1.moli NaOH
C=n/V(litri); 0,5=n/0,6 ---->0,3molNaOH
moli HCl 1=n/0,4-----> 0,4molHCl
NaOH +HCl--> NaCl +H2O
1mol.......1mol
0,3mol.....0,3mol---------------------> exces de 0,1molHCl in (0,6+0,4)_l
pH=-lg[H⁺]=-lg10⁻¹----> pH=1
2.mol HNO3 2=n/0,2--.>n=0,4molHNO3
mol Ca(OH)2 0,25=n/0,8--> n=0,2mol Ca(OH)2
Ca(OH)2 +2HNO3---->Ca(NO3)2 +2H20
1mol 2mol
0,2mol........0,4mol---------------se consuma ambii reactanti, SOLUTIE NEUTRA---> pH=7