Chemistry Physical Chemistry → Class 11 Chemical & Ionic Equilibrium

Chemical & Ionic Equilibrium: Le Chatelier’s Principle, Buffer Solutions & Solubility Products

By Dr. Evelyn Reed, Molecular Chemistry Specialist • 13 min read • Published: September 2026

Master equilibrium constants (Kp vs Kc), Le Chatelier stress perturbations, buffer action mechanics, pH of salts, and sparingly soluble salt precipitation.

Chemical equilibrium represents dynamic balance: reactant molecules continue converting into products, and products re-form reactants at precisely equal forward and reverse rates.

Le Chatelier’s Principle dictates that when an equilibrium system undergoes stress (changes in concentration, pressure, volume, or temperature), the position of equilibrium shifts in the direction that counteracts the applied stress.

Ionic equilibrium addresses weak electrolytes, self-ionization of water ($K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at $25\,^\circ\text{C}$), and buffer solutions capable of resisting $\text{pH}$ changes upon addition of small amounts of strong acid or base.

The Henderson-Hasselbalch equation quantifies buffer $\text{pH}$: $\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}\right)$. For sparingly soluble salts, precipitation occurs whenever the ionic reaction quotient $Q_{\text{sp}}$ exceeds the solubility product constant $K_{\text{sp}}$.

Key Conceptual Takeaways

  • Equilibrium constant $K$ depends exclusively on temperature; catalysts, concentrations, and pressure do not alter $K$.
  • Adding an inert gas at constant volume causes NO shift in equilibrium (partial pressures remain unchanged).
  • Precipitation occurs if and only if ionic product $Q_{\text{sp}} > K_{\text{sp}}$.

1. Le Chatelier’s Principle and Thermodynamic Equilibrium

For an exothermic reaction ($\Delta H < 0$), increasing temperature favors the endothermic reverse direction, lowering equilibrium constant $K$. For endothermic reactions ($\Delta H > 0$), raising temperature increases $K$.

Increasing total pressure (by decreasing container volume) shifts equilibrium toward the side with fewer gas moles ($\Delta n_g < 0$). If $\Delta n_g = 0$ (e.g., $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$), pressure changes have zero effect on equilibrium position.

Relationship Between Kp and Kc for Gaseous Equilibria
K_p = K_c (R T)^{\Delta n_g} \quad \text{where} \quad \Delta n_g = \sum n_{\text{products(g)}} - \sum n_{\text{reactants(g)}}
K_p, K_c: Equilibrium constants expressed in partial pressures (atm/bar) and molar concentrations (mol/L)
\Delta n_g: Net change in moles of gaseous species in balanced stoichiometric equation
R, T: Universal gas constant ($0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$) and absolute temperature (Kelvin)
Worked Problem:

For the synthesis of ammonia: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, calculate $\Delta n_g$ and predict the effect of (a) increasing pressure, and (b) adding argon gas at constant volume.

Solution:

$\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$. (a) Increasing pressure shifts equilibrium toward fewer gas moles ($\Delta n_g < 0$), thus shifting the reaction forward toward $\text{NH}_3$ production. (b) Adding argon at constant volume increases total pressure but does NOT alter the partial pressures or molar concentrations of $\text{N}_2$, $\text{H}_2$, or $\text{NH}_3$; hence there is zero shift in equilibrium.

Rule: Inert gas at constant volume = NO SHIFT. Inert gas at constant pressure shifts equilibrium toward greater number of gas moles.

2. Buffer Solutions and the Henderson-Hasselbalch Equation

An acidic buffer consists of a weak acid and its conjugate salt (e.g., $\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$). A basic buffer consists of a weak base and its conjugate salt (e.g., $\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$).

Buffers resist $\text{pH}$ fluctuations through conjugate acid-base neutralization: added $\text{H}^+$ ions are soaked up by conjugate base anions, while added $\text{OH}^-$ ions are neutralized by weak acid molecules.

