Chapter 3: Ionic Equilibria
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What are electrolytes and non-electrolytes? Give one example of each.
Electrolytes: Substances which give rise to ions when dissolved in water or in their molten state and conduct electricity are called electrolytes. Example: $NaCl$ (Sodium chloride).
Non-electrolytes: Substances which do not ionize and do not conduct electricity in their aqueous solution or molten state are called non-electrolytes. Example: Urea, Sucrose.
2. Differentiate between strong and weak electrolytes.
| Strong Electrolytes | Weak Electrolytes |
|---|---|
| Ionize completely (almost 100%) in aqueous solution. | Ionize only partially in aqueous solution. |
| High electrical conductivity. | Low electrical conductivity. |
| Degree of dissociation ($\alpha$) is approx 1. | Degree of dissociation ($\alpha$) is much less than 1. |
| Examples: Strong acids ($HCl$), Strong bases ($NaOH$), most salts ($NaCl$). | Examples: Weak acids ($CH_3COOH$), Weak bases ($NH_4OH$). |
3. Define degree of dissociation ($\alpha$). State its formula.
Degree of dissociation ($\alpha$): It is defined as the fraction of the total number of moles of an electrolyte that dissociates into its ions when equilibrium is attained.
$$ \alpha = \frac{\text{Number of moles dissociated}}{\text{Total number of moles dissolved}} $$
Percent dissociation = $\alpha \times 100$
4. State Arrhenius theory of acids and bases.
According to Arrhenius (1887):
- Acid: A substance that contains hydrogen and gives rise to hydrogen ions ($H^+$) in aqueous solution. Example: $HCl \rightleftharpoons H^+_{(aq)} + Cl^-_{(aq)}$
- Base: A substance that contains a hydroxyl group and produces hydroxide ions ($OH^-$) in aqueous solution. Example: $NaOH \rightleftharpoons Na^+_{(aq)} + OH^-_{(aq)}$
5. What are the limitations of the Arrhenius theory?
- It explains the acidic/basic behavior only in aqueous solutions. It cannot explain the behavior in non-aqueous solvents (like liquid ammonia).
- It cannot explain the basicity of substances that do not contain an $OH^-$ group, such as ammonia ($NH_3$) or sodium carbonate ($Na_2CO_3$).
- It doesn't recognize that the bare $H^+$ ion doesn't exist independently in water; it exists as the hydronium ion ($H_3O^+$).
6. State Bronsted-Lowry theory of acids and bases.
According to J.N. Bronsted and T.M. Lowry (1923):
- Acid: A substance (molecule or ion) that has a tendency to donate a proton ($H^+$) to another substance. (Proton donor).
- Base: A substance (molecule or ion) that has a tendency to accept a proton ($H^+$) from another substance. (Proton acceptor).
Example: $HCl + NH_3 \rightleftharpoons NH_4^+ + Cl^-$. Here $HCl$ donates a proton (acid) and $NH_3$ accepts it (base).
7. What are conjugate acid-base pairs? Explain with an example.
A pair of an acid and a base that differ from each other by a single proton ($H^+$) is called a conjugate acid-base pair.
Example: $HCl (acid_1) + H_2O (base_2) \rightleftharpoons H_3O^+ (acid_2) + Cl^- (base_1)$
- $HCl$ and $Cl^-$ are a conjugate acid-base pair. ($Cl^-$ is the conjugate base of $HCl$).
- $H_2O$ and $H_3O^+$ are a conjugate acid-base pair. ($H_3O^+$ is the conjugate acid of $H_2O$).
8. State Lewis theory of acids and bases.
According to G.N. Lewis (1923):
- Lewis Acid: Any species (atom, ion, or molecule) that can accept a pair of electrons to form a coordinate covalent bond. (Electron pair acceptor). Examples: $BF_3$, $AlCl_3$, $H^+$, $Cu^{2+}$.
- Lewis Base: Any species (atom, ion, or molecule) that can donate a pair of electrons to form a coordinate covalent bond. (Electron pair donor). Examples: $NH_3$, $H_2O$, $Cl^-$.
