pH

Submitted by plavka994 on Mon, 07/06/2015 - 06:45
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Izracunaj pH 0,5%-tne otopine HCN-a ako je K(HCN)= 7,9*10^-10

buduci_student's picture

buduci_student - Mon, 07/06/2015 - 08:38

HCN----->H + CN-

K=[H][CN]/[HCN]

K=(x)(X)/{HCN]

K=x2/[HCN}

x2=K*[HCN]

x=sqrt(7,9*10-10*0,5)

x=sqrt(3,95*10-10)

x=1,987*10-5

ph=-log(x)=4,7

Ako mozes molim te samo reci rezultat ako ga imas u zbirci,jer sam ovu koncentraciju samo predpostavio

 

buduci_student's picture

buduci_student - Mon, 07/06/2015 - 08:45

.......HCN(aq) + H2O(l) <=> H3O(+)(aq) + CN(-)(aq)
..Poc. 7,9x10-10 m... -- ......... 0 m ............. 0 m
..P........ -x .......... -- ......... +x ............... +x
..E.. 7,9x10-10 - x. -- ........... x ................. x

K=[H3O][CN]/[HCN]

K=(x)(X)/{HCN]

K=x2/[HCN}

x2=K*[HCN]

E sada nisam siguran cemu je{HCN] jednako, ili 0,5 ili 0,0005

milicaradenkovic's picture

milicaradenkovic - Mon, 07/06/2015 - 14:32

(HCN)=0.005

buduci_student's picture

buduci_student - Mon, 07/06/2015 - 14:37

E da da 0,5/100=0,005

.......HCN(aq) + H2O(l) <=> H3O(+)(aq) + CN(-)(aq)
..Poc. 7,9x10-10 m... -- ......... 0 m ............. 0 m
..P........ -x .......... -- ......... +x ............... +x
..E.. 7,9x10-10 - x. -- ........... x ................. x

K=[H3O][CN]/[HCN]

K=(x)(X)/{HCN]

K=x2/[HCN}

x2=K*[HCN]

x=sqrt(7,9*10-10*0,005)

x=sqrt(3,95*10-12)

x=3,95*10-6

ph=-log(x)=5,4