Chapter 3: Ionic Equilibria
Complete Theory, Derivations, Numericals & Fully Solved Board PYQs
1. Introduction, Electrolytes & Degree of Dissociation
The equilibrium established between unionized molecules and ions in the solution of weak electrolytes is called ionic equilibrium.
Types of Electrolytes
- Strong Electrolytes: Substances which ionize completely or almost completely in their aqueous solutions. Examples: Strong acids ($HCl, HNO_3, H_2SO_4$), strong bases ($NaOH, KOH$), and most salts ($NaCl, KCl$).
- Weak Electrolytes: Substances which ionize to a small extent in their aqueous solutions. Examples: Weak acids ($CH_3COOH, HCN$), weak bases ($NH_4OH$).
Degree of Dissociation ($\alpha$)
It is defined as the fraction of total number of moles of an electrolyte that dissociate into its ions at a state of equilibrium.
$$ \alpha = \frac{\text{Number of moles dissociated}}{\text{Total number of moles initially present}} $$Percent dissociation = $\alpha \times 100$
2. Theories of Acids and Bases
To define acids and bases, three major theories are proposed:
I. Arrhenius Theory
Acid: A substance which contains hydrogen and gives rise to hydrogen ions ($H^+$) in aqueous solution. Example: $HCl \rightarrow H^+ + Cl^-$
Base: A substance which contains the hydroxyl group and gives rise to hydroxyl ions ($OH^-$) in aqueous solution. Example: $NaOH \rightarrow Na^+ + OH^-$
Limitation: It is restricted to aqueous solutions only and cannot explain the basic nature of $NH_3$ (which lacks $OH^-$).
II. Bronsted-Lowry Theory (Proton Theory)
Acid: A substance (molecule or ion) which donates a proton ($H^+$) to another substance. (Proton donor)
Base: A substance (molecule or ion) which accepts a proton ($H^+$) from another substance. (Proton acceptor)
Conjugate Acid-Base Pair:
A pair of an acid and a base differing by a single proton is called a conjugate acid-base pair.
$HCl + H_2O \rightleftharpoons H_3O^+ + Cl^-$
Here, ($HCl, Cl^-$) is one conjugate pair and ($H_2O, H_3O^+$) is the other.
Amphoteric Nature of Water: Water can act as both an acid and a base. It acts as an acid with $NH_3$ and as a base with $HCl$.
III. Lewis Theory (Electron Theory)
Acid: Any species that can accept a share in an electron pair. (Electron pair acceptor). Examples: $BF_3, AlCl_3, H^+$.
Base: Any species that can donate a share in an electron pair. (Electron pair donor). Examples: $NH_3, H_2O, Cl^-$.
3. Ostwald's Dilution Law
This law expresses the quantitative relationship between the degree of dissociation ($\alpha$) of a weak electrolyte and its concentration ($c$).
For a Weak Acid ($HA$):
Consider the dissociation of a weak acid $HA$: $HA \rightleftharpoons H^+ + A^-$
If $c$ is initial concentration and $\alpha$ is degree of dissociation, at equilibrium:
$[H^+] = \alpha c$, $[A^-] = \alpha c$, $[HA] = c(1 - \alpha)$
Acid dissociation constant $K_a$: $$ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(\alpha c)(\alpha c)}{c(1 - \alpha)} = \frac{\alpha^2 c}{1 - \alpha} $$
For a very weak acid, $\alpha <<< 1$, so $(1 - \alpha) \approx 1$. Thus, $K_a = \alpha^2 c$.
$$ \alpha = \sqrt{\frac{K_a}{c}} = \sqrt{K_a \cdot V} $$Statement: The degree of dissociation of a weak electrolyte is inversely proportional to the square root of its concentration (or directly proportional to the square root of the volume containing 1 mole).
