Vrednost pH u rastvoru CH3CH2COOK jednaka je sa pH vreednoscu u rastvoru HCOOK i iznosi 9,22.Koja od soli ima vecu koncentraciju i koliko puta?k(CH3CH2COOK)=1.3x10^-5K(HCOOK)=1,8x10^-4
Da li se ove konstante odnose na kiseline ili na soli jer ja mislim da su ove konstante konstante kiselina a ne soli.
Da u pravu si,na kiseline se odnosi K (CH3CH2COOH)=1.3x10^-5 K(HCOOH)= 1,8x10^-4
Jel resenje 13.5?
Ne znam,ne pise..jel mozes da ga uradis?
CH3CH2COOH+H2O---------------->CH3CH2COO- + H3O+
CH3CH2COO- + H2O--------------->CH3CH2COOH+OH-
Kb={CH3CH2COOH}*{OH-}/{CH3CH2COO-}
{CH3CH2COOH}={OH-}=x
pH=9.22 a pOH=14-9.22=4.78pa je {OH-}=10(na pOH)=1.66*10(-5) mol/l
Kb=Kw/Ka=1*10(-14)/1.3*10(-5)=7.69*10(-10)
{CH3CH2COO-}=(1.66*10(-5)*2)/7.69*10(-10)
{CH3CH2COO-}=0.36 mol/l
CH3CH2COOK--------------->CH3CH2COO- + K+
{CH3CH2COOK}={CH3CH2COO-}=0.36 mol/l
HCOOH-------------->HCOO- + H+
HCOO- + H2O--------------->HCOOH+ OH-
{HCOOH}={OH-}=x
{HCOO-}=x*2/Kb
Kb=Kw/Ka=1*10(-14)/1.8*10(-4)=5.55*10(-11)
{HCOO-}=(1.66*10(-5))*2/5.55*10(-11) OH- joni su jednaki kao i u prvom slucaju jer je pH jednako
{HCOO-}=4.96 mol/l
HCOOK--------------->HCOO- + K+
{HCOOK}={HCOO-}=4.96 mol/l
{HCOOK}/{CH3CH2COOK}=4.96/0.36=13.5
{HCOOK}=13.5*{CH3CH2COOK}
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soli
Da li se ove konstante odnose na kiseline ili na soli jer ja mislim da su ove konstante konstante kiselina a ne soli.
Da u pravu si,na kiseline se
Da u pravu si,na kiseline se odnosi K (CH3CH2COOH)=1.3x10^-5 K(HCOOH)= 1,8x10^-4
Jel resenje 13.5?
Jel resenje 13.5?
Ne znam,ne pise..jel mozes
Ne znam,ne pise..jel mozes da ga uradis?
soli
CH3CH2COOH+H2O---------------->CH3CH2COO- + H3O+
CH3CH2COO- + H2O--------------->CH3CH2COOH+OH-
Kb={CH3CH2COOH}*{OH-}/{CH3CH2COO-}
{CH3CH2COOH}={OH-}=x
pH=9.22 a pOH=14-9.22=4.78pa je {OH-}=10(na pOH)=1.66*10(-5) mol/l
Kb=Kw/Ka=1*10(-14)/1.3*10(-5)=7.69*10(-10)
{CH3CH2COO-}=(1.66*10(-5)*2)/7.69*10(-10)
{CH3CH2COO-}=0.36 mol/l
CH3CH2COOK--------------->CH3CH2COO- + K+
{CH3CH2COOK}={CH3CH2COO-}=0.36 mol/l
HCOOH-------------->HCOO- + H+
HCOO- + H2O--------------->HCOOH+ OH-
{HCOOH}={OH-}=x
{HCOO-}=x*2/Kb
Kb=Kw/Ka=1*10(-14)/1.8*10(-4)=5.55*10(-11)
{HCOO-}=(1.66*10(-5))*2/5.55*10(-11) OH- joni su jednaki kao i u prvom slucaju jer je pH jednako
{HCOO-}=4.96 mol/l
HCOOK--------------->HCOO- + K+
{HCOOK}={HCOO-}=4.96 mol/l
{HCOOK}/{CH3CH2COOK}=4.96/0.36=13.5
{HCOOK}=13.5*{CH3CH2COOK}