`HN_(3)` (hydroazic acid) is a weak acid dissociating as: `HN_(3) hArr H^(o+) + N_(3)^(Θ)`. Find the concentration of `Ag^(Θ)` ions, if excess of solid `AgN_(3)` is added to a solution maintained at `pH = 4`. The ionisation constant `K_(a)` of `HN_(3)` is `2.0 xx 10^(-5)`. The solubility of `AgN_(3)` in pure water is found to be `5.4 xx 10^(-3)` at `25^(@)C`.

Explanation

I method (using direct formula for the solubility of salt of weak acid in `H^(o+)` ion
`S (at pH = 4) = sqrt(K_(sp)) = [1+(H^(o+))/(K_(a))]^(1//2)`
`= sqrt(K_(sp))[1+(10^(-4))/(2.0 xx 10^(-5))]^(1//2)`
(since `sqrt(K_(sp)) = S = 5.4 xx 10^(-3)M)`
`= (5.4 xx 10^(-3)) [1+0.5 xx 10]^(1//2)`
`= (5.4 xx 10^(-3)) (6)^(1//2) = 5.4 xx 10^(-3) xx 2.449`
`=13.2246 xx 10^(-3) = 0.0132M`
`S_(Ag^(o+))` in buffer `= 0.0132M`
Second method:
`AgN_(3)` is a sparingly soluble salt, dissociating in water as:
`AgN_(3) hArr Ag^(o+) + N_(3)^(Θ)`
`K_(sp) = [Ag^(o+)] [N_(3)^(Θ)]`
Since solubility of `AgN_(3)` in water is `5.4 xx 10^(-3)`
`K_(sp) = (5.4 xx 10^(-3))^(2) = 2.92 xx 10^(-5)`
Now we have to find the solubility of `AgN_(3)` in solution having `pH = 4`.
Let solubility of `AgN_(3)` be `x M` at `pH = 4`.
`rArr [Ag^(o+)] = x M`, but `[N_(3)^(Θ) != xM`
`[N_(3)^(Θ)]` will be decided by dissociation of `HN_(3))`.
let `[N_(3)^(Θ)] = yM`
First, assume that whole of `N_(3)^(Θ)( x M)` formed from `AgN_(3)` reacts with `H^(o+)` ions to form `HN_(3)`.
`rArr [HN_(3)] = xM`
Now `HN_(3)` dissociates as follows:
`{:(HN_(3)hArr,H^(o+),N_(3)^(Θ),,),(x,10^(-4),-,,),(x-y,10^(-4)+y~~10^(-4),y,,):}`
Now we have a simultaneous equilibrium in aqueous solution involving dissociation of `AgN_(3)` and `HN_(3)`.
`K_(sp) of AgN_(3) = [Ag^(o+)] [N_(3)^(Θ)] = xy ...(i)`
`K_(sp)` of `HN_(3) = ([H^(o+)][N_(3)^(Θ)])/([HN_(3)]) = ([H^(o+)])/((x-y)) ...(ii)`
Solving equaitons (i) and (ii) simultaneously, we have:
`x = [Ag^(o+)] = sqrt((K_(sp))/(K_(a))([H^(o+)]+K_(a)))`
Substitute the values of `K_(sp),K_(a)`, and `[H^(o+)]` to get `x = 0.0132 mol L^(-1)`.
Note: In this example, since the dissociation constant of acid is quite high and `pH` of the solution is quite low, we can not assume that `x - y ~~ x`. You can check the same by comparing `[H^(o+)]` and `K_(a)` in the expression for `x`.