MHT-CET Chemistry · Teaching notes

Ionic Equilibria — MHT-CET Chemistry

The most heavily tested MHT-CET Chemistry chapter (127 PYQs) — and almost entirely a calculation chapter built on one idea: weak electrolytes only partly ionise, and a handful of equilibrium constants (Ka, Kb, Kw, Ksp) let you predict everything from that. It teaches in six movements, foundations first: (1) theories of acids and bases — Arrhenius, Bronsted-Lowry and Lewis; (2) ionic equilibrium — Ka, Kb, degree of dissociation and Ostwald's dilution law; (3) pH, pOH and the ionic product of water Kw; (4) salt hydrolysis — the four salt types and their solution pH; (5) buffer solutions and the Henderson-Hasselbalch equation; (6) the solubility product Ksp — solubility, the common-ion effect and precipitation. Formula concepts carry the computational core; the salt-type and Ksp-stoichiometry tables carry the recall. Every PYQ tagged.

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Subtopic notes

PYQ weightage by concept

24 concepts · 127 PYQs — where the marks actually sit, so you know what to drill first

Theories of Acids and Bases13 PYQs · 10%

Concept	PYQs	Share
The three theories: Arrhenius, Bronsted-Lowry and Lewis	7	6%
Conjugate acid-base pairs	4	3%
Amphoteric species	2	2%

Ionic Equilibrium: Ka, Kb and Degree of Dissociation24 PYQs · 19%

Concept	PYQs	Share
Ostwald's dilution law: Ka and Kb from alpha and concentration	12	9%
Degree of dissociation and percent dissociation	6	5%
Ion concentration of a weak acid or base	4	3%
Ka x Kb = Kw, relative strength and the effect of dilution	2	2%

pH, pOH and the Ionic Product of Water24 PYQs · 19%

Concept	PYQs	Share
pH of strong acids and strong bases	9	7%
pH, pOH and the relation pH + pOH = 14	7	6%
pH of weak acids and weak bases	6	5%
Ionic product of water, Kw	2	2%

Salt Hydrolysis17 PYQs · 13%

Concept	PYQs	Share
The four salt types and their solution pH	8	6%
Which ion hydrolyses — classifying a given salt	8	6%
Hydrolysis constant, degree of hydrolysis and pH	1	1%

Buffer Solutions and the Henderson-Hasselbalch Equation19 PYQs · 15%

Concept	PYQs	Share
Henderson-Hasselbalch equation — pH of an acidic buffer	12	9%
What a buffer is and how to recognise one	4	3%
Basic buffers — the pOH form and converting to pH	2	2%
Equal salt and acid — pH equals pKa	1	1%

Solubility Product (Ksp)30 PYQs · 24%

Concept	PYQs	Share
Solubility of a 1:1 (AB) salt: Ksp = S squared	15	12%
Solubility of AB2, A2B and A2B3 salts	8	6%
Ksp in terms of solubility, by salt type	3	2%
Ksp from pH and from mass solubility	3	2%
Solubility product expression	1	1%
Common ion effect on solubilityfoundation	—	—

Formula & revision sheet

18 formulas · 6 reference tables · 47 gotchas across all subtopics — the exam-eve cheat-sheet

Theories of Acids and Bases

Formulas (1)

Conjugate acid-base pairs · Conjugate base from an acid $\text{Acid} \;\rightleftharpoons\; \text{Conjugate base} + \text{H}^+$

Reference tables (2)

The three theories: Arrhenius, Bronsted-Lowry and Lewis3 rows

Theory	Acid is	Base is	Example acid / base
Arrhenius	Gives $\text{H}^+$ in water	Gives $\text{OH}^-$ in water	$\text{HCl}$ / $\text{NaOH}$
Bronsted-Lowry	Proton $(\text{H}^+)$ donor	Proton $(\text{H}^+)$ acceptor	$\text{HCl}$ / $\text{NH}_3$ A Bronsted base ACCEPTS a proton — this is why $\text{NH}_3$ 'acts as a base when reacted with water' (it takes an $\text{H}^+$ to become $\text{NH}_4^+$ ).
Lewis	Electron-pair acceptor	Electron-pair donor	$\text{BF}_3$ / $\text{NH}_3$ $\text{BCl}_3$ is a Lewis acid but NOT a Bronsted acid — it accepts an electron pair yet has no proton to donate.

Each later theory contains the earlier one; a Lewis acid is the most general kind.

