Titracije

Submitted by plavka994 on Mon, 04/25/2016 - 22:46
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Za titraciju 0,2000 g smjese Na2CO3 i K2CO3 troši se 33,34 cm3 rastvora HCl koncetracije 0,100 mol/dm3. Izračunati sastav smjese.

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milicaradenkovic - Wed, 04/27/2016 - 19:05

K2CO3+ 2HCl----------->2KCl+H2O+CO2

Na2CO3+2HCl----------->2NaCl+H2O+CO2

n(HCl)u.=c*V=0.03334*0.1=0.003334 mol

m(smese)=m(Na2CO3)+m(K2CO3)

n(HCl)u=n(HCl)1+n(HCl)2

n(HCl)1=2*n(K2CO3)

n(HCl)2=2*(Na2CO3)

0.003334=2*m(Na2CO3)/Mr(Na2CO3)+2*m(K2CO3)/Mr(K2CO3)

m(Na2CO3)=0.2-m(K2CO3)

0.003334=2*(0.2-m(K2CO3))/138+2*m(K2CO3)/106

138*106*0.003334/2=106*0.2-106*m(K2CO3)+138*m(K2CO3)

32*m(K2CO3)=24.3848-21.2

m(K2CO3)=3.1848/32=0.099525 g

m(Na2CO3)=0.2-0.099525=0.100475 g

w(Na2CO3)=m(Na2CO3)/m(smese)=0.100475/0.2=0.5023=50.23 %

w(K2CO3)=m(K2CO3)/m(smese)=0.099525/0.2=0.4977=49.77 %

 

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plavka994 - Sun, 05/01/2016 - 14:08

Hvalaa punooo :*