Question: i need help with finding the concentration in both side Fitter paper 0.229 filter paper + sol =0.38 1Ba(OH)2(as)+1H2SO4(aOC)1BaSO4(s)+2H2O(L)Ba2+2OH+2H++SO42BaSO4(s)+2H2O(L)Calculation #1: moles BaCOH)2= moles H2SO4(MV) [Ba(OH)2]=

i need help with finding the concentration in both side

$0.229 filter paper + sol =0.38 1Ba(OH)2(as)+1H2SO4(aOC)1BaSO4(s)+2H2O(L)Ba2+2OH+2H++SO42BaSO4(s)+2H2O(L)Calculation \#1: moles BaCOH)2= moles H2SO4(MV)$

[Ba(OH)2]= moles (Ba(OH)) H2: moles BaSO4(s)= moles of Ba(OH)2 [Ba(OH)2]= moles of

Fitter paper 0.229 filter paper + sol =0.38 1Ba(OH)2(as)+1H2SO4(aOC)1BaSO4(s)+2H2O(L)Ba2+2OH+2H++SO42BaSO4(s)+2H2O(L)Calculation \#1: moles BaCOH)2= moles H2SO4(MV) [Ba(OH)2]= moles (Ba(OH)) H2: moles BaSO4(s)= moles of Ba(OH)2 [Ba(OH)2]= moles of Ba(OH)2/vBa(OH)2 H2SO4(0.01798M) 3 graph Titration Results volume of Ba1(OH)2 used:- 10.0mL=0.01L Concentration of H2SO4 : - 0.01798M Equation for decroasing line:- BaSO4+2H2OBa24+2OH4+2H7+Sa42 Equation for increasing line: - 3a2+2+OH+2H4+5O4+SasO4+2H2O calculated equivalence volume: 210 concentration of unknown Ba(OH)2 solution (by titration): Precipitation Results mass of filler paper:- 0.22g Mass of filter Paper AND Basoy Precipitate:- 0.38g Mass of Basoy Precipitate: 0.22g0.38g=0.16g concentration of unknown Ba(OH)2 solution: (by precipitation)

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