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!
Calculating pH
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Re: Calculating pH
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.
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.
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Re: Calculating pH
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.
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