M, fenol= 94g/mol
M, baza = 40g/mol
c/1000=. md/ms---> md=80g NaOH
ecuatia reactiei:
1mol.........1mol 1mol
C6H5OH +NaOH ----> C6H5ONa + H2O
94g............m=40g........n=.1mol
rezulta ca (80-40)gNaOH, adica 1 mol ,sunt in exces
1mol 1mol
NaOH+HCl---> NaOH+H2O
1mol 1mol
C6H5ONa+HCl ---> c6H5OH +NaCl
total 2 molHCl
c= n/V---> 1=2/V---> V= 2l