"2 A Bromate formation during chlorination of bromide containing water is a slowprocessHOCl +Br- =========== HOBr + Cl=K = {HOBr}{Cl-}/{HOCl} [ Br-}HOBr +HOBr----------BrO2- +Br- + 2H+K = {HOBr}{HOBr}/{BrO2-} [ Br-}[2H+]HOBr+ BrO2- --------- BrO3- + Br- + H+K = {HOBr}{ BrO2}/{BrO3-} [ Br-}[H+]Bromate is formed by stepwise oxidation by ozone of bromide to hypobromiteand then to bromate.O3 + Br- --------------O2 + OBr-K = {O3}{ Br}/{O2} [ OBr-}2O3 + OBr- ---------------2O2 + BrO3-K = {2O3}{OBr}/{2O2} [ BrO3-}B,I, 0.05mg of bromate is formed for 300mg of chlorideIi,0.007mg of bromate is formed for 20mg of ozoneC 1.chlorine residual is 0.7mg/ml Bromate formed = 3 x 10^-5– 0.7= 2.3x 10^-5 grams2. Residual ozone concentration =0.2mg/mlBromate formed = 2 x 10^-7– 0.2=1.8 10^-7 gramsBromide oxidation is favorable in both the casesd. Bromate formation by ozone is most favourableAdvantage is the bromate formed pollute the water less and the water can beused for drinking purpose.3AMass of H3PO4 =97.995181 / 10 x 1 = 9.7995181g for PH– 1For PH -6.2 = 1.5gMass of H3PO4 =97.995181 /25x 1 = 3.919g for PH– 1For PH 9.1 = 0.42gmass of NaH2Po4 = 119.9 g/mol/10 X1 =11.99g for PH -1For PH 6.2 = 1.9gmass of NaH2Po4 = 119.9 g/mol/25 X1 =4.796g for PH -1For PH 9.1 = 0.5gmass of Na2HPo4 = 141.9 g/mo/10 X1 =14.19g for PH -1For PH 6.2 = 2.2gmass of Na2HPo4 = 141.9 g/mo/25 X1 =5.67g for PH -1For PH 9.1 = 0.6gMass of Na3PO4 = 163.9 /10 X 1 = 16.39 g For PH 6.2 = 2.6gMass of Na3PO4 = 163.9 /25 X 1 = 6.556 gFor PH 9.1 =0.72gB, Titration curveC, 10mM phosphate buffer 25mM phosphate bufferd. Buffer strength = change in concentration/change in PH =0.1+10/9.1+0.05=1.10 "