Ionization Of Acids And Bases

Q: How is pKa related to acid strength?

pKa is minus log of Ka. Smaller pKa (larger Ka) means a stronger acid. Strong acids like HCl have pKa around minus 6; weak acids like acetic acid have pKa around 4.74; very weak acids like HCN have pKa around 9.3. A difference of one unit in pKa corresponds to a factor of 10 in Ka.

Q: Why does the second ionization of a polyprotic acid always have a smaller Ka?

Removing a positive H plus from a neutral H3PO4 molecule is easier than removing another H plus from the negatively charged H2PO4 minus ion, and much harder from the doubly negative HPO4 2 minus. Each successive removal fights an increasing negative charge that holds the remaining protons more tightly, so Ka1 is much larger than Ka2 which is much larger than Ka3, typically by factors of 10000 to a million.

Q: Why is HClO4 a stronger acid than HClO?

In HClO4, chlorine is bonded to four oxygens which withdraw electron density from the O-H bond and stabilize the resulting ClO4 minus anion by delocalizing negative charge over four oxygens. In HClO, chlorine is bonded to just one oxygen, so much less electron withdrawal and charge delocalization. This makes HClO4 much stronger.

Q: Why is CH3COONa solution basic?

Sodium acetate dissociates completely into Na plus and CH3COO minus. Na plus is the cation of a strong base and does not hydrolyze. But CH3COO minus is the conjugate base of the weak acid CH3COOH; it is basic enough to accept a proton from water, producing OH minus. The excess OH minus makes the solution basic.

Q: Why does the pH of CH3COONH4 not depend on concentration?

Both ions hydrolyze. The overall hydrolysis constant Kh equals Kw divided by the product Ka times Kb, which leads to a ratio h over 1 minus h equal to root Kh - this has no concentration term. The final pH formula pH equals one half of pKw plus pKa minus pKb is independent of concentration. Dilution shifts both hydrolysis equilibria equally, so their competing effects cancel out.

Q: Why is NaHCO3 solution slightly basic?

HCO3 minus is amphiprotic: it can donate H plus (giving CO3 2 minus, using Ka2 of H2CO3) or accept H plus (giving H2CO3, using Kb related to Ka1). Its pH is one half of pKa1 plus pKa2, which is about 8.35 - slightly basic. The formula reflects a balance between the anion's acid and base tendencies.

Q: Why is phenolphthalein a suitable indicator for weak acid vs strong base titration but not for weak base vs strong acid?

Phenolphthalein has pKIn around 9.4, so its useful range is 8.3 to 10.0. In a weak acid vs strong base titration, the equivalence pH is basic which falls within phenolphthalein's transition zone. In weak base vs strong acid titration, equivalence pH is acidic, where phenolphthalein remains colourless throughout and cannot signal the endpoint. For that titration methyl orange or methyl red is used.

ChemistryEquilibriumFor JEE aspirants

The ionization of acids and bases in water is quantified by their dissociation constants $K_{a}$ and $K_{b}$ . For a weak acid $H A ⇌ H^{+} + A^{-}$ , $K_{a} = C α^{2} / (1 - α)$ measures how far ionization proceeds. Larger $K_{a}$ (smaller $p K_{a}$ ) means a stronger acid. For any conjugate acid-base pair, $K_{a} \times K_{b} = K_{w} = 1 0^{- 14}$ . This concept covers weak-acid and weak-base pH calculations, mixtures (strong + weak, weak + weak), polyprotic acids like $H_{3} P O_{4}$ and $H_{2} S$ , the factors that determine acid strength (bond strength, atomic size, electronegativity, oxoacid trends), salt hydrolysis for all four salt types, and the Ostwald theory of indicators.

