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

Explanation

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) - pH or (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))`