Help- analitička hemija
Submitted by plavka994 on Mon, 05/30/2016 - 19:55

Forums:
Analizom 1,000 g neke stijene dobijeno je 0,2840 g smjese NaCl i KCl. Za titraciju hlorida u ovoj smjesi potrošeno je 42,95 cm3 rastvora AgNO3 koncentracije 0,100 mol/dm3. Koliki su maseni udjeli K2O i Na2O u uzorku?
Rješenje: w(K2O)= 9,64% i w(Na2O)=6,97%
Molim postupak. :D
m(smese)=0.2840 g
m(smese)=0.2840 g
AgNO3+NaCl------------->AgCl+NaNO3
AgNO3+KCl----------------->AgCl+KNO3
n(AgNO3)=c*V=0.1*0.04295=0.004295 mol
n(AgNO3)=n(NaCl)+n(KCl)
0.004295-=m(NaCl)/Mr+m(KCl)/Mr
0.004295*74.5*58.5=74.5*m(NaCl)+58.5*m(KCl)
m(smese)=m(NaCl)+m(KCl)
m(NaCl)=m(smese)-m(KCl)
18.71=74.5*(0.2840-m(KCl))+58.5*m(KCl)
-2.458= -16*m(KCl)
m(KCl)=2.458/16=0.1536 g
n(KCl)=m(KCl)/Mr=0.1536/74.5=0.00206 mol
Na2O+2HCl------------>2NaCl+H2O
K2O+2HCl----------------->2KCl+H2O
n(K2O)=n)KCl)/2=0.00206/2=0.00103 mol
m(K2O)=n(K2O)*Mr=0.00103*94=0.09682 g
w(K2O)=m(K2O)/m(smese)=0.09682/1=0.09682=9.682 %
m(Na2O)=m(smese)-m(K2O)=0.284-0.1536=0.1304 g
n(NaCl)=m(NaCl)/Mr=0.1304/58.5=0.00223 mol
n(Na2O)=n(NaCl)/2=0.00223/2=0.00115 mol
m(Na2O)=n(Na2O)*Mr=0.001115*62=0.06913 g
w(Na2O)=m(Na2O)/Mr=0.06913/1=0.06913=6.913 %