(A)Calculate the PH at the equivalence point in titrating 0.100M SOLUTION OF CHL

Question

(A)Calculate the PH at the equivalence point in titrating 0.100M SOLUTION OF CHLOROUS ACID(HCLO2) WITH 0.080MNaOH. (B)CALCULATE THE PH AT THE EQUIVALENCE POINT IN TITRATING 0.100M SOLUTION OF BENZOIC ACID( C6H5COOH) WITH 0.080MNaOH only those who know the question well will get all the pointS, NO COPY AND PASTE PLEASE, I WILL AWARD LOTS OF POINTS FOR SHOWING ALL THE STEPS. wILLING TO GIVE MAX kARMA POINTS

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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