Ionic Equilibrium Class 11 Notes for IIT JEE & NEET

Ionic equilibrium is the second half of the NCERT chapter Equilibrium, which is Chapter 6 of the current Class 11 Chemistry book (Chapter 7 in older books). The first half, on Kc, Kp and Le Chatelier’s principle, is in eSaral’s Chemical Equilibrium Class 11 notes. In the JEE Main syllabus, both halves are one unit, Equilibrium. These Ionic Equilibrium Class 11 Notes follow the NCERT chapter and add the IIT JEE and NEET problem types.

Ionic Equilibrium Class 11 Notes: Chapter Overview and Syllabus

Check the latest CBSE, NTA (JEE Main), JEE Advanced and NMC (NEET) syllabi for the exact lists. The number of questions changes every year, so use previous years’ papers to see the trend.

Tip from the eSaral Chemistry team: In every numerical, first decide which species controls the pH: a strong acid or base, a weak acid or base, a buffer or a salt. Each case has one formula, and most mistakes come from using the wrong one.

Ionic Equilibrium Class 11 Notes for IIT JEE and NEET

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

Electrolytes and the Degree of Ionisation in Ionic Equilibrium Class 11 Notes

A strong electrolyte ionises completely in water (HCl, NaOH, NaCl), and a weak electrolyte ionises only partly (CH$_3$COOH, NH$_3$). The degree of ionisation is $\alpha = \dfrac{\text{amount ionised}}{\text{amount dissolved}}$.

For a weak acid HA of concentration $C$:

$$K_a = \frac{C\alpha^{2}}{1 - \alpha} \approx C\alpha^{2} \quad (\alpha \ll 1), \qquad \alpha = \sqrt{\frac{K_a}{C}}$$

This is Ostwald’s dilution law: $\alpha$ increases as the solution is diluted, and $[\mathrm{H^+}] = C\alpha = \sqrt{K_aC}$. The ionisation constant $K_a$ changes only with the temperature.

Acid–Base Theories in Ionic Equilibrium Class 11 Notes

• Conjugate pairs: HA and $\mathrm{A^-}$ differ by one proton. A strong acid has a very weak conjugate base, and a weak acid has a stronger conjugate base. For a conjugate pair, $K_a \times K_b = K_w$.
• Amphiprotic species (water, $\mathrm{HCO_3^-}$, $\mathrm{HS^-}$) can act as an acid and as a base.
• The acid strength depends on $\mathrm{p}K_a = -\log K_a$: the smaller the pK$_a$, the stronger the acid.

Ionisation of Water and the pH Scale in Ionic Equilibrium Class 11 Notes

$$\mathrm{H_2O \rightleftharpoons H^+ + OH^-}, \qquad K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 1.0\times10^{-14}\ \text{at } 298\ \text{K}$$ $$\mathrm{pH} = -\log[\mathrm{H^+}], \qquad \mathrm{pOH} = -\log[\mathrm{OH^-}], \qquad \mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w = 14\ \text{at } 298\ \text{K}$$

• $K_w$ rises with temperature, so at a higher temperature, pure water has $\mathrm{pH} < 7$, but it is still neutral because $[\mathrm{H^+}] = [\mathrm{OH^-}]$.
• A strong acid of concentration $C$ has $\mathrm{pH} = -\log C$, and a strong base has $\mathrm{pOH} = -\log C$.
• Very dilute solutions (around $10^{-7}$ M or less): the $\mathrm{H^+}$ of water cannot be ignored. For $10^{-8}$ M HCl, solve $[\mathrm{H^+}]^{2} - 10^{-8}[\mathrm{H^+}] - 10^{-14} = 0$, which gives $\mathrm{pH} \approx 6.98$, and not 8.

Weak Acids, Weak Bases and Polyprotic Acids in Ionic Equilibrium Class 11 Notes

Polyprotic acids (H$_2$S, H$_2$CO$_3$, H$_3$PO$_4$) ionise in steps, and $K_{a1} \gg K_{a2} \gg K_{a3}$. The first step gives almost all the $\mathrm{H^+}$, so the pH is found from $K_{a1}$. For a solution of H$_2$A with no other source of $\mathrm{A^{2-}}$, $[\mathrm{A^{2-}}] \approx K_{a2}$.

Common Ion Effect in Ionic Equilibrium Class 11 Notes

The ionisation of a weak electrolyte falls when a strong electrolyte with a common ion is added (Le Chatelier’s principle). For a weak acid HA with its salt NaA added:

$$[\mathrm{H^+}] = K_a\,\frac{[\mathrm{HA}]}{[\mathrm{A^-}]}$$

For example, sodium acetate suppresses the ionisation of acetic acid, and sodium chloride lowers the solubility of silver chloride.

