If $K_a > K_b \rightarrow$ Acidic
If $K_b > K_a \rightarrow$ Basic
If $K_a \approx K_b \rightarrow$ Neutral (or nearly neutral)

Hydrolysis constant: $K_h = \frac{K_w}{K_a K_b}$

6. Buffer Solutions

Definition: A solution that resists significant changes in its pH upon the addition of small amounts of strong acid or strong base.
Typically consists of:
- A weak acid and its conjugate base (e.g., $CH_3COOH/CH_3COONa$) $\rightarrow$ Acidic Buffer
- A weak base and its conjugate acid (e.g., $NH_3/NH_4Cl$) $\rightarrow$ Basic Buffer
Henderson-Hasselbalch Equation: Used to calculate the pH of a buffer solution.
- For Acidic Buffer: (weak acid $HA$ and its salt $A^-$) $$\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$$ or $$\text{pH} = \text{p}K_a + \log_{10} \frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}$$ Example: For a buffer of $CH_3COOH$ and $CH_3COONa$.
- For Basic Buffer: (weak base $B$ and its salt $BH^+$) $$\text{pOH} = \text{p}K_b + \log_{10} \frac{[\text{Salt}]}{[\text{Base}]}$$ or $$\text{pOH} = \text{p}K_b + \log_{10} \frac{[\text{Conjugate Acid}]}{[\text{Weak Base}]}$$ Example: For a buffer of $NH_3$ and $NH_4Cl$.

7. Solubility Product ($K_{sp}$)

Definition: For a sparingly soluble ionic compound, the solubility product is the product of the molar concentrations of its constituent ions, each raised to the power of its stoichiometric coefficient in the balanced dissolution equilibrium equation.
For a generic sparingly soluble salt $A_x B_y (s)$, its dissolution equilibrium is: $A_x B_y (s) \rightleftharpoons xA^{y+} (aq) + yB^{x-} (aq)$
The Solubility Product is: $K_{sp} = [A^{y+}]^x [B^{x-}]^y$
Relationship between Solubility ($s$) and $K_{sp}$: (where $s$ is the molar solubility in mol/L)
- For $AB$ type salt (e.g., $AgCl$): $K_{sp} = s \cdot s = s^2$
- For $A_2B$ or $AB_2$ type salt (e.g., $CaF_2$, $Ag_2S$): $K_{sp} = (s)^1 (2s)^2 = 4s^3$
- For $A_x B_y$ type salt: $K_{sp} = (xs)^x (ys)^y = x^x y^y s^{(x+y)}$
Example: For $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$, if $s$ is solubility, then $[Ag^+]=s$, $[Cl^-]=s$. $K_{sp} = s^2$. Example: For $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$, if $s$ is solubility, then $[Ca^{2+}]=s$, $[F^-]=2s$. $K_{sp} = (s)(2s)^2 = 4s^3$.
Condition for Precipitation:
- Ionic Product ($Q_{sp}$): Calculated in the same way as $K_{sp}$ but using current (non-equilibrium) ion concentrations.
- If $Q_{sp} > K_{sp}$: The solution is supersaturated; precipitation occurs until $Q_{sp} = K_{sp}$.
- If $Q_{sp} < K_{sp}$: The solution is unsaturated; no precipitation, more solid can dissolve.
- If $Q_{sp} = K_{sp}$: The solution is saturated; system is at equilibrium.
Common Ion Effect: The solubility of a sparingly soluble salt is significantly decreased when a solution containing a common ion is added. This is due to Le Chatelier's principle, as the added common ion shifts the dissolution equilibrium of the salt to the left, reducing its solubility. Example: Adding $NaCl$ to a saturated solution of $AgCl$ will decrease the solubility of $AgCl$ because of the common $