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get ph 2 a2 ka2 regardless of concentration 3 ha

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Unformatted text preview: r polyprotic acids, there will be n+1 terms in the denominator (for an acid that loses “n” H+) and the terms will be made up of K values and [H+] raised to various powers ….note the pattern [H on table 8.7. This is plotted in Figure 8.9 Assumptions About Many Diprotic Systems Assumptions (1) When Ka1 >>> Ka2, most all of the H+ comes from Ka1 —so treat this like a monoprotic acid….. Get pH (2) [A2–] ~ Ka2 (regardless of concentration) (3) [HA–] ~ [H+] (If Ka1>>>Ka2) WITH DIPROTIC SYSTEMS, There are Three WITH DIFFERENT SPECIES THAT COULD START A DIFFERENT SOLUTION: SOLUTION: (1) H2A… use approximations to get [H+]; (2) HA….Since HA– (Or HPro in the amino acid example) HA could act as an acid or as a base, there are “competing Exerciese 8.2 equilibr...
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