pH vrednost

Submitted by zzz555 on Sun, 02/09/2014 - 13:01
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Izracunajte kolika je pH vrednost u 100 cm3 rastvora 0,0010 M CH3NH3+Cl- ( pKa, CH3NH3+ 10,6).

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milicaradenkovic - Sun, 02/09/2014 - 13:17

CH3NH3+Cl-+H2O ---------------------->CH3NH3+ +Cl-

CH3NH3+ + H2O---------------->CH3NH2+H3O+

n(CH3NH3+Cl-)=c*V=0.1*0.001=0.0001 mol

c(CH3NH3+Cl-)=n(CH3NH3+Cl-)/V=0.0001 mol/l

{CH3NH3+}=c(CH3NH3+Cl-)=0.0001 mol/l

Ka={CH3NH2}*{H3O+}/{CH3NH3+}

{CH3NH2}={H3O+}=x

2.51*10(-11)=x*2/0.0001

x=(koren(2.5*10(-11)*0.0001))=5*10(-8)

{H3O+}=5*10(-8) mol/l

pH= -log{H3O+}=7.3

 

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zzz555 - Sun, 02/09/2014 - 13:21

Tako sam i ja radio, ali u resenju pise da je pH=6,7

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milicaradenkovic - Sun, 02/09/2014 - 14:52

Onda je greska u jednacini

CH3NH3+ +  H2O----------------->CH3NH42+ + OH-

Ka={CH3NH42+}*{OH-}/{CH3NH3+}

{CH3NH42+}={OH-}=x

pa se dobije da je {OH-}=5*10(-8) mol/l

pOH= -log{OH-}=7.3

pH=14-pOH=14-7.3=6.7

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zzz555 - Sun, 02/09/2014 - 15:47

Ok, hvala