Similarly, for a weak base ($BOH$): $K_b = \alpha^2 c \implies \alpha = \sqrt{\frac{K_b}{c}}$
4. Autoionization of Water and pH Scale
Pure water ionizes to a very small extent: $H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$
Ionic Product of Water ($K_w$): The product of molar concentrations of hydrogen (hydronium) ions and hydroxyl ions in water.
$$ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \text{ (at 298 K)} $$In pure water, $[H_3O^+] = [OH^-] = 1.0 \times 10^{-7} \text{ M}$.
pH and pOH Scale (Sorensen)
pH: Negative base-10 logarithm of hydrogen ion concentration.
$$ pH = -\log_{10}[H^+] $$pOH: Negative base-10 logarithm of hydroxyl ion concentration.
$$ pOH = -\log_{10}[OH^-] $$Relationship between pH and pOH
Since $[H^+][OH^-] = 10^{-14}$
Taking $-\log_{10}$ on both sides:
$$ pH + pOH = 14 \text{ (at 298 K)} $$5. Hydrolysis of Salts
Hydrolysis: The reaction in which the cation, anion, or both of a salt react with water to produce acidity or alkalinity in the solution is called hydrolysis.
| Type of Salt | Examples | Nature of Solution | pH at 298K |
|---|---|---|---|
| Salt of Strong Acid & Strong Base | $NaCl, KNO_3, Na_2SO_4$ | Neutral (No hydrolysis occurs) | pH = 7 |
| Salt of Strong Acid & Weak Base | $NH_4Cl, CuSO_4, NH_4NO_3$ | Acidic (Cationic hydrolysis) | pH < 7 |
| Salt of Weak Acid & Strong Base | $CH_3COONa, K_2CO_3, NaCN$ | Basic/Alkaline (Anionic hydrolysis) | pH > 7 |
| Salt of Weak Acid & Weak Base | $CH_3COONH_4, (NH_4)_2CO_3$ | Depends on relative strengths of Acid ($K_a$) and Base ($K_b$) |
If $K_a > K_b$: Acidic If $K_a < K_b$: Basic If $K_a = K_b$: Neutral |
6. Buffer Solutions
A Buffer Solution is defined as a solution which resists drastic changes in its pH upon the addition of a small amount of strong acid, strong base, or water.
Types of Buffer Solutions
-
1. Acidic Buffer: A solution containing a weak acid and its salt with a strong base.
Example: $CH_3COOH + CH_3COONa$.
Henderson-Hasselbalch Equation for Acidic Buffer: $$ pH = pK_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]} $$ Where $pK_a = -\log_{10}K_a$ -
2. Basic Buffer: A solution containing a weak base and its salt with a strong acid.
Example: $NH_4OH + NH_4Cl$.
Henderson-Hasselbalch Equation for Basic Buffer: $$ pOH = pK_b + \log_{10} \frac{[\text{Salt}]}{[\text{Base}]} $$ Where $pK_b = -\log_{10}K_b$
Applications of Buffer Solutions
- In biochemical systems: The pH of human blood is maintained at 7.36 - 7.42 by the $H_2CO_3 / HCO_3^-$ buffer system.
- In agriculture: Soils are buffered. Addition of fertilizers alters pH, so soils are tested to choose correct fertilizers.
- In industry: Used in the manufacture of paper, dyes, inks, paints, and drugs (like penicillin preparations).
- In analytical chemistry: Buffer of $NH_4OH + NH_4Cl$ is used in qualitative analysis for precipitation of Group IIIA cations.
7. Solubility Product ($K_{sp}$)
For a sparingly soluble salt $A_xB_y$, the equilibrium in a saturated solution is:
$A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$
The Solubility Product ($K_{sp}$) is defined as the product of the molar concentrations of constituent ions, each raised to the power of its stoichiometric coefficient in the balanced equilibrium equation at a given temperature.
$$ K_{sp} = [A^{y+}]^x [B^{x-}]^y $$Relationship between Solubility ($S$) and $K_{sp}$
Let $S$ be the molar solubility (mol/L). Then $[A^{y+}] = xS$ and $[B^{x-}] = yS$.
$$ K_{sp} = (xS)^x (yS)^y = x^x y^y S^{(x+y)} $$
- For 1:1 type salt ($AgCl, BaSO_4$): $x=1, y=1 \implies K_{sp} = S^2$
- For 1:2 type salt ($PbCl_2, CaF_2$): $x=1, y=2 \implies K_{sp} = 4S^3$
- For 1:3 type salt ($Al(OH)_3$): $x=1, y=3 \implies K_{sp} = 27S^4$
Condition for Precipitation
Ionic Product (IP): It is the product of ion concentrations at any state (not necessarily equilibrium).
