Question

300 g sol NaOH cp=20% se neutralizeaza cu 730 g HCl de cp=36,5%.
Concentratia finala=?
Apa=?
Nimeni

Answer 1

C = md/ms×100 => md = C×ms/100 = 20×300/100 = 60g NaOH

m H2O = ms-md = 300-60 = 240g

md = 36,5×730/100 = 266,45g HCl

m H2O = ms-md = 730-266,45 = 463,55g

n NaOH = 60/40 = 1,5 moli
n HCl = 266,45/36,5 = 7,3 moli

n HCl exces = 7,3-1,5 = 5,8 moli

m HCl exces = 5,8×36,5 = 211,7g

NaOH + HCl => NaCl + H2O

n NaOH = n NaCl = 1,5 moli

mdf = 1,5×58,5 + 211,7 = 299,45g
msf = ms1 + ms2 = 300+730 = 1030g
m H2O final = 240+463,55 = 703,55g

Cf = mdf/msf×100 = 299,45/1030×100 = 29%