Amphoteric species5 rows

Species	Amphoteric?	Why
$\text{H}_2\text{O}$	Yes	Gives $\text{OH}^-$ (acid) and takes $\text{H}^+$ to form $\text{H}_3\text{O}^+$ (base) Water is the bank's default answer for 'which species is amphoteric'.
$\text{HCO}_3^-$	Yes	Loses $\text{H}^+$ to $\text{CO}_3^{2-}$ or gains $\text{H}^+$ to $\text{H}_2\text{CO}_3$
$\text{HCl}$	No	Only donates a proton (acid only)
$\text{NaOH}$	No	Only gives $\text{OH}^-$ (base only)
$\text{CH}_3\text{COOH}$	No	Acts only as an acid (donates a proton)

Amphoteric = can be either acid or base; water and

\text{HCO}_3^-

are the standard examples.

Watch out for (5)

Match the activity to the right theory
→ The three theories: Arrhenius, Bronsted-Lowry and Lewis
Lewis acid vs Bronsted acid
→ The three theories: Arrhenius, Bronsted-Lowry and Lewis
Conjugate base of a strong acid is weak
→ Conjugate acid-base pairs
Pick species differing by ONE proton — not a random pair
→ Conjugate acid-base pairs
Water is amphoteric — acetic acid is not
→ Amphoteric species

Ionic Equilibrium: Ka, Kb and Degree of Dissociation

Formulas (4)

Degree of dissociation and percent dissociation · Percent dissociation and alpha from Ka $\text{\% dissociation} = \alpha \times 100 \qquad \alpha = \sqrt{\dfrac{K_a}{c}}$
Ostwald's dilution law: Ka and Kb from alpha and concentration · Ostwald's dilution law $K_a = \dfrac{c\alpha^2}{1-\alpha} \approx c\alpha^2 \qquad (K_b \approx c\alpha^2 \text{ for a base})$
Ion concentration of a weak acid or base · Hydrogen / hydroxide ion concentration $[\text{H}^+] = c\alpha = \sqrt{K_a\,c} \qquad [\text{OH}^-] = c\alpha = \sqrt{K_b\,c}$
Ka x Kb = Kw, relative strength and the effect of dilution · Conjugate-pair relation and relative strength $K_a \times K_b = K_w \qquad \dfrac{\alpha_1}{\alpha_2} = \sqrt{\dfrac{K_{a1}}{K_{a2}}}$

Watch out for (8)

Percent is 100 times the fraction
→ Degree of dissociation and percent dissociation
alpha from Ka needs the DIVISION form
→ Degree of dissociation and percent dissociation
Square the alpha, not just alpha
→ Ostwald's dilution law: Ka and Kb from alpha and concentration
Use the small-alpha approximation only when it is small
→ Ostwald's dilution law: Ka and Kb from alpha and concentration
c times alpha, not c times alpha squared
→ Ion concentration of a weak acid or base
Multiply under the root for [H+]
→ Ion concentration of a weak acid or base
Ka times Kb equals Kw — a product, not a sum
→ Ka x Kb = Kw, relative strength and the effect of dilution
Dilution raises alpha but leaves Ka fixed
→ Ka x Kb = Kw, relative strength and the effect of dilution

pH, pOH and the Ionic Product of Water

Formulas (4)

Ionic product of water, Kw · Ionic product of water $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \quad (25\,^{\circ}\text{C})$
pH, pOH and the relation pH + pOH = 14 · pH, pOH and their sum $\text{pH} = -\log[\text{H}^+], \quad \text{pOH} = -\log[\text{OH}^-], \quad \text{pH} + \text{pOH} = 14$
pH of strong acids and strong bases · Strong acid / strong base $[\text{H}^+] = Z\,c \;\Rightarrow\; \text{pH} = -\log(Z\,c); \qquad \text{pH} = 14 - \text{pOH}$
pH of weak acids and weak bases · Weak acid / base $[\text{H}^+] = \alpha c = \sqrt{K_a\,c} \;\Rightarrow\; \text{pH} = \tfrac12\!\left(pK_a - \log c\right)$

Watch out for (9)

Kw = 10 to the minus 14 only at 25 degrees C
→ Ionic product of water, Kw
Divide into Kw, do not subtract
→ Ionic product of water, Kw
pH + pOH = 14 only at 25 degrees C
→ pH, pOH and the relation pH + pOH = 14
Higher pH means LOWER concentration
→ pH, pOH and the relation pH + pOH = 14
Double for dibasic / diacidic
→ pH of strong acids and strong bases
For a base, do not forget the 14 - pOH step
→ pH of strong acids and strong bases
Apply the degree of dissociation before the log
→ pH of weak acids and weak bases
sqrt(Ka c), not Ka c
→ pH of weak acids and weak bases
A weak dibasic acid still furnishes 2 H+
→ pH of weak acids and weak bases

