IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - ionic eq-1

joyfrancis

Calculate (i)degree of hydrolysis of decinormal KCN solution at 25^o. The dissosiation const of HCN is 7.2*10^-10 and ionic product of water is 10^-14. Also find pH value of this sol.

(Give full sol WITH the ans)

amitsingh

Look KCN is salt of a weak acidand strong base.

So just use the formulae-

PH = 7 + 1/2PK_a + 1/2 logC

And degree of hydrolysis, h = root ( K_h/C)

Use this u will get the ans

Tell me if something is wrong.

magiclko

ofcourse the formula is correct, nd it will be gr8 if u rember the formula, but even if u dnt(like i cant

), here's the full solution, and also the proof of the formula

KCN <====> K ⁺+ CN ^-

in an aqueous solution of KCN, the following equlibrium will exist

KCN + H₂O <=====> HCN + KOH .............(1)

or it can be written as

CN^-+ H₂O <====> HCN + OH^-

t=0 c 0 0

t=

c(1-h) ch ch

thus the hydrolysis constant K_h = [OH^-] [HCN ]

[CN^-]

frm here we'll get h =

(K_h/ C) for h<<<<1

also HCN <====> H⁺+ CN^- K_a(dissociation constant)

thus we have K_h X K_a = K_w

=> K_h = K_w/ K_a

thrfore h =

(K_h/ C)

=

{ K_w/ ( K_a C) }

thus [OH^-] = ch =

{ (CK_w)/ K_a }

takingnegative log of noth sides we have

- log [OH] = - log [

{ (CK_w)/ K_a } ]

=> pOH = 1/2 [pK_w - pK_a - log C ]

=> pH = 1/2 [pK_w + pK_a + log C ]

u have the values, u'll get the answer

and even if u have any problem do knock back :)

ramya_kapoor

degree of hyd. (h)=root kh/c

and kh=kw/ka....

rest evrythg is correct

magiclko

i have proved the same thing ramya

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