Buffer Solutions and the Henderson Equation in Ionic Equilibrium Class 11 Notes

A buffer resists a change in pH on the addition of a small amount of an acid or a base. An acidic buffer is a weak acid with its salt (CH$_3$COOH and CH$_3$COONa), and a basic buffer is a weak base with its salt (NH$_4$OH and NH$_4$Cl).

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}, \qquad \mathrm{pOH} = \mathrm{p}K_b + \log\frac{[\text{salt}]}{[\text{base}]}$$

• The buffer capacity is maximum when $[\text{salt}] = [\text{acid}]$, so that $\mathrm{pH} = \mathrm{p}K_a$.
• An effective buffer works in the range $\mathrm{pH} = \mathrm{p}K_a \pm 1$, which is for a ratio $[\text{salt}]/[\text{acid}]$ between $0.1$ and $10$.
• When a strong acid or base is added, find the new moles of the acid and the salt first (the strong acid converts the salt into the weak acid), and then use the equation.
• Blood is buffered by $\mathrm{H_2CO_3}/\mathrm{HCO_3^-}$.

Hydrolysis of Salts in Ionic Equilibrium Class 11 Notes

The degree of hydrolysis is $h = \sqrt{\dfrac{K_h}{C}}$ for the first two cases. For a salt of a weak acid and a weak base, $h = \sqrt{K_h}$, so it does not depend on the concentration, and the pH is independent of the concentration.

Indicators and Titration in Ionic Equilibrium Class 11 Notes

An acid–base indicator is a weak acid or base whose acid and base forms have different colours. It changes colour in a range of about $\mathrm{p}K_{In} \pm 1$.

The pH at the equivalence point is $7$ for a strong acid with a strong base, above $7$ for a weak acid with a strong base (use phenolphthalein), and below $7$ for a strong acid with a weak base (use methyl orange).

Solubility Product and Precipitation in Ionic Equilibrium Class 11 Notes

For a sparingly soluble salt $\mathrm{A}_x\mathrm{B}_y$ in a saturated solution:

$$\mathrm{A}_x\mathrm{B}_y(s) \rightleftharpoons x\,\mathrm{A}^{y+} + y\,\mathrm{B}^{x-}, \qquad K_{sp} = [\mathrm{A}^{y+}]^{x}[\mathrm{B}^{x-}]^{y} = x^{x}y^{y}s^{x+y}$$

• Precipitation: compare the ionic product $Q$ with $K_{sp}$. If $Q > K_{sp}$, the salt precipitates. If $Q = K_{sp}$, the solution is saturated, and if $Q < K_{sp}$, the salt dissolves.
• Common ion effect on solubility: for an AB salt in a solution with the common ion at the concentration $c$, $s \approx \dfrac{K_{sp}}{c}$.
• The solubility of a salt of a weak acid (for example a sulphide or a carbonate) is higher in an acidic solution, and a hydroxide is more soluble in an acid.

IIT JEE and NEET Problem Tips in Ionic Equilibrium Class 11 Notes

• Decide the type first: strong acid or base, weak acid or base, buffer, hydrolysis or $K_{sp}$.
• Mixing problems: find the moles of $\mathrm{H^+}$ and $\mathrm{OH^-}$, subtract, and divide by the total volume.
• Check the approximation $\alpha \ll 1$ (it is valid when $\alpha < 5\%$). If it is not valid, solve the quadratic.
• Buffers: use moles, and not concentrations, after the addition of an acid or a base, because the volume is common to both the salt and the acid.
• Hydrolysis: write the reaction, find $K_h$, and then use the formula for the pH.
• $K_{sp}$: write the formula of the salt first, and use $x^xy^ys^{x+y}$.

Key Formulas in Ionic Equilibrium Class 11 Notes

Solved Examples from Ionic Equilibrium Class 11 Notes

Example 1 (degree of ionisation): Find $\alpha$, $[\mathrm{H^+}]$ and the pH of 0.1 M ethanoic acid ($K_a = 1.8\times10^{-5}$).

$\alpha = \sqrt{\dfrac{1.8\times10^{-5}}{0.1}} \approx 1.34\times10^{-2}$ (1.34%), $[\mathrm{H^+}] = C\alpha \approx 1.34\times10^{-3}$ M, and $\mathrm{pH} \approx 2.87$.

Example 2 (strong acid and base): Find the pH of 0.01 M HCl and of 0.01 M NaOH.

HCl: $\mathrm{pH} = -\log 0.01 = 2$. NaOH: $\mathrm{pOH} = 2$, so $\mathrm{pH} = 14 - 2 = 12$.

Example 3 (a very dilute acid): Find the pH of $10^{-8}$ M HCl.

$[\mathrm{H^+}] = \dfrac{10^{-8} + \sqrt{10^{-16} + 4\times10^{-14}}}{2} \approx 1.05\times10^{-7}$ M, so $\mathrm{pH} \approx 6.98$.

Example 4 (mixing): Find the pH when 100 mL of 0.1 M HCl is mixed with 50 mL of 0.1 M NaOH.

