[kylanjin]

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。

[Benjamin Nguyen 1J]

You have the correct Ka2 value to use but the ICE table should be set up for the equation HSO4- (aq) + H2O (l) --> SO4^2- (aq) + H3O+ (aq) with the initial concentrations being 0.15 M for HSO4- and H3O+ and 0 M for SO4^2-