Salt Hydrolysis

Formulas (1)

Hydrolysis constant, degree of hydrolysis and pH · Hydrolysis constant, degree and salt pH $K_h = \dfrac{K_w}{K_a}\qquad h = \sqrt{\dfrac{K_h}{c}}\qquad \text{pH} = 7 + \tfrac{1}{2}\left(pK_a + \log c\right)$

Reference tables (2)

The four salt types and their solution pH4 rows

Salt type	Example salt	Ion that hydrolyses	Solution / pH
Strong acid + strong base	$\text{NaCl}$ , $\text{KNO}_3$	None	Neutral, $\text{pH} = 7$ These salts are NOT hydrolysed — both ions come from strong parents and do not react with water.
Strong acid + weak base	$\text{NH}_4\text{Cl}$ , $\text{CuSO}_4$	Cation	Acidic, $\text{pH} < 7$
Weak acid + strong base	$\text{CH}_3\text{COONa}$ , $\text{Na}_2\text{CO}_3$	Anion	Basic, $\text{pH} > 7$
Weak acid + weak base	$\text{CH}_3\text{COONH}_4$ , $\text{NH}_4\text{CN}$	Both ions	Depends on $K_a$ vs $K_b$ $\text{NH}_4\text{CN}$ is basic because HCN ( $K_a \approx 4\times10^{-10}$ ) is a much weaker acid than $\text{NH}_4\text{OH}$ ( $K_b \approx 1.8\times10^{-5}$ ) is a base, so $K_b > K_a$ .

The pH is set by the WEAKER parent: weak base → acidic, weak acid → basic, both strong → neutral.

Which ion hydrolyses — classifying a given salt5 rows

Salt	Weak parent	Ion that hydrolyses	Litmus effect
$\text{CuCl}_2$	Weak base $\text{Cu(OH)}_2$	$\text{Cu}^{2+}$ (cation)	Acidic — blue litmus turns red
$\text{NH}_4\text{NO}_3$	Weak base $\text{NH}_4\text{OH}$	$\text{NH}_4^{+}$ (cation)	Acidic — blue litmus turns red
$\text{CH}_3\text{COONa}$	Weak acid $\text{CH}_3\text{COOH}$	$\text{CH}_3\text{COO}^{-}$ (anion)	Basic — red litmus turns blue
$\text{KCN}$	Weak acid HCN	$\text{CN}^{-}$ (anion)	Basic — red litmus turns blue
$\text{NaNO}_3$	None (both strong)	Neither ion	Neutral — no litmus change $\text{NaNO}_3$ , NaCl and KCl are neutral — they are classic 'no change' distractors in litmus questions.

Weak-base cation → acidic (blue→red); weak-acid anion → basic (red→blue).

Watch out for (6)

A strong-acid + strong-base salt does NOT hydrolyse
→ The four salt types and their solution pH
Match the salt to the RIGHT parents
→ The four salt types and their solution pH
Only the ion of the WEAKER partner hydrolyses
→ Which ion hydrolyses — classifying a given salt
Weak-base cation → acidic, not basic
→ Which ion hydrolyses — classifying a given salt
Divide $K_w$ by the WEAK parent's constant
→ Hydrolysis constant, degree of hydrolysis and pH
The degree of hydrolysis carries a square root
→ Hydrolysis constant, degree of hydrolysis and pH

Buffer Solutions and the Henderson-Hasselbalch Equation

Formulas (3)

Henderson-Hasselbalch equation — pH of an acidic buffer · Henderson-Hasselbalch (acidic buffer) $\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$
Equal salt and acid — pH equals pKa · Buffer with equal salt and acid $[\text{salt}] = [\text{acid}] \;\Rightarrow\; \text{pH} = pK_a = -\log K_a$
Basic buffers — the pOH form and converting to pH · Henderson-Hasselbalch (basic buffer) $\text{pOH} = pK_b + \log\dfrac{[\text{salt}]}{[\text{base}]} \qquad \text{pH} = 14 - \text{pOH}$

Reference tables (1)

What a buffer is and how to recognise one3 rows

Buffer type	Components	Example
Acidic buffer (pH < 7)	Weak acid + salt of that acid with a strong base	$\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$ The salt supplies the conjugate base (acetate). A strong acid + salt is NOT a buffer.
Basic buffer (pH > 7)	Weak base + salt of that base with a strong acid	$\text{NH}_4\text{OH} + \text{NH}_4\text{Cl}$ The salt supplies the conjugate acid (ammonium). Note the components: weak base + its salt with a strong acid.
Blood buffer	Carbonic acid + its salt (bicarbonate)	$\text{H}_2\text{CO}_3 / \text{HCO}_3^-$ The bicarbonate buffer holds human blood pH near 7.4 — a frequently asked recall item.

