6E.1A textbook problem

[Hanadi Kamal 1J]

The question is "calculate the Ph of 0.05M H2SO4 solution"

I understand there is a short cut of doing H2SO4 = 2H+ + SO4- thus 2(0.05) = 0.1M and plugging in 0.1M into -log(.1)= 1. However, I am unsure how to do it the more accurate way because the textbook answer says 1.23 for the pH. I tried using an ICE box but I getting confused because my answers don't match what the textbook answer is. Thank you.

[Erika Patel 3I]

Hi, so for this problem H2SO4 is going to first de-protonate in water into H3O+ and HSO4-. This would be a complete deprotonation, so the H3O+ concentration would equal the original concentration of H2SO4, which is 0.05 M. Next, you would use the equation for the second deprotonation (which is a partial deprotonation that establishes equilibrium): HSO₄^- + H₂O H₃O⁺ + SO₄^2-. Your ICE table would have the concentration of HSO4- initial equal to 0.05, H3O+ initial equal to 0.05 and SO4^2- initial equal to 0. From there you would set up the Ka2 equation as normal and plug the values 0.05-x, 0.05+x and x in for HSO4-, H3O+ and SO4^2- respectively. Solving using the quadratic equation will give you the value for x, and then from there you can calculate the pH using the equilibrium concentration of H3O+.