9. Justify: $BF_3$ is a Lewis acid.
In Boron trifluoride ($BF_3$), the central Boron atom has only 6 electrons in its valence shell (it has an incomplete octet). Therefore, it is an electron-deficient molecule and has a strong tendency to accept a lone pair of electrons to complete its octet. Since it acts as an electron-pair acceptor, it is a Lewis acid.
10. Prove the amphoteric nature of water with suitable examples.
Water is amphoteric because it can act as both a Bronsted acid (proton donor) and a Bronsted base (proton acceptor).
- As a base: When reacted with a strong acid ($HCl$), water accepts a proton.
$HCl + H_2O \rightarrow H_3O^+ + Cl^-$ (Water acts as a base) - As an acid: When reacted with a weak base ($NH_3$), water donates a proton.
$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$ (Water acts as an acid)
11. Write the equation for the auto-ionization of water. What is the ionic product of water ($K_w$)?
Water undergoes self-ionization to a very small extent:
$$ H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} $$
Ionic Product of Water ($K_w$): It is the product of the molar concentrations of hydronium ions ($H_3O^+$) and hydroxyl ions ($OH^-$) in water or in any aqueous solution at a given temperature.
$$K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \text{ at 298 K}$$
12. How does the ionic product of water ($K_w$) change with temperature? Why?
The ionic product of water ($K_w$) increases with an increase in temperature.
Reason: The auto-ionization of water is an endothermic process. According to Le Chatelier's principle, an increase in temperature shifts the equilibrium to the right, favoring greater ionization. This increases the concentration of both $[H_3O^+]$ and $[OH^-]$, thereby increasing $K_w$.
13. Define pH and pOH of a solution.
pH: It is defined as the negative logarithm to the base 10 of the molar concentration of hydrogen ions ($H^+$ or $H_3O^+$) in the solution.
$$ \text{pH} = -\log_{10}[H_3O^+] $$
pOH: It is defined as the negative logarithm to the base 10 of the molar concentration of hydroxyl ions ($OH^-$) in the solution.
$$ \text{pOH} = -\log_{10}[OH^-] $$
14. Derive the relationship between pH and pOH for an aqueous solution.
The ionic product of water is given by: $[H_3O^+][OH^-] = K_w$
At 298 K, $K_w = 1.0 \times 10^{-14}$. Thus, $[H_3O^+][OH^-] = 10^{-14}$
Taking $\log_{10}$ on both sides:
$\log_{10}[H_3O^+] + \log_{10}[OH^-] = \log_{10}(10^{-14})$
$\log_{10}[H_3O^+] + \log_{10}[OH^-] = -14$
Multiplying by -1:
$(-\log_{10}[H_3O^+]) + (-\log_{10}[OH^-]) = 14$
Since $\text{pH} = -\log_{10}[H_3O^+]$ and $\text{pOH} = -\log_{10}[OH^-]$, we get:
$$ \text{pH} + \text{pOH} = 14 $$
15. How do you classify aqueous solutions as acidic, basic, or neutral based on pH at 298 K?
- Neutral solution: $[H_3O^+] = [OH^-] = 10^{-7} \text{ M}$. Therefore, pH = 7.
- Acidic solution: $[H_3O^+] > [OH^-]$, meaning $[H_3O^+] > 10^{-7} \text{ M}$. Therefore, pH < 7.
- Basic solution: $[H_3O^+] < [OH^-]$, meaning $[H_3O^+] < 10^{-7} \text{ M}$. Therefore, pH > 7.
16. Calculate the pH of 0.01 M HCl solution.
HCl is a strong acid, so it dissociates completely.
$HCl \rightarrow H^+ + Cl^-$
Therefore, $[H^+] = \text{Concentration of } HCl = 0.01 \text{ M} = 10^{-2} \text{ M}$.
$\text{pH} = -\log_{10}[H^+] = -\log_{10}(10^{-2})$
$\text{pH} = -(-2\log_{10}10) = 2$
The pH of the solution is 2.