Key Formulas - Quick Reference

Weak monoprotic acid: $K_{a} = \frac{C α ^{2}}{1 - α} \approx C α^{2}$ ; $α = K_{a} / C$ ; $[H^{+}] = K_{a} \cdot C$ ; $p H = \frac{1}{2} (p K_{a} - lo g C)$
Weak base: $[O H^{-}] = K_{b} \cdot C$ ; $pO H = \frac{1}{2} (p K_{b} - lo g C)$
Two weak acids together: $[H^{+}] = C_{1} K_{a_{1}} + C_{2} K_{a_{2}}$
Conjugate pair relation: $K_{a} \times K_{b} = K_{w}$ ; $p K_{a} + p K_{b} = 14$ at $25 ° C$
Polyprotic acid: $K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}$ ; $[H^{+}] \approx K_{a_{1}} \cdot C$ (first-step dominates)
Salt of WA + SB (anionic hydrolysis): $K_{h} = K_{w} / K_{a}$ ; $p H = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$
Salt of SA + WB (cationic hydrolysis): $K_{h} = K_{w} / K_{b}$ ; $p H = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$
Salt of WA + WB: $K_{h} = K_{w} / (K_{a} K_{b})$ ; $p H = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ - independent of concentration
Amphiprotic anion (e.g., $H C O_{3}^{-}$ ): $p H = \frac{1}{2} (p K_{a_{1}} + p K_{a_{2}})$

1. Dissociation Constants $K_{a}$ and $K_{b}$

For a weak monoprotic acid $H A$ at concentration $C$ in water:

$H A ⇌ H^{+} + A^{-}$

$H A H^{+} A^{-} initial C 00 eqbm C (1 - α) C α C α$

Applying the law of mass action:

$K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} = \frac{C α ^{2}}{1 - α}$

For weak acids with $α ≪ 1$ : $K_{a} \approx C α^{2} \Rightarrow α = K_{a} / C$ , giving:

$[H^{+}] = C α = K_{a} \cdot C \Rightarrow p H = \frac{1}{2} (p K_{a} - lo g C)$

Similarly for a weak base $B O H$ : $K_{b} = C α^{2} / (1 - α)$ , and $[O H^{-}] = K_{b} \cdot C$ .

Relative strengths of two weak acids (or two weak bases) at equal concentrations:

\frac{α _{1}}{α _{2}} = \frac{K _{a_{1}}}{K _{a_{2}}}

. Greater

K_{a}

(or smaller

p K_{a}

) means stronger acid.

2. $p K_{a}$ and $p K_{b}$

Just as pH avoids handling tiny $[H^{+}]$ values, $p K_{a}$ avoids handling tiny $K_{a}$ values:

$p K_{a} = - lo g K_{a}, p K_{b} = - lo g K_{b}$

Interpretation: smaller $p K_{a}$ = stronger acid; smaller $p K_{b}$ = stronger base.

Acid	$K_{a}$	$p K_{a}$	Strength
$H C l O_{4}$	$> 1 0^{7}$	$< - 7$	Very strong
$H C l$	$\sim 1 0^{6}$	$\sim - 6$	Strong
$H F$	$6.8 \times 1 0^{- 4}$	$3.17$	Weak
$C H_{3} C O O H$	$1.8 \times 1 0^{- 5}$	$4.74$	Weak
$H_{2} C O_{3}$	$4.3 \times 1 0^{- 7}$	$6.37$	Very weak
$H C N$	$4.9 \times 1 0^{- 10}$	$9.31$	Very weak

3. Relation Between $K_{a}$ and $K_{b}$ for a Conjugate Pair

For a conjugate acid-base pair $H A / A^{-}$ :

$H A ⇌ H^{+} + A^{-}; K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}$

$A^{-} + H_{2} O ⇌ H A + O H^{-}; K_{b} = \frac{[ H A ] [ O H ^{-} ]}{[ A ^{-} ]}$

Multiplying:

$K_{a} \cdot K_{b} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} \cdot \frac{[ H A ] [ O H ^{-} ]}{[ A ^{-} ]} = [H^{+}] [O H^{-}] = K_{w}$

$K_{a} \cdot K_{b} = K_{w} \Rightarrow p K_{a} + p K_{b} = p K_{w} = 14 at 25 ° C$

This relation is one of the most useful in ionic equilibrium. It tells us that a stronger acid necessarily has a weaker conjugate base, and vice versa. If

K_{a}

H C N

4.9 \times 1 0^{- 10}

, then

K_{b}

of its conjugate base

C N^{-}

1 0^{- 14} / (4.9 \times 1 0^{- 10}) \approx 2 \times 1 0^{- 5}

- a moderately strong base.

