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shefali sharma Grade: 12
        

degree of dissociation of weak monoprotic  acid can be given as 


(A) a= 1/[1+10(pKa+pH)]


(B) a=1/[1+10(pKa-pH)]


(C) a=1/[1+10(-pKa+pH)]


(D) a=1/[1+10(pKa/pH)]


please give detailed solution

7 years ago

Answers : (1)

AskiitianExpert Pramod-IIT-R
47 Points
										Dear Student MS. shefali,


HA == H+ + A-
c(1-a) ca ca

Ka = ca2/(1-a)

c= (1-a)/a2 * Ka...............(1)

H+ = ca

c = H+/a....................(2)

by equation 1 and 2
H+/a == (1-a)/a2 * Ka

H+ == (1-a)/a * Ka

H+/Ka == (1-a)/a

H+/Ka == (1/a)-1

(H+/Ka) +1 == (1/a)

[10(-pH)]/[10(-pKa)] + 1 == (1/a)

[10(pKa-pH) + 1] == (1/a)

a=1/[1+10(pKa-pH)]

So B option is correct.
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Pramod Kumar
IITR Alumni






7 years ago
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