A 100.0 ml sample of 0.10 m ca(oh)2 is titrated with 0.10 m hbr. determine the ph of the solution after the addition of 400.0 ml hbr. the chemical equation is below. ca(oh)2 (aq) + 2hbr (aq) → cabr2 (aq) + h20 (l)
the balanced equation for the reaction is as follows; Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O stoichiometry of Ca(OH)₂ to HBr is 1:2 number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol 1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation therefore 0.010 mol of Ca(OH)₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol then pH of the medium can be calculated using the excess H⁺ ions HBr is a strong acid therefore complete ionization [HBr] = [H⁺] [H⁺] = 0.020 mol / (100.0 + 400.0 mL) = 0.020 mol / 0.5 L = 0.040 mol/L pH = -log[H⁺] pH = - log [0.040 M] pH = 1.40 pH of the medium is 1.40