4. pH of a Weak Monoprotic Acid

Solved Example 1

Calculate the pH of (a)

1 0^{- 1} M

, (b)

1 0^{- 3} M

, and (c)

1 0^{- 6} M

acetic acid solutions. Take

K_{a} = 2 \times 1 0^{- 5}

Solution:

(a) $C = 1 0^{- 1}$ M: $α = K_{a} / C = 2 \times 1 0^{- 5} /1 0^{- 1} = 2 \times 1 0^{- 4} \approx 1.4 \times 1 0^{- 2}$ . Since $α ≪ 0.1$ , approximation valid.

$[H^{+}] = C α = 1 0^{- 1} \times 1.4 \times 1 0^{- 2} = 1.4 \times 1 0^{- 3}$ . $p H = 3 - \frac{1}{2} lo g 2 \approx 2.85$ .

(b) $C = 1 0^{- 3}$ M: $α = 2 \times 1 0^{- 5} /1 0^{- 3} = 2 \times 1 0^{- 2} \approx 0.14$ . This is $> 0.1$ - approximation fails.

Solve exactly: $K_{a} = C α^{2} / (1 - α)$ , so $2 \times 1 0^{- 5} = 1 0^{- 3} α^{2} / (1 - α)$ . This gives $α \approx 0.131$ .

$[H^{+}] = 1 0^{- 3} \times 0.131 = 1.31 \times 1 0^{- 4}$ . $p H \approx 3.88$ .

(c) $C = 1 0^{- 6}$ M: $α \approx 2 \times 1 0^{- 5} /1 0^{- 6} = 20 > 1$ - clearly meaningless. Solve exactly: $2 \times 1 0^{- 5} = 1 0^{- 6} α^{2} / (1 - α)$ , giving $α \approx 0.95$ . So at very low concentration the weak acid is $\sim 95%$ ionized - it behaves nearly like a strong acid.

$[H^{+}] \approx 9.5 \times 1 0^{- 7} M$ . $p H \approx 6.02$ .

Key insight: At infinite dilution, even weak acids are essentially fully ionized. So the pH of

1 0^{- 6} M

acetic acid approaches that of

1 0^{- 6} M

HCl. This is a direct consequence of Ostwald's dilution law.

5. pH of a Mixture of Two Weak Acids

Let two weak monoprotic acids $H A_{1}$ and $H A_{2}$ (concentrations $C_{1}, C_{2}$ ; dissociation constants $K_{a_{1}}, K_{a_{2}}$ ) be present together. Both contribute $H^{+}$ ; $[H^{+}]$ is common in solution.

For each acid: $K_{a_{1}} = \frac{[ H ^{+} ] [ A _{1}^{-} ]}{C _{1}}$ and $K_{a_{2}} = \frac{[ H ^{+} ] [ A _{2}^{-} ]}{C _{2}}$ (since $α ≪ 1$ , $[H A_{i}] \approx C_{i}$ ).

Charge balance (neglecting water): $[H^{+}] = [A_{1}^{-}] + [A_{2}^{-}]$ . Substituting:

$[H^{+}] = \frac{C _{1} K _{a_{1}}}{[ H ^{+} ]} + \frac{C _{2} K _{a_{2}}}{[ H ^{+} ]} \Rightarrow [H^{+}]^{2} = C_{1} K_{a_{1}} + C_{2} K_{a_{2}}$

$[H^{+}] = C_{1} K_{a_{1}} + C_{2} K_{a_{2}}$

Solved Example 2

Calculate the pH of a solution obtained by mixing equal volumes of

0.02 M H O C l

(

K_{a} = 2 \times 1 0^{- 4}

) and

0.2 M C H_{3} C O O H

(

K_{a} = 2 \times 1 0^{- 5}

Solution:

After mixing, concentrations halve: $C_{1} = 0.01 M (H O C l), C_{2} = 0.1 M (C H_{3} C O O H)$ .

$[H^{+}] = C_{1} K_{a_{1}} + C_{2} K_{a_{2}} = (0.01) (2 \times 1 0^{- 4}) + (0.1) (2 \times 1 0^{- 5}) = 2 \times 1 0^{- 6} + 2 \times 1 0^{- 6} = 4 \times 1 0^{- 6} = 2 \times 1 0^{- 3}$ .