- If $IP < K_{sp}$: Solution is unsaturated (No precipitation).
- If $IP = K_{sp}$: Solution is exactly saturated.
- If $IP > K_{sp}$: Solution is supersaturated (Precipitation occurs).
8. Common Ion Effect
Definition: The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion is called the common ion effect.
Example: The dissociation of weak acid $CH_3COOH$ is suppressed by adding strong electrolyte $CH_3COONa$ (common ion is $CH_3COO^-$). According to Le Chatelier's principle, the common ion shifts the equilibrium of the weak electrolyte backwards.
Application: In salt analysis, to precipitate Group II basic radicals as sulfides, $H_2S$ gas is passed in the presence of dilute $HCl$. The common $H^+$ ion from $HCl$ suppresses the dissociation of weak acid $H_2S$, decreasing $[S^{2-}]$, ensuring only Group II radicals (which require low $K_{sp}$) precipitate.
9. Solved Textbook Numericals
Problem 1: pH Calculation
Calculate the pH and pOH of $0.01 \text{ M}$ $HCl$ solution.
Solution:
$HCl$ is a strong acid, so it dissociates completely.
$HCl \rightarrow H^+ + Cl^-$
Therefore, $[H^+] = c = 0.01 \text{ M} = 10^{-2} \text{ M}$.
Formula: $pH = -\log_{10}[H^+]$
$pH = -\log_{10}(10^{-2}) = 2\log_{10}10 = 2 \times 1 = 2$
We know, $pH + pOH = 14$
$pOH = 14 - 2 = 12$
Answer: pH = 2, pOH = 12.
Problem 2: Solubility Product
The solubility of $AgCl$ in water at 298 K is $1.06 \times 10^{-5} \text{ mol dm}^{-3}$. Calculate its solubility product.
Solution:
Solubility ($S$) = $1.06 \times 10^{-5} \text{ mol dm}^{-3}$
$AgCl$ dissociates as: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$
This is a 1:1 type salt. Let solubility be $S$.
$[Ag^+] = S$, and $[Cl^-] = S$
$K_{sp} = [Ag^+][Cl^-] = (S)(S) = S^2$
$K_{sp} = (1.06 \times 10^{-5})^2$
$K_{sp} = 1.1236 \times 10^{-10}$
Answer: Solubility product of AgCl is $1.1236 \times 10^{-10}$.
Problem 3: Ostwald's Dilution Law
A weak monobasic acid is 0.05% dissociated in 0.02 M solution. Calculate the dissociation constant of the acid.
Solution:
Given: Percentage dissociation = 0.05%
Degree of dissociation, $\alpha = \frac{0.05}{100} = 5 \times 10^{-4}$
Concentration, $c = 0.02 \text{ M} = 2 \times 10^{-2} \text{ M}$
Formula for weak acid: $K_a = \alpha^2 c$ (since $\alpha$ is very small, $1-\alpha \approx 1$)
$K_a = (5 \times 10^{-4})^2 \times (2 \times 10^{-2})$
$K_a = (25 \times 10^{-8}) \times (2 \times 10^{-2})$
$K_a = 50 \times 10^{-10} = 5.0 \times 10^{-9}$
Answer: The dissociation constant ($K_a$) is $5.0 \times 10^{-9}$.
10. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Define: Lewis acid. (Oct 2013, March 2022)
Answer: A Lewis acid is any species (molecule or ion) that can accept a share in an electron pair.