17. Calculate the pH of $10^{-3}$ M NaOH solution.
NaOH is a strong base and dissociates completely.
$NaOH \rightarrow Na^+ + OH^-$
$[OH^-] = 10^{-3} \text{ M}$
$\text{pOH} = -\log_{10}[OH^-] = -\log_{10}(10^{-3}) = 3$
We know, $\text{pH} + \text{pOH} = 14$
$\text{pH} + 3 = 14 \Rightarrow \text{pH} = 11$
The pH of the solution is 11.
18. Define dissociation constant of a weak acid ($K_a$) and a weak base ($K_b$).
For a weak acid $HA \rightleftharpoons H^+ + A^-$, the dissociation constant ($K_a$) is the equilibrium constant for its dissociation in water.
$$ K_a = \frac{[H^+][A^-]}{[HA]} $$
For a weak base $BOH \rightleftharpoons B^+ + OH^-$, the dissociation constant ($K_b$) is the equilibrium constant for its dissociation.
$$ K_b = \frac{[B^+][OH^-]}{[BOH]} $$
They are a measure of the strength of the weak acid or base.
19. State Ostwald's dilution law.
Ostwald's Dilution Law: It states that for a weak electrolyte, the degree of dissociation ($\alpha$) is inversely proportional to the square root of its concentration ($C$), or directly proportional to the square root of the volume ($V$) of the solution containing 1 mole of the electrolyte.
Mathematically: $\alpha \propto \frac{1}{\sqrt{C}}$ or $\alpha \propto \sqrt{V}$
20. Derive Ostwald's dilution law for a weak acid ($HA$).
Consider a weak acid $HA$ with initial concentration $C$ mol/L and degree of dissociation $\alpha$.
Reaction: $HA \rightleftharpoons H^+ + A^-$
Initial conc: $C \quad \quad \quad 0 \quad \quad 0$
Eq. conc: $C(1-\alpha) \quad C\alpha \quad \quad C\alpha$
The dissociation constant $K_a = \frac{[H^+][A^-]}{[HA]}$
$$ K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C^2\alpha^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} $$
For a weak acid, $\alpha$ is very small, so $1 - \alpha \approx 1$.
Therefore, $K_a = C\alpha^2 \Rightarrow \alpha^2 = \frac{K_a}{C} \Rightarrow \alpha = \sqrt{\frac{K_a}{C}}$
Since $C = 1/V$, $\alpha = \sqrt{K_a \cdot V}$. This proves the law.
21. What is the formula for $[H^+]$ and $[OH^-]$ of weak electrolytes using Ostwald's law?
From the derivation of Ostwald's law:
For a weak acid, $[H^+] = C\alpha$. Substituting $\alpha = \sqrt{\frac{K_a}{C}}$:
$$ [H^+] = C\sqrt{\frac{K_a}{C}} = \sqrt{K_a \cdot C} $$
Similarly, for a weak base, $[OH^-] = C\alpha = \sqrt{K_b \cdot C}$.
22. What are the four types of salts based on the strength of their parent acids and bases?
- Salt of Strong Acid and Strong Base (e.g., $NaCl, KNO_3$).
- Salt of Strong Acid and Weak Base (e.g., $NH_4Cl, CuSO_4$).
- Salt of Weak Acid and Strong Base (e.g., $CH_3COONa, Na_2CO_3$).
- Salt of Weak Acid and Weak Base (e.g., $CH_3COONH_4, (NH_4)_2CO_3$).
23. Define salt hydrolysis.
Salt Hydrolysis: It is defined as a reaction in which the cation or the anion, or both the ions of a salt react with water to produce acidity, basicity, or sometimes neutrality in the aqueous solution. It is the reverse of neutralization.
24. Why does an aqueous solution of NaCl have a pH of 7?
$NaCl$ is a salt of a strong acid ($HCl$) and a strong base ($NaOH$).
