## Equilibrium

# 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} = Cα^{2} , 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}

### View the Topic in this Video from 0:10 to 10:39

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1. For acidic buffer, pH = pK_{a} + log\frac{\left[salt\right]}{\left[acid\right]}

2. For basic buffer, pOH = pK_{b} + log\frac{\left[salt\right]}{\left[base\right]}

3. \tt Buffer \ capacity\left(\phi\right) = \frac{number \ of \ moles \ of \ acids \ or \ base \ added \ to \ 1 L \ of \ buffer}{change \ in \ pH}

4. For\ A_{x}B_{y} \rightleftharpoons xA^{y+} + y B^{x-}

K_{sp} = \left[A^{y+}\right]^{x} \left[B^{x-}\right]^{y}