Hydrolysis of Salts and Solubility and Buffer Solutions

Degree of hydrolysis :
K_{h} = \frac{C\alpha.C\alpha}{C - C\alpha} = \frac{C\alpha^{2}}{1 - \alpha}
K_h = Cα² , 1 − α ≅ 1
\alpha = \sqrt{\frac{K_{h}}{C}}

pH of weak acid and strong base :
[OH^{-}] = \sqrt{\frac{K_{w}.C}{K_{a}}}
pOH = 7 - \frac{1}{2}pk_{a} - \frac{1}{2}logC
pH = 14 - \left(7 - \frac{1}{2}pk_{a} - \frac{1}{2}logC\right)
pH = 7 - \frac{1}{2}pk_{a} + \frac{1}{2}logC

Salt of strong acid and weak base :
NH_4^+ + H_{2}O \rightleftharpoons NH_{4}OH + H^{+}
k_{h} = \frac{k_{w}}{k_{b}}
\alpha = \sqrt{\frac{k_{h}}{C}}
pH = 7 - \frac{1}{2}pk_{a} - \frac{1}{2}logC

Salt of weak acid and weak base :
k_{h} = \frac{k_{w}}{k_{a}k_{b}}
k_{h} = \frac{C\alpha.C\alpha}{(C - \alpha) (C - \alpha)} = \alpha^{2}
\alpha = \sqrt{k_{h}}
pH = 7 + \frac{1}{2}pk_{a} - \frac{1}{2}pk_{a}

Types of buffers :

Acidic buffer :
Weak acid and its salt
CH_{3}COOH + CH_{3}COONa
HCN + KCN

Basic buffer :
Weak base and its salt
NH_{3} + NH_{4}Cl
NH_{4}OH + NH_{4}Cl

Henderson equation pH of acidic buffer :
pH = pk_{a} + log\left[\frac{salt}{Acid}\right]
Basic buffer − Henderson equation
pOH = pk_{b} + log\frac{\left[B^{+}\right]}{Base}
pOH = pk_{b} + log\frac{N_{s}V_{s}}{N_{b}V_{b}}

Buffer capacity (φ) :
\tt \phi = \frac{no. of \ moles \ of \ SA \ or \ SB \ added \ to \ 1L \ of \ buffer \ solution}{change \ in \ pH(\triangle pH)}
pH = pk_a, [salt] = [Acid]
pOH = Pk_b, [salt] = [Base]

Solubility product :
A_{x}.B_{y} \rightleftharpoons xA^{y+} + yB^{x-}
k_{sp} = [A^{y+}]^{x} [B^{x-}]^{y}
Relation between solubility and solubility product.
k_{sp} = (xs)^{x}.(ys)^{y} \Rightarrow x^{x}.y^{y}.s^{x + y}
s = \sqrt{k_{sp}}
s =\left(\frac{k_{sp}}{4}\right)^{1/3}; s =\left(\frac{k_{sp}}{27}\right)^{1/4}; s =\left(\frac{k_{sp}}{108}\right)^{1/5}