Question

0,2 moli de NaOH sunt neutralizati de o solutie de H2SO4 de concentratie 49%. Sa se calculeze concentratia procentuala a solutiei finale.
Multumesc!

Answer 1

2moli.......98g...........142g..........36g
2NaOH + H2SO4 -> Na2SO4 + 2H2O
0,2moli....a=9,8g.....14,2g.........3,6g
ms=md*100/c=9,8*100/49=20g solutie
insa mapa=ms-md=20-9,8=10,2g apa ramase in solutia finala
mdfinal=14,2g Na2SO4
msfinal=14,2+3,6+10,2=28g solutie
cfinal=mdf*100/msf=14,2*100/28=50,71%