When dissolved in water, it dissociates entirely into $Na^+$ and $Cl^-$ ions. The $Na^+$ ion has no tendency to react with $OH^-$ (as it would form strong base $NaOH$, which dissociates completely). Similarly, $Cl^-$ does not react with $H^+$.
Since neither ion reacts with water (no hydrolysis occurs), the concentrations of $[H_3O^+]$ and $[OH^-]$ remain equal to those in pure water. Thus, the solution is neutral (pH = 7).
25. Explain why an aqueous solution of $CH_3COONa$ is basic in nature.
$CH_3COONa$ is a salt of a weak acid ($CH_3COOH$) and a strong base ($NaOH$).
It completely dissociates: $CH_3COONa \rightarrow CH_3COO^- + Na^+$
The $Na^+$ ion does not undergo hydrolysis. However, the acetate ion ($CH_3COO^-$) undergoes hydrolysis because it is the conjugate base of a weak acid:
$$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$$
This hydrolysis reaction produces excess $OH^-$ ions in the solution, making the solution basic (pH > 7).
26. Explain why an aqueous solution of $NH_4Cl$ is acidic in nature.
$NH_4Cl$ is a salt of a strong acid ($HCl$) and a weak base ($NH_4OH$).
It completely dissociates: $NH_4Cl \rightarrow NH_4^+ + Cl^-$
The $Cl^-$ ion does not hydrolyze. The $NH_4^+$ ion undergoes hydrolysis because it is the conjugate acid of a weak base:
$$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$$
This hydrolysis produces excess $H^+$ ($H_3O^+$) ions in the solution, making the solution acidic (pH < 7).
27. What determines the pH of a salt of a weak acid and a weak base?
For a salt of a weak acid and a weak base (like $CH_3COONH_4$), both cation and anion undergo hydrolysis. The nature of the solution depends on the relative strength of the acid ($K_a$) and base ($K_b$):
- If $K_a > K_b$, the solution is slightly acidic.
- If $K_a < K_b$, the solution is slightly basic.
- If $K_a = K_b$, the solution is neutral (e.g., $CH_3COONH_4$ since $K_a \approx K_b$).
28. Define a Buffer Solution.
Buffer Solution: A solution which resists drastic changes in its pH upon the addition of a small amount of strong acid, a strong base, or upon dilution, is called a buffer solution.
29. What are the two types of buffer solutions? Give an example of each.
- Acidic Buffer: A mixture containing a weak acid and its salt with a strong base.
Example: Acetic acid ($CH_3COOH$) + Sodium acetate ($CH_3COONa$). - Basic Buffer: A mixture containing a weak base and its salt with a strong acid.
Example: Ammonium hydroxide ($NH_4OH$) + Ammonium chloride ($NH_4Cl$).
30. Write the Henderson-Hasselbalch equation for an acidic buffer.
The pH of an acidic buffer is given by:
$$ \text{pH} = pK_a + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]} $$
Where $pK_a = -\log_{10} K_a$, $[Salt]$ is the concentration of the salt, and $[Acid]$ is the concentration of the weak acid.
31. Write the Henderson-Hasselbalch equation for a basic buffer.
The pOH of a basic buffer is calculated first:
$$ \text{pOH} = pK_b + \log_{10} \frac{[\text{Salt}]}{[\text{Base}]} $$
Where $pK_b = -\log_{10} K_b$, $[Salt]$ is the concentration of the salt, and $[Base]$ is the concentration of the weak base.
The pH is then found using: $\text{pH} = 14 - \text{pOH}$
32. Explain the buffer action of an acidic buffer ($CH_3COOH + CH_3COONa$).
The buffer contains a large amount of undissociated $CH_3COOH$ and acetate ions $CH_3COO^-$ (from completely dissociated salt).
- Addition of strong acid ($H^+$): The added $H^+$ is consumed by the conjugate base ($CH_3COO^-$) to form weakly ionizing acetic acid:
$CH_3COO^- + H^+ \rightarrow CH_3COOH$. Thus, pH does not drop. - Addition of strong base ($OH^-$): The added $OH^-$ is neutralized by the weak acid:
$CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$. Thus, pH does not rise.
