6E.3b

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Daniela Shatzki 2E
Posts: 53
Joined: Sat Aug 24, 2019 12:16 am

6E.3b

Postby Daniela Shatzki 2E » Tue Jan 21, 2020 2:38 pm

I was kinda confused on how to solve this problem: find the pH of 0.10M (COOH)2(aq) at 25 °C. I solved it one way and got 1.25 as the pH but the answer key says 1.28 so I didn't know if I was close enough or if I did it wrong. I wasn't really sure about how the equation would look for (COOH)2 and H2O.

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Re: 6E.3b

Postby Brian Tangsombatvisit 1C » Tue Jan 21, 2020 3:16 pm

Since Ka1 is much larger than Ka2, the second dissociation reaction can be ignored. We can just focus on the first dissociation reaction, where (COOH)2 donates one of its protons to water (like any other acid), forming H3o+ and (COOH)COO-. You would set up your ICE table like normal and solve for x by setting the equilibrium expression equal to the Ka1. Since K is not < 10^-3, you would not be able to approximate x as 0. The difference in your answer and the solution manual's answer is negligible so it was probably just a rounding error.


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