Solubility Product & Salt Hydrolysis
Master the final pillar of Ionic Equilibrium. Decode the $K_{sp}$ shortcuts, conquer the Common Ion Effect on precipitates, and calculate the exact pH of hydrolyzed salts.
Module Focus
Not all salts are neutral, and not all salts fully dissolve. Salt Hydrolysis explores how the ions of a salt react with water to make a solution acidic or basic. Solubility Product ($K_{sp}$) deals with sparingly soluble salts, quantifying exactly how much of a solid can dissolve before it starts to precipitate. In NEET, you must quickly identify the type of salt and instantly deploy the correct mathematical shortcut.
1. Salt Hydrolysis
Salt hydrolysis is the reverse of neutralization. It is the reaction of the cation, anion, or both of a salt with water, which alters the concentration of $H^+$ or $OH^-$ ions, thereby changing the pH of the solution.
Examples: $NaCl, KNO_3, Na_2SO_4$
- No Hydrolysis takes place.
- Solution remains neutral.
- $\mathbf{pH = 7}$ (at $25^\circ \text{C}$).
Examples: $CH_3COONa, NaCN, Na_2CO_3$
- Anionic Hydrolysis (The weak anion reacts).
- Solution is basic ($\mathbf{pH > 7}$).
- $K_h = \frac{K_w}{K_a}$
- $pH = 7 + \frac{1}{2}(pK_a + \log c)$
Examples: $NH_4Cl, CuSO_4, AgNO_3$
- Cationic Hydrolysis (The weak cation reacts).
- Solution is acidic ($\mathbf{pH < 7}$).
- $K_h = \frac{K_w}{K_b}$
- $pH = 7 - \frac{1}{2}(pK_b + \log c)$
Examples: $CH_3COONH_4, (NH_4)_2CO_3$
- Both Cationic & Anionic Hydrolysis.
- pH depends on relative strength of $K_a$ vs $K_b$.
- $K_h = \frac{K_w}{K_a \cdot K_b}$
- $pH = 7 + \frac{1}{2}(pK_a - pK_b)$
- (pH is strictly INDEPENDENT of concentration 'c'!)
2. Solubility & Solubility Product ($K_{sp}$)
For a sparingly soluble salt, an equilibrium exists between the undissolved solid and its dissolved ions in a saturated solution.
The General Formula Shortcut
Let the salt be $A_x B_y$ and its molar solubility be '$s$' mol/L.
$\mathbf{K_{sp} = x^x \cdot y^y \cdot s^{(x+y)}}$
$s = \sqrt{K_{sp}}$
$s = \left(\frac{K_{sp}}{4}\right)^{1/3}$
$s = \left(\frac{K_{sp}}{27}\right)^{1/4}$
3. Common Ion Effect on Solubility
According to Le Chatelier's Principle, adding a strong electrolyte that shares a common ion with the sparingly soluble salt will push the equilibrium backward. The solubility of a salt is always DECREASED in the presence of a common ion.
Example: Calculate the solubility ($s'$) of $AgCl$ in $0.1\text{ M } NaCl$ solution. (Given $K_{sp}$ of $AgCl = 10^{-10}$)
2. The weak salt ($AgCl$) dissolves slightly to give '$s'$' amount of $Ag^+$ and $Cl^-$.
3. Total $[Cl^-] = 0.1 + s'$. Since $s'$ is tiny, we approximate total $[Cl^-] \approx \mathbf{0.1\text{ M}}$.
4. Apply formula: $K_{sp} = [Ag^+][Cl^-]$
$10^{-10} = (s')(0.1)$
$s' = \frac{10^{-10}}{10^{-1}} = \mathbf{10^{-9}\text{ M}}$.
Notice how solubility dropped from $10^{-5}\text{ M}$ (in pure water) to $10^{-9}\text{ M}$!
4. Condition for Precipitation
Just like the Reaction Quotient ($Q$) helps predict the direction of a reaction, the Ionic Product ($Q_{sp}$) helps predict whether a precipitate will form when two solutions are mixed.
Unsaturated Solution
Saturated Solution
Supersaturated
If a solution contains multiple ions that can precipitate with an added reagent, the salt that requires the lowest concentration of the precipitating ion (based on its $K_{sp}$) will precipitate FIRST.
NEET Grand Test: Solubility & Hydrolysis
15 High-Order Thinking Questions testing common ion approximations, precipitation criteria, and pH equations.
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