$\mathrm{H^+}$: 10 mmol. $\mathrm{OH^-}$: 5 mmol. The excess is 5 mmol in 150 mL, so $[\mathrm{H^+}] = 0.0333$ M and $\mathrm{pH} \approx 1.48$.

Example 5 (a buffer): Find the pH of a buffer with 0.1 M ethanoic acid and 0.2 M sodium ethanoate ($K_a = 1.8\times10^{-5}$, $\mathrm{p}K_a = 4.745$).

$\mathrm{pH} = 4.745 + \log\dfrac{0.2}{0.1} = 4.745 + 0.301 \approx 5.05$.

Example 6 (adding acid to a buffer): One litre of a buffer has 0.1 mol of the acid and 0.1 mol of the salt ($\mathrm{p}K_a = 4.74$). Find the pH after the addition of 0.01 mol of HCl.

The acid becomes $0.11$ mol, and the salt becomes $0.09$ mol. $\mathrm{pH} = 4.74 + \log\dfrac{0.09}{0.11} \approx 4.74 - 0.087 = 4.65$. The pH changes by only 0.09.

Example 7 (hydrolysis, weak acid and strong base): Find the pH of 0.1 M sodium ethanoate.

$\mathrm{pH} = \tfrac{1}{2}\left(14 + 4.745 + \log 0.1\right) = \tfrac{1}{2}(14 + 4.745 - 1) \approx 8.87$.

Example 8 (hydrolysis, strong acid and weak base): Find the pH of 0.1 M ammonium chloride ($K_b = 1.8\times10^{-5}$).

$\mathrm{pH} = \tfrac{1}{2}\left(14 - 4.745 - \log 0.1\right) = \tfrac{1}{2}(14 - 4.745 + 1) \approx 5.13$.

Example 9 ($K_{sp}$ from the solubility): The solubility of AgCl is $1.3\times10^{-5}$ mol/L. Find $K_{sp}$.

$K_{sp} = s^{2} = (1.3\times10^{-5})^{2} \approx 1.7\times10^{-10}$.

Example 10 (solubility from $K_{sp}$): $K_{sp}$ of PbCl$_2$ is $1.6\times10^{-5}$. Find its solubility.

$4s^{3} = 1.6\times10^{-5}$, so $s^{3} = 4\times10^{-6}$ and $s \approx 1.59\times10^{-2}$ mol/L.

Example 11 (common ion effect): Find the solubility of AgCl ($K_{sp} = 1.8\times10^{-10}$) in 0.1 M NaCl.

$s \approx \dfrac{K_{sp}}{[\mathrm{Cl^-}]} = \dfrac{1.8\times10^{-10}}{0.1} = 1.8\times10^{-9}$ mol/L, which is about 7000 times lower than in pure water.

Example 12 (precipitation): Equal volumes of $2\times10^{-4}$ M AgNO$_3$ and $2\times10^{-4}$ M NaCl are mixed. Does AgCl precipitate ($K_{sp} = 1.8\times10^{-10}$)?

After mixing, each ion is $1\times10^{-4}$ M, so $Q = 10^{-8} > K_{sp}$, and AgCl precipitates.

Example 13 ($K_a$ from the pH): A 0.01 M solution of a weak acid has a pH of 3.5. Find $K_a$.

$[\mathrm{H^+}] = 10^{-3.5} \approx 3.16\times10^{-4}$ M, so $K_a = \dfrac{[\mathrm{H^+}]^{2}}{C} = \dfrac{(3.16\times10^{-4})^{2}}{0.01} \approx 1.0\times10^{-5}$.

Example 14 (a basic buffer): Find the pH of a solution that has 0.2 M NH$_3$ and 0.1 M NH$_4$Cl ($\mathrm{p}K_b = 4.745$).

$\mathrm{pOH} = 4.745 + \log\dfrac{0.1}{0.2} = 4.745 - 0.301 = 4.444$, so $\mathrm{pH} = 14 - 4.444 \approx 9.56$.

Common Mistakes in Ionic Equilibrium Class 11 Notes Problems

How to Study Ionic Equilibrium Class 11 Notes

1. Read the NCERT chapter on equilibrium first, and download the textbook from eSaral’s NCERT books for Class 11 page.
2. Learn the first half of the chapter with eSaral’s Chemical Equilibrium Class 11 notes, because Le Chatelier’s principle is the basis of the common ion effect. The link $\Delta G^\circ = -RT\ln K$ is in eSaral’s Thermodynamics short notes.
3. Learn the formula for each case (strong, weak, buffer, hydrolysis and $K_{sp}$), and practise one problem of each type.
4. Revise with eSaral’s Class 11 Chemistry notes and JEE Main Chemistry revision notes.
5. Test yourself with eSaral’s JEE Main chapter-wise PYQs and NEET chapter-wise PYQs.