Polyprotic Acids and Bases 12.79

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Semi Yoon
Posts: 60
Joined: Fri Sep 28, 2018 12:27 am

Polyprotic Acids and Bases 12.79

Postby Semi Yoon » Sun Jan 20, 2019 10:19 pm

How would you calculate the pH of 0.15 M H2SO4(aq) at 25 C?

AditiSax
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Re: Polyprotic Acids and Bases 12.79

Postby AditiSax » Sun Jan 20, 2019 10:45 pm

Set up an ICE table as Dr. Lavelle has demonstrated in lecture, with the initial concentration of sulfuric acid as 0.15M and the initial concentrations of hydronium ions and conjugate base as both 0. The change in the concentration of sulfuric acid would be -x and the changes in H30+ and HSO4- would be +x. Using the Ka value for sulfuric acid and solving using the expression, you should end up with the equilibrium concentration of hydronium ions in the system. Take the negative log of this value and you should be able to calculate the pH.

Griffin Carter 2I
Posts: 16
Joined: Wed Nov 14, 2018 12:23 am

Re: Polyprotic Acids and Bases 12.79

Postby Griffin Carter 2I » Sun Jan 20, 2019 11:07 pm

For the first set of calculations you simply do the strong acid reaction of H2SO4(aq). This will fully dissociate because it is a strong acid. Find your concentration of H2O+ and HSO4- for this first section. For the second set of calculations, you will need to set up an ice table with HSO4-,SO4-2, and H30+. The most important part is to remember that the initial value of H30+ will not be 0 because of that first strong acid reaction. You then can simply do the same problem solving techniques taught in class using the ICE table to solve the remainder of the problem. Though,you are unable to use approximations for this problem because the error is too high. Find the final hydronium concentration by adding the x value found in the second part to the initial hydronium concentration for a total final concentration. Then take the -log of that value to find pH.


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