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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

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


1 Answers

AskiitianExpert Pramod-IIT-R
47 Points
11 years ago
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. Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Pramod Kumar IITR Alumni

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