calculate the ph at the equivalence point in titrating 0.100M solution of chloro
ID: 760336 • Letter: C
Question
calculate the ph at the equivalence point in titrating 0.100M solution of chlorous acid(HCLO2) WITH 0.080M NaOH.Explanation / Answer
(a) hydrobromic acid (HBr) Ph= 7.0 b) Na HCrO4 + NaOH --> Na2 CrO4 CrO4= does a hydrolysis with water @ equiv point--> CrO4= & H2O --> HCrO4- & OH- (it is a basic solution) Khydrolysis = Kwater / Kacid = 1e-14 / 3.0e-7 = 3.33e-8 3.33e-8 = [HCrO4-] [OH- ] / [CrO4=] CrO4 is not 0.1 molar, it was diluted with the addition of the 0.039 molar NaOH. it requires 2.56 times as much 0.039 molar naoh to destroy the HCrO4. there has been a 3.56 fold dilution by the end of titration .... CrO4 is now 0.0281 Molar so 3.33e-8 = [HCrO4-] [OH- ] / [CrO4=] becomes 3.33e-8 = [x][x ] / [0.0281] 1.185e-6 = [x][x ] x= [OH] = 1.09e-3 pOH = 2.96 14 - pOH = pH pH= 11.04
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