33. Mention four properties of a good buffer solution.
- It has a definite and relatively constant pH.
- Its pH does not change upon standing for a long time.
- Its pH does not change significantly on dilution.
- Its pH does not change drastically upon the addition of a small amount of strong acid or base.
34. What is buffer capacity?
Buffer capacity ($\beta$): It is a quantitative measure of the buffer's resistance to pH change. It is defined as the ratio of the number of moles of acid or base added per liter of the buffer solution to the resulting change in pH.
$$ \beta = \frac{db}{dpH} $$
Where $db$ is the small amount of strong base/acid added in moles/liter, and $dpH$ is the change in pH.
35. State any two applications of buffer solutions.
- In Biology: Human blood is naturally buffered to a pH of about 7.36 to 7.42 by the bicarbonate ($H_2CO_3 / HCO_3^-$) buffer system. A change of just 0.2 pH units can be fatal.
- In Industry: The manufacture of paper, dyes, inks, paints, and drugs requires strict pH control maintained by industrial buffers.
36. What is a sparingly soluble salt? Give an example.
A salt that dissolves in water only to a very small extent is called a sparingly soluble salt. Its saturated solution is highly dilute.
Examples: Silver chloride ($AgCl$), Barium sulfate ($BaSO_4$), Lead(II) iodide ($PbI_2$).
37. Define Solubility ($S$). What are its units?
Solubility ($S$): It is defined as the maximum amount in moles of a solute that dissolves in a given volume of solvent to form a saturated solution at a specific temperature. It represents the molar concentration of the saturated solution.
Units: $\text{mol L}^{-1}$ (Molar) or $\text{g L}^{-1}$.
38. Define Solubility Product ($K_{sp}$).
Solubility Product ($K_{sp}$): In a saturated solution of a sparingly soluble salt, the product of the molar concentrations of its constituent ions, with each concentration term raised to the power equal to its stoichiometric coefficient in the balanced equilibrium equation, is constant at a given temperature.
39. Write the general expression for $K_{sp}$ for a salt $B_x A_y$.
For a sparingly soluble salt $B_x A_y$ dissociating as:
$$ B_x A_{y(s)} \rightleftharpoons xB^{y+}_{(aq)} + yA^{x-}_{(aq)} $$
The solubility product is written as:
$$ K_{sp} = [B^{y+}]^x [A^{x-}]^y $$
40. Derive the relationship between Solubility ($S$) and Solubility Product ($K_{sp}$) for AgCl (AB type salt).
The dissociation of $AgCl$ is:
$$ AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)} $$
If $S$ is the solubility in mol/L, then $[Ag^+] = S$ and $[Cl^-] = S$.
$$ K_{sp} = [Ag^+][Cl^-] $$
$$ K_{sp} = (S)(S) = S^2 $$
Therefore, $S = \sqrt{K_{sp}}$.
41. Derive the relationship between Solubility ($S$) and Solubility Product ($K_{sp}$) for $PbCl_2$ (AB2 type salt).
The dissociation of $PbCl_2$ is:
$$ PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)} $$
If $S$ is the solubility, then $[Pb^{2+}] = S$ and $[Cl^-] = 2S$.
$$ K_{sp} = [Pb^{2+}][Cl^-]^2 $$
$$ K_{sp} = (S)(2S)^2 = (S)(4S^2) = 4S^3 $$
Therefore, $S = \sqrt[3]{\frac{K_{sp}}{4}}$.
42. Write the $K_{sp}$ expression and its relation with $S$ for $Ca_3(PO_4)_2$.
Dissociation: $Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO_4^{3-}_{(aq)}$
Concentrations: $[Ca^{2+}] = 3S$, $[PO_4^{3-}] = 2S$
$$ K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 $$
$$ K_{sp} = (3S)^3(2S)^2 = (27S^3)(4S^2) = 108S^5 $$
43. What is Ionic Product (IP)? How does it differ from $K_{sp}$?
Ionic Product (IP): It is the product of the concentration of the ions of a sparingly soluble salt at any given concentration (not necessarily saturated), with each concentration raised to its stoichiometric power.
