a)0,0001 M = 10 la -4 M
pH=-lg[HCl]=-lg 10 la -4 =>pH=4
b)cM=nr moli/volum=30*10-3/0,3=>cM=0,1 M
pH=-lg[HCl]=-lg 10 la -1 =>pH=1
c)cM=m HCl/masa molara HCl*volum=1,825/36,5*50=>cM=0,001 M
pH=-lg[HCl]=-lg 10 la -3 =>pH=1
a)pOH=-lg[KOH]=-lg 10 la -2 =>pOH = 2
pH=14-pOH=14-2=>pH=12
b)pOH=-lg[NaOH]=-lg 10 la -2 =>pOH = 2
pH=14-pOH=14-2=>pH=12