A buffer always pairs a weak partner with its conjugate (from the salt).

Watch out for (7)

A strong acid + its salt is NOT a buffer
→ What a buffer is and how to recognise one
Match the salt to the right partner
→ What a buffer is and how to recognise one
Ratio is salt over acid — don't invert it
→ Henderson-Hasselbalch equation — pH of an acidic buffer
Use concentrations directly — no volume conversion
→ Henderson-Hasselbalch equation — pH of an acidic buffer
Equal concentrations means the log term is zero
→ Equal salt and acid — pH equals pKa
Find pOH first, then subtract from 14
→ Basic buffers — the pOH form and converting to pH
Use pKb for a base, pKa for an acid
→ Basic buffers — the pOH form and converting to pH

Solubility Product (Ksp)

Formulas (5)

Solubility product expression · General solubility product $K_{sp} = [A^{y+}]^x\,[B^{x-}]^y \qquad (A_x B_y \rightleftharpoons x\,A^{y+} + y\,B^{x-})$
Solubility of a 1:1 (AB) salt: Ksp = S squared · AB salt: solubility and solubility product $K_{sp} = S^2 \qquad\Longleftrightarrow\qquad S = \sqrt{K_{sp}}$
Solubility of AB2, A2B and A2B3 salts · AB2 / A2B salt: solubility and solubility product $K_{sp} = 4S^3 \qquad\Longleftrightarrow\qquad S = \sqrt[3]{\dfrac{K_{sp}}{4}}$
Ksp from pH and from mass solubility · Molar solubility from mass solubility $S = \dfrac{m}{M} \qquad\text{and}\qquad [\text{OH}^-] = 10^{-\text{pOH}},\ \ \text{pOH} = 14 - \text{pH}$
Common ion effect on solubility · Solubility in presence of a common ion $S = \dfrac{K_{sp}}{C}$

Reference tables (1)

Ksp in terms of solubility, by salt type4 rows

Salt type	Dissociation	Ksp in terms of S	Example salt
AB (1:1)	$AB \rightleftharpoons A^+ + B^-$	$K_{sp} = S^2$	AgCl, AgBr, CaCO3, NiS Most common type in the bank. Recover S by a single square root: $S = \sqrt{K_{sp}}$ .
AB2 or A2B (1:2)	$AB_2 \rightleftharpoons A^{2+} + 2B^-$	$K_{sp} = 4S^3$	PbI2, PbCl2, Ag2CrO4, Ba(OH)2 Recover S by $S = \sqrt[3]{K_{sp}/4}$ — divide by 4 first, then take the cube root.
AB3 or A3B (1:3)	$AB_3 \rightleftharpoons A^{3+} + 3B^-$	$K_{sp} = 27S^4$	Fe(OH)3-type, AlCl3-type Recover S by $S = \left(\dfrac{K_{sp}}{27}\right)^{1/4}$ .
A2B3 or A3B2 (2:3)	$A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$	$K_{sp} = 108\,S^5$	Ca3(PO4)2, Al2(SO4)3 Factor is $2^2 \times 3^3 = 4 \times 27 = 108$ . Recover S by $S = \left(\dfrac{K_{sp}}{108}\right)^{1/5}$ .

The numerical factor is the product of each coefficient raised to its own power; the exponent on S is the total number of ions produced.

Watch out for (12)

Raise each ion to its own coefficient
→ Solubility product expression
The solid is left out
→ Solubility product expression
AB2 is 4S cubed, not S squared
→ Ksp in terms of solubility, by salt type
Match the root to the exponent on S
→ Ksp in terms of solubility, by salt type
Take the square root — do not report Ksp as the solubility
→ Solubility of a 1:1 (AB) salt: Ksp = S squared
Handle the power correctly under the root
→ Solubility of a 1:1 (AB) salt: Ksp = S squared
Divide by the factor BEFORE taking the root
→ Solubility of AB2, A2B and A2B3 salts
Group the power of ten into a multiple of the root
→ Solubility of AB2, A2B and A2B3 salts
Metal-ion concentration is HALF the hydroxide in M(OH)2
→ Ksp from pH and from mass solubility
Convert grams to moles before using Ksp
→ Ksp from pH and from mass solubility
A common ion LOWERS solubility
→ Common ion effect on solubility
Use the common-ion concentration, not the square root
→ Common ion effect on solubility