Help- analitička hemija

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

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milicaradenkovic - Tue, 05/31/2016 - 08:10

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 %