Difference: $K_{sp}$ is applicable only for a saturated solution (at equilibrium). IP is applicable at any state of the solution.
44. State the conditions for the precipitation of a salt.
Precipitation of a salt occurs depending on the relation between Ionic Product (IP) and Solubility Product ($K_{sp}$):
- If IP < $K_{sp}$: The solution is unsaturated. No precipitation occurs. More salt can dissolve.
- If IP = $K_{sp}$: The solution is exactly saturated. Equilibrium exists, no precipitation.
- If IP > $K_{sp}$: The solution is supersaturated. Precipitation will occur until IP becomes equal to $K_{sp}$.
45. Define Common Ion Effect.
Common Ion Effect: It is defined as the suppression of the degree of dissociation of a weak electrolyte upon the addition of a strong electrolyte containing a common ion.
46. Explain the common ion effect with the example of Acetic acid and Sodium acetate.
Acetic acid is a weak acid: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
Sodium acetate is a strong electrolyte: $CH_3COONa \rightarrow CH_3COO^- + Na^+$
When $CH_3COONa$ is added to $CH_3COOH$, the concentration of the common ion, $CH_3COO^-$, increases drastically. According to Le Chatelier's principle, the equilibrium of the weak acid shifts to the left (backward direction) to consume the excess $CH_3COO^-$. This suppresses the dissociation of $CH_3COOH$.
47. How does the common ion effect influence the solubility of a sparingly soluble salt?
The addition of a strong electrolyte containing a common ion decreases the solubility of a sparingly soluble salt.
Example: To a saturated solution of $AgCl_{(s)} \rightleftharpoons Ag^+ + Cl^-$, if $NaCl$ (strong electrolyte) is added, the concentration of the common ion $Cl^-$ increases. The equilibrium shifts backward, precipitating more $AgCl_{(s)}$, thus reducing its solubility.
48. State one application of the common ion effect in qualitative analysis.
In Group III analysis of cations ($Fe^{3+}, Al^{3+}, Cr^{3+}$), they are precipitated as hydroxides using $NH_4OH$. To prevent the precipitation of Group IV cations ($Zn^{2+}, Ni^{2+}$, etc.), $NH_4Cl$ is added first.
The common ion ($NH_4^+$) from the strong electrolyte $NH_4Cl$ suppresses the ionization of the weak base $NH_4OH$. This lowers the $[OH^-]$ concentration just enough to exceed the $K_{sp}$ of Group III hydroxides, but not enough to exceed the $K_{sp}$ of Group IV hydroxides.
49. Calculate the pH of a buffer solution containing 0.05 M $NH_4OH$ and 0.1 M $NH_4Cl$. ($K_b$ for $NH_4OH$ = $1.8 \times 10^{-5}$)
This is a basic buffer.
$pK_b = -\log_{10}(1.8 \times 10^{-5}) = 5 - \log_{10}(1.8) = 5 - 0.2553 = 4.7447$
Using Henderson's equation for basic buffer: $\text{pOH} = pK_b + \log_{10} \frac{[\text{Salt}]}{[\text{Base}]}$
$\text{pOH} = 4.7447 + \log_{10} \frac{0.1}{0.05} = 4.7447 + \log_{10} 2$
$\text{pOH} = 4.7447 + 0.3010 = 5.0457$
$\text{pH} = 14 - \text{pOH} = 14 - 5.0457 = 8.9543$
50. The solubility of $AgCl$ is $1.3 \times 10^{-5}$ M. Calculate its solubility product.
$AgCl$ is an AB type salt.
$AgCl \rightleftharpoons Ag^+ + Cl^-$
Formula: $K_{sp} = S^2$
Given $S = 1.3 \times 10^{-5}$ M
$K_{sp} = (1.3 \times 10^{-5})^2 = 1.69 \times 10^{-10}$
The solubility product $K_{sp}$ is $1.69 \times 10^{-10}$.
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