Question

3,9 g potasiu se adauga peste o sol. de KOH si rezulta 80 g sol KOH de concentratie c=11,2%. Calculati compozitia procentuala a solutiei initiale !! URGEEEEEENT !!!!

Answer 1

C=md/ms*100 => md=C*ms/100 = 11,2*80/100 = 8,96g KOH

md i = mdf - m adaugat = 8,96-3,9 = 5,06g KOH

m sol i =msf-m adaugat = 80-3,9 = 76,1g sol

Ci = mdi/msi*100 = 5,06/76,1*100 = 6,649 %

Answer 2

solutia initiala : ms1    c1      md1 = ms1·c1/100 
nK = 3,9/39 = 0,1 moli = nKOH  (din reactia K + H2O)
K + H2O ⇒ KOH + 1/2 H2 
solutia finala : ms = 80 g = ms1 + 3,9 - 1/2 ·0,1   (H2 degajat)
c = 11,2%    md = 0,112·80 = 8,96 g = md1 + 0,1·56 
ms1 = 80 - 3,9 + 0,05 = 76,15 g 
md1 = 8,96 - 5,6 = 3,36g      c1 = 3,36·100/76,15 = 4,41%