Q2. Define pH. (March 2017)
Answer: pH is defined as the negative base-10 logarithm of the molar concentration of hydrogen (or hydronium) ions in a solution. ($pH = -\log_{10}[H^+]$)
Q3. Write the relationship between solubility and solubility product for $PbI_2$. (Oct 2015)
Answer: $PbI_2$ is a 1:2 type salt. It dissociates as $PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$. If solubility is $S$, $[Pb^{2+}]=S, [I^-]=2S$. Therefore, $K_{sp} = [Pb^{2+}][I^-]^2 = (S)(2S)^2 = 4S^3$. The relationship is $K_{sp} = 4S^3$.
2 Mark Questions (SA-I)
Q4. State and explain Ostwald's dilution law for weak acid. (March 2014, March 2019)
Statement: The degree of dissociation of a weak acid is inversely proportional to the square root of its concentration (or directly proportional to square root of dilution).
Explanation: Let a weak acid HA have initial concentration $c$ and degree of dissociation $\alpha$. At equilibrium, $[H^+] = \alpha c$, $[A^-] = \alpha c$, $[HA] = c(1-\alpha)$.
The dissociation constant $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{\alpha^2 c^2}{c(1-\alpha)} = \frac{\alpha^2 c}{1-\alpha}$.
For a weak acid, $\alpha$ is very small, $1-\alpha \approx 1$. Thus, $K_a = \alpha^2 c \implies \alpha = \sqrt{\frac{K_a}{c}}$.
Q5. What is a buffer solution? Give one application of it in biological systems. (March 2016, Oct 2020)
Answer: A buffer solution is a solution whose pH remains practically constant upon the addition of a small amount of strong acid, strong base, or water.
Application: Human blood is a natural buffer solution maintained at a pH of 7.36 to 7.42 by the $H_2CO_3$ / $HCO_3^-$ buffer system. This is crucial for the functioning of enzymes and biological processes.
3 Mark Questions (SA-II)
Q6. Explain the common ion effect with a suitable example. How is it applied in qualitative analysis? (Oct 2014, March 2018)
Common Ion Effect: The phenomenon in which the degree of dissociation of a weak electrolyte is suppressed by the addition of a strong electrolyte containing a common ion is called the common ion effect. It is an application of Le Chatelier's principle.
Example: Addition of sodium acetate ($CH_3COONa$, a strong electrolyte) to acetic acid ($CH_3COOH$, a weak electrolyte) suppresses the ionization of acetic acid due to the common acetate ion ($CH_3COO^-$).
Application in Qualitative Analysis: To precipitate Group II basic radicals as sulfides, $H_2S$ gas is passed through the solution in the presence of dilute $HCl$. The strong acid $HCl$ provides a common $H^+$ ion which suppresses the ionization of weak acid $H_2S$. This decreases the $[S^{2-}]$ concentration to a level where only the solubility product of Group II sulfides (which is very low) is exceeded, preventing the precipitation of Group IV sulfides.
Q7. Calculate the pH of a buffer solution composed of 0.1 M weak base $BOH$ and 0.2 M of its salt $BA$. ($K_b$ for $BOH = 1.8 \times 10^{-5}$) (March 2021)
Given:
$[\text{Base}] = 0.1 \text{ M}$
$[\text{Salt}] = 0.2 \text{ M}$
$K_b = 1.8 \times 10^{-5}$
Step 1: Calculate $pK_b$
$pK_b = -\log_{10}(K_b) = -\log_{10}(1.8 \times 10^{-5})$
$pK_b = 5 - \log_{10}(1.8) = 5 - 0.2553 = 4.7447$
Step 2: Use Henderson-Hasselbalch equation for Basic Buffer
$pOH = pK_b + \log_{10}\frac{[\text{Salt}]}{[\text{Base}]}$
$pOH = 4.7447 + \log_{10}\frac{0.2}{0.1}$
$pOH = 4.7447 + \log_{10}(2)$
$pOH = 4.7447 + 0.3010 = 5.0457$
Step 3: Calculate pH
$pH + pOH = 14$
$pH = 14 - 5.0457 = 8.9543$
Final Answer: The pH of the buffer solution is 8.95.
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