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Chemistry-acid base equilibria
found ka2 to be 3.39*10^-10.- chromeX
- Post #5
- Forum: Biology and Chemistry Homework Help
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Chemistry-acid base equilibria
i figured out how to get ka1, but I'm still having trouble with ka2. how i got ka1- use ICE with H2A + H2O -> HA- + H3O+, using the values for halfway to the first equivalence point. Ka1=([HA-][H3O+]/[H2A]) [H2A]=[HA-]=.0049. (cancels out), leaving Ka1 = [H3O+] = 10^-4.57=2.69*10^-5...- chromeX
- Post #4
- Forum: Biology and Chemistry Homework Help
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Chemistry-acid base equilibria
Here's what I've tried- the moles of H2A at the beginning is .01624L times 0.0200M. You also need 32.48mL of base to reach 2nd equiv point. the part that trips me up is that HA-is an amphoteric species. I used ICE with HA- + H20 -> A2- + H3O+, finding the conc. of [H3O+] as 10^-7.02 with...- chromeX
- Post #3
- Forum: Biology and Chemistry Homework Help
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Chemistry-acid base equilibria
p-Hydroxybenzoic acid, HOC6H4COOH, is a weak diprotic acid. A 25.00 mL aliquot of a dilute solution of HOC6H4COOH is titrated with a 0.0200 M NaOH solution. The first equivalence point was reached after 16.24 mL of the NaOH solution was added. a. If the values of the pH after 8.12 and 16.24 mL...- chromeX
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- Base Equilibria
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- Forum: Biology and Chemistry Homework Help