Prove that the degree of dissociation of a weak monoprotic acid is given by alpha = (1)/(1+ 10^((pK_(a)-pH))) where K_(a) is the dissociation constant of the acid.

Prove that the degree of dissociation of a weak monoprotic acid is giv

1.	Prove that the degree of dissociation of a weak monoprotic acid is given by alpha = (1)/(1+ 10^((pK_(a)-pH))) where K_(a) is the dissociation constant of the acid.
Answer» Solution :Suppose we start with C mol `L^(-1)` of the weak monoprotic ACID HA. Then ` {:(,HA,HARR,H^(+),+,A^(-)),("INITIAL molar conc.",C,,,,),("Molar conc.",C-C alpha,,C alpha,,C alpha),("after dissociation",=C (1-alpha),,,,),(,,,,,):}` Thus, `K_(a) = (C alpha. C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha) or C = (K_(a)(1-alpha))/(alpha^(2)) ...(i)` Also, `[H^(+)]=C alpha` ...(ii) Substituting the value of C from eqn. (i) , we get `[H^(+)]=(K_(a)(1-alpha))/(alpha^(2)) xx alpha=(K_(a)(1-alpha))/(alpha)` `:. -log (H^(+)]=-[log K_(a) + log (1-alpha)-log alpha]or pH = pK_(a) - log (1-alpha) + log alpha` or `log.(1-alpha)/(alpha) = pK_(a) - pHor (1-alpha)/(alpha) = 10^(pK_(a)-pH) or (1)/(alpha)-1=10^(pK_(a)-pH)` or `(1)/(alpha) = 1 + 10^(pK_(a)-pH) or alpha= (1)/(1+10^(pK_(a)-pH))`