Henderson-Hasselbalch Equations for Acidic and Basic Buffers
\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Conjugate Base}]}{[\text{Weak Acid}]}\right) \quad \text{and} \quad \text{pOH} = \text{p}K_b + \log_{10}\left(\frac{[\text{Conjugate Acid}]}{[\text{Weak Base}]}\right)
\text{p}K_a: $-\log_{10}(K_a)$, negative logarithm of acid dissociation constant
[\text{Conjugate Base}]: Molar concentration of salt anion (e.g., $[\text{CH}_3\text{COO}^-]$)
[\text{Weak Acid}]: Molar concentration of un-dissociated weak acid (e.g., $[\text{CH}_3\text{COOH}]$)
Worked Problem:

Calculate the pH of a buffer solution containing $0.1\text{ M } \text{CH}_3\text{COOH}$ and $0.2\text{ M } \text{CH}_3\text{COONa}$. ($\text{p}K_a$ of acetic acid = $4.74$; $\log_{10} 2 = 0.301$).

Solution:

Using Henderson-Hasselbalch: $\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = 4.74 + \log_{10}\left(\frac{0.2}{0.1}\right) = 4.74 + \log_{10}(2) = 4.74 + 0.301 = 5.04$.

Rule: A buffer possesses maximum buffering capacity when $[\text{Salt}] = [\text{Acid}]$, at which point $\text{pH} = \text{p}K_a$.

Common Misconceptions & Examination Traps

Trap: Calculating Ksp for AxBy salts without multiplying by stoichiometric powers.
Why it is wrong: For salt $A_x B_y$ dissolving to give $x A^{y+} + y B^{x-}$, solubility $s$ yields $[A] = xs$ and $[B] = ys$. $K_{\text{sp}} = (xs)^x (ys)^y = (x^x y^y) s^{x+y}$.
First-Principles Approach: For $\text{Ca}_3(\text{PO}_4)_2$: $K_{\text{sp}} = (3s)^3 (2s)^2 = (27 s^3)(4 s^2) = 108 s^5$.

Interactive Practice Checkpoints

Question 1 (Medium)

Which of the following mixtures forms an effective buffer solution?

A. 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaCl
B. 100 mL of 0.1 M CH3COOH + 50 mL of 0.1 M NaOH ✔ (Correct)
C. 100 mL of 0.1 M HNO3 + 100 mL of 0.1 M NH4OH
D. 50 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH
Explanation: $0.1\text{ M } \text{CH}_3\text{COOH}$ ($10\text{ mmol}$) reacts with $0.1\text{ M } \text{NaOH}$ ($5\text{ mmol}$) to produce $5\text{ mmol}$ of $\text{CH}_3\text{COONa}$, leaving $5\text{ mmol}$ of unreacted $\text{CH}_3\text{COOH}$. This equimolar mixture of weak acid and its salt forms an ideal buffer with $\text{pH} = \text{p}K_a$.

Previous-Year Exam Questions (PYQ Vault)

NEET 2022 +4 Marks

The solubility product of AgCl is 1.6 x 10^-10 at 298 K. The solubility of AgCl in 0.1 M NaCl solution is:

Verified Answer: Option B (1.6 x 10^-9 M)
$\text{NaCl}$ is a strong electrolyte providing $[\text{Cl}^-] = 0.1\text{ M}$. Due to common ion effect, solubility $s$ of $\text{AgCl}$ is small, so total $[\text{Cl}^-] = s + 0.1 \approx 0.1\text{ M}$. $K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] \implies 1.6 \times 10^{-10} = s \times (0.1) \implies s = 1.6 \times 10^{-9}\text{ M}$.

2-Minute High-Yield Exam Revision

Governing Equation:
\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \quad \Big| \quad K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \quad \Big| \quad K_p = K_c (RT)^{\Delta n_g}
Key Recall Checkpoints:
  • Le Chatelier: system shifts to oppose applied stress.
  • Inert gas at constant V causes no shift; at constant P shifts toward more gas moles.
  • Common ion effect suppresses degree of dissociation of weak electrolytes.
Academic Peer Review Certification
Reviewed by Dr. Evelyn Reed, Ph.D.
Molecular Chemistry Specialist • Royal Society of Chemistry Academic Network
Verification Date: September 2026

Frequently Asked Questions

Why does adding common ion suppress solubility of a salt?

By Le Chatelier’s Principle, introducing an ion already present in the equilibrium expression increases the reaction quotient Q_sp above K_sp, forcing the dissolution equilibrium backward into solid precipitate form.

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