2,3 g a g b g
2Na + 2H2O => 2NaOH + H2
2x23 2x18 2x40
avem initial 1 L apa = 1000 g apa
=>
a = 2x18x2,3/2x23 = 1,8 g apa consumata in reactie
=> m.apa ramasa = 1000 - a = 1000-1,8 = 998,2 g apa
b = 2x40x2,3/2x23 = 4 g NaOH
=> md = 4 g
Cm = md/miuxVs = 4/40x1 = 0,1 mol/L
NaOH fiind baza tare stim ca Cm = [HO-]
pOH = -lg[HO-] = -lg[Cm]
= -lg[10^-1] = -(-1)x1 = 1
pH + pOH = 14 => pH = 14-pOH
=> pH = 13
