A100.0 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

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na