CHEMCA
EXAM MASTER FORMULA SHEET
Chemical & Ionic Equilibrium
1. Chemical Equilibrium & Thermodynamics
For $aA + bB \rightleftharpoons cC + dD$
\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]Pure solids and pure liquids are taken as unity (1).
$\Delta n_g = (\text{moles of gaseous prod.}) - (\text{moles of gaseous react.})$
At Eq ($Q = K, \Delta G = 0$):
\[ \Delta G^\circ = -2.303 RT \log K_{eq} \]Temperature dependence of Equilibrium Constant
Endothermic ($\Delta H>0$): $T \uparrow \implies K \uparrow$
Exothermic ($\Delta H<0$): $T \uparrow \implies K \downarrow$
- Pressure Increase: Shifts equilibrium towards fewer moles of gas ($\Delta n_g$).
- Volume Increase: Pressure drops, shifts towards more moles of gas.
- Inert Gas added at Constant Volume: No shift (partial pressures remain same).
- Inert Gas added at Constant Pressure: Volume increases, shifts towards more gaseous moles.
- Catalyst: Reaches equilibrium faster, but NO effect on the value of $K_{eq}$ or final yield.
2. Acids, Bases & Ionic Equilibrium
At $90^\circ C$, $K_w \approx 10^{-12}$. Therefore, at $90^\circ C$, neutral $pH = 6$. (pH 7 is BASIC at 90°C!).
Degree of Dissociation ($\alpha$):
\[ \alpha = \sqrt{\frac{K_a}{C}} \quad (\text{Valid if } \alpha < 0.05) \]Concentration & pH:
\[ [H^+] = C\alpha = \sqrt{K_a \cdot C} \] \[ pH = \frac{1}{2}[pK_a - \log C] \]3. Buffer Solutions
Solutions which resist change in pH upon addition of small amounts of strong acid or strong base.
E.g., $CH_3COOH + CH_3COONa$
E.g., $NH_4OH + NH_4Cl$
Moles of strong acid/base added per liter to change pH by 1 unit. Maximum when $[Salt] = [Acid]$.
Buffer works best when ratio of Salt to Acid is between 1/10 and 10.
4. Salt Hydrolysis Summary
Reaction of cation or anion (or both) of a salt with water to produce acidity or basicity.
| Type of Salt & Nature | Hydrolysis Const. ($K_h$) | Degree of Hyd. ($h$) | pH Formula (at 25°C) |
|---|---|---|---|
| Strong Acid + Strong BaseNeutral ($NaCl, KNO_3$) | No Hydrolysis | $pH = 7$ | |
| Strong Acid + Weak BaseAcidic ($NH_4Cl$) | $\frac{K_w}{K_b}$ | $\sqrt{\frac{K_w}{K_b \cdot C}}$ | $7 - \frac{1}{2}(pK_b + \log C)$ |
| Weak Acid + Strong BaseBasic ($CH_3COONa$) | $\frac{K_w}{K_a}$ | $\sqrt{\frac{K_w}{K_a \cdot C}}$ | $7 + \frac{1}{2}(pK_a + \log C)$ |
| Weak Acid + Weak BaseDepends on Ka/Kb ($CH_3COONH_4$) | $\frac{K_w}{K_a \cdot K_b}$ | $\sqrt{\frac{K_w}{K_a \cdot K_b}}$ | $7 + \frac{1}{2}(pK_a - pK_b)$ (Independent of Concentration C) |
5. Solubility Product ($K_{sp}$) & Precipitation
For sparingly soluble salt $A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$:
Where $s$ is the solubility in mol/L.
- $Q_{sp} < K_{sp}$: Unsaturated, no precipitate.
- $Q_{sp} = K_{sp}$: Saturated, at equilibrium.
- $Q_{sp} > K_{sp}$: Supersaturated, Precipitation occurs!
Addition of a common ion suppresses the dissociation of a weak electrolyte or solubility of a sparingly soluble salt.
Solubility of $AgCl$ in $0.1M\ NaCl$ is $s' = \frac{K_{sp}}{0.1}$
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This is really helpful, thank you so much
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