$p H = - lo g (2 \times 1 0^{- 3}) = 3 - lo g 2 \approx 2.70$ .

6. Polyprotic Acids JEE Advanced

A polyprotic acid can donate more than one $H^{+}$ per molecule, in successive steps. For phosphoric acid $H_{3} P O_{4}$ :

$H_{3} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}; K_{a_{1}} = 7.5 \times 1 0^{- 3}$

$H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}; K_{a_{2}} = 6.2 \times 1 0^{- 8}$

$H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}; K_{a_{3}} = 3.6 \times 1 0^{- 13}$

Universally, $K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}$ (typically by factors of $1 0^{4} - 1 0^{6}$ ).

Why does each step get harder? Removing a positive $H^{+}$ from a neutral molecule ( $H_{3} P O_{4}$ ) is easier than removing it from a negatively charged ion ( $H_{2} P O_{4}^{-}$ ), which is easier still than removing it from a doubly negative ion ( $H P O_{4}^{2 -}$ ). The increasing negative charge holds the remaining protons more tightly.

Simplification: because

K_{a_{1}} ≫ K_{a_{2}}

, essentially all

[H^{+}]

in a polyprotic acid solution comes from the first ionization. Moreover, the

H^{+}

from the first step suppresses the second and third steps via common-ion effect. So

[H^{+}] \approx K_{a_{1}} \cdot C

works to good approximation.

Solved Example 3

Calculate

[H^{+}]

[H S^{-}]

, and

[S^{2 -}]

in a solution that is

0.1 M

H C l

and

0.1 M

H_{2} S

. Given

K_{a_{1}} (H_{2} S) = 1 0^{- 7}

and

K_{a_{2}} (H_{2} S) = 1.3 \times 1 0^{- 13}

Solution:

Nearly all $[H^{+}]$ comes from $H C l$ : $[H^{+}] \approx 0.1 M$ , so $p H \approx 1$ .

From the first step: $K_{a_{1}} = \frac{[ H ^{+} ] [ H S ^{-} ]}{[ H _{2} S ]}$

$1 0^{- 7} = \frac{( 0.1 ) [ H S ^{-} ]}{0.1} \Rightarrow [H S^{-}] = 1 0^{- 7} M$ .

From the second step: $K_{a_{2}} = \frac{[ H ^{+} ] [ S ^{2 -} ]}{[ H S ^{-} ]}$

$1.3 \times 1 0^{- 13} = \frac{( 0.1 ) [ S ^{2 -} ]}{1 0 ^{- 7}} \Rightarrow [S^{2 -}] = 1.3 \times 1 0^{- 19} M$ .

The extreme suppression of $[S^{2 -}]$ in acidic solution is the basis of selective sulphide precipitation in qualitative analysis (Group II vs Group IV separation).

7. Factors Affecting Acid Strength

(a) H-A bond strength

The easier it is to break the H-A bond, the stronger the acid. Down a group, bond strength decreases (larger atoms overlap poorly with hydrogen), so acid strength increases: $H F < H C l < H B r < H I$ . Iodine is the largest, so H-I is weakest and $H I$ is the strongest hydrohalic acid.

(b) Electronegativity

Across a period, electronegativity increases and the polar $H - A$ bond becomes easier to ionize heterolytically: $C H_{4} < N H_{3} < H_{2} O < H F$ . Both bond polarity and stability of the conjugate base ( $F^{-}$ is more stable than $O H^{-}$ etc.) contribute.

(c) Stability of the conjugate base

The more stable the conjugate base $A^{-}$ , the more the equilibrium $H A ⇌ H^{+} + A^{-}$ favours ionization. Stability is enhanced by (i) larger size, (ii) electron delocalization by resonance, and (iii) electron-withdrawing groups.

Example: $C C l_{3} C O O H$ ( $p K_{a} = 0.7$ ) is much stronger than $C H_{3} C O O H$ ( $p K_{a} = 4.74$ ), because the three chlorines withdraw electron density and stabilize the carboxylate.

(d) Oxoacid trends

For oxoacids of the form $H O - X - (= O)_{n}$ :

Same central