·CuSO4 + 2NaOH ⇒ Na2SO4 + Cu(OH)2
ms1 = ms2 = m
ms final = 2m
initial : n CuSO4 = (c1·m/100)/160 nNaOH =( c2·m/100)/40
c2·m /40 =2·c1·m/160 c2 = c1/2
n Na2SO4 = n CuSO4 = nCu(OH)2 = [m·c1/100]/160 = nNaOH /2 = =[m·c2/100]/80
m sare = [m·c1/160]100 ·142 = 0,8875m·c1/100
[0,8875·m·c1]/100 / (2m) ·100 = 25 0,8875·c1 = 50 c1 = 56,338%
m baza ={ [m·c2/100] /80}·98
[(mc2/80)·98]/2m = 10 1,225c2= 20 c2 = 16/326%