Izracunati koliko grama soli je rastvoreno ako pH vrednost u rastvoru C6H5COOK iznosi 6,2Kk(C6H5COOH)=6,5x 10^-5
pH=6.2 pa je {H+}=6.31*10(-7) mol/l
pOH=7.8 pa je {OH-}=1.58*10(-8) mol/l
C6H5COOH------------------>C6H5COO- + H+
C6H5COO- +H2O---------------->C6H5COOH+OH-
{C6H5COOH}={OH-}=1.58*10(-8) mol/l
Ka={C6H5COO-}*{H+}/{C6H5COOH}
6.5*10(-5)=6.31*10(-7)*{C6H5COO-}/1.58*10(-8)
1.03*10(-12)={C6H5COO-}*6.31*10(-7)
{C6H5COO-}=1.63*10(-6) mol/l
C6H5COOK------------>C6H5COO- + K+
{C6H5COOK}={C6H5COO-}=1.63*10(-6) mol/l
n(C6H5COOK)=1.63*10(-6) mol
m(soli)=n(C6H5COOK)*Mr=1.63*10(-6)*160=2.608*10(-4) g
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Grama soli
pH=6.2 pa je {H+}=6.31*10(-7) mol/l
pOH=7.8 pa je {OH-}=1.58*10(-8) mol/l
C6H5COOH------------------>C6H5COO- + H+
C6H5COO- +H2O---------------->C6H5COOH+OH-
{C6H5COOH}={OH-}=1.58*10(-8) mol/l
Ka={C6H5COO-}*{H+}/{C6H5COOH}
6.5*10(-5)=6.31*10(-7)*{C6H5COO-}/1.58*10(-8)
1.03*10(-12)={C6H5COO-}*6.31*10(-7)
{C6H5COO-}=1.63*10(-6) mol/l
C6H5COOK------------>C6H5COO- + K+
{C6H5COOK}={C6H5COO-}=1.63*10(-6) mol/l
n(C6H5COOK)=1.63*10(-6) mol
m(soli)=n(C6H5COOK)*Mr=1.63*10(-6)*160=2.608*10(-4) g