HCl + KOH ---> KCl + H2O
c%=md/ms * 100=>md=c% * ms/100=2,8*40/100=>md=1,12 g KOH
pH=-lg[HCl]=-lg 10⁻¹=1 =>cM=10-1 moli/litru=0,1 moli/litru
cM=nr moli/volum=masa HCl/masa molara HCl * volum=>masa HCl=cM * masa molara HCl * volum=>masa =0,1*36,5*0,5 = 1.825 g HCl
1,825>1,12=>exces de acid=>pH mai acid=>rosu