a)cM=m acid/masa molara acid * volum=0,63/63*0,1=0,1 mol/litru
pH=-lg[cM]=-lg 10 la -1=>pH=1
b)HNO3 + KOH ---> KNO3 + H2O
md=(ms * c)/100=(100*2)/100=2g KOH
la 63 g HNO3.......56 g KOH
la x g HNO3.........2 g KOH x =2,25 g HNO3
2,25 > 2 => solutia este in exces de acid => pH acid => solutie incolora