Calculating pH

[205992763]

I am struggling with this problem, could someone help walk me through it please?

Calculate the pH of 0.15M H2SO4 (aq) at 25C. H2SO4(aq), Ka2 = 1.2 x 10^-2.

Thank you in advance!

[Max Kaminsky [3L]]

The first thing to take note of is that H₂SO₄ is a strong and diprotic acid, meaning that the conjugate of the first dissociation is also acidic and will contribute to decreasing the pH. Since it is strong, all of the H₂SO₄ is converted to H⁺ and HSO₄^-, meaning you have 0.15M [H⁺]. However, you need to make another ICE table for the next dissociation, with the initials being 0.15M of HSO₄^- in the reactants, and 0.15M of H⁺ and 0M of SO₄^- in the products. After this ICE table, you have
1.2*10^-2 = [H⁺][SO₄^-]/[HSO₄^-], with all of your knowns and unknowns as x filled in. Once you solve for x, you know the H⁺ concentration by the end of the reaction, and you can take the negative log of that to get your pH.

[Bryan Lee 2L]

An initial observation is that H2SO4 is a very strong acid and all of it will dissociate. Given its strength, all H2SO4 dissociates into H+ and HSO4-, resulting in a concentration of 0.15M H+ and HSO4-. However, another ICE table must be constructed, starting with 0.15M of HSO4- in the reactants and 0.15M of H+ and 0M of SO4- in the products. Using this ICE table, we make the equilibrium expression: 1.2*10-2 = [H+][SO4-]/[HSO4-]. Solving for x yields the final concentration of H+, and from there, you can do the calculation of pH by taking the negative logarithm of this concentration.