moli acid: c=n/V 4=n/0,05---> n=0,2mol acid
moli baza 1= n/0,1---> n=0,1mol baza
ecuatia reactiei
CH3COOH+KOH---> CH3COOK+H2O
1mol 1mol
deduc ca in reactie intra cate 0,1mol si din acid si din baza--< 0,1mol acid ramane neconsumat(exces), care se afla in 25ml solutie