-calculez ms si md din solutia de acid
ms= Vsxro= 16,6cmcubx1,1g.cmcub=18,26 g sol.
c=mdx100/ms---> md=20x18,26/100=3,65gHCl
m,apa= ms-md= 18,26g-3,65g=14,61g
-deduc moli si mase HCl, si moli de substante ce participa la reactie
niu,HCl=m/M= 3,65g/36,5g/mol=0,1mol
1mol.......1mol.......1mol...........1mol
HCl + NaOH ---> NaCl + H2O
0,1mol...x=0,1mol....y=0,1mol,,,z=0,1mol
m,NaOH=0,1X40--> 4gNaOH
m,NaCl= 0,1x58,5--> 5,85gNaCl
m,apa= 0,1x18--> 1,8g apa
-deduc ms de solutie de baza ,stiind md=m,NaOH= 4g
c=mdx100/ms-----> ms= 4x100/25--> 16gsol.NaOH
m,apa= ms-md= 16-4--> 12g
-solutia finala contine dizolata sare NaCl in apa rezultata din reactie adunata cu masele de apa ramase din solutiile de acid si baza
md= 5,85g
m,apa= 14,61+1,8+12-->28,41g apa
ms= md+m,apa= 5,85+28,41=>34,26g
c,f= mdx100/ms= 5,85x100/ 34,26=.......................calculeaza !!!
