The number of mols of HBr initially add to the so.pdf
- 1. The number of mols of HBr initially add to the solution of Zn(OH)2 = 0.350 L *0.5 M = 0.175 mol Even after addition of HBr to the Zn(OH)2 solution, the solution is acidic, The number of mols of H+ present after the reaction completed is calculated by using the titration with NaOH. the number of mols of NaOH consumed = (0.5 M*88.5 mL*1.0 L) /1000 mL = 0.04425 mol therefore the number of mols of HBr remained = 0.04425 mol Therefore number of mols of HBr consumed with the Zn(OH)2 = 0.175 -0.04425 = 0.1307 mol The balnced equation between then Zn(OH)2 and HBr is - Zn(OH)2 (Aq) + 2 HBr (aq ) -------------> ZnBr2 (Aq) + 2 H2O(l) 0.1307 mol Therfore - number of moles of Zn(OH)2 reacted with HBr = 0.1307 /2 = 0.06535 mol therfore mass of Zn(OH)2 added to the solution = 99.38 g/mol*0.06535 = 6.5 g Solution The number of mols of HBr initially add to the solution of Zn(OH)2 = 0.350 L *0.5 M = 0.175 mol Even after addition of HBr to the Zn(OH)2 solution, the solution is acidic, The number of mols of H+ present after the reaction completed is calculated by using the titration with NaOH. the number of mols of NaOH consumed = (0.5 M*88.5 mL*1.0 L) /1000 mL = 0.04425 mol therefore the number of mols of HBr remained = 0.04425 mol Therefore number of mols of HBr consumed with the Zn(OH)2 = 0.175 -0.04425 = 0.1307 mol The balnced equation between then Zn(OH)2 and HBr is - Zn(OH)2 (Aq) + 2 HBr (aq ) -------------> ZnBr2 (Aq) + 2 H2O(l) 0.1307 mol Therfore - number of moles of Zn(OH)2 reacted with HBr = 0.1307 /2 = 0.06535 mol therfore mass of Zn(OH)2 added to the solution = 99.38 g/mol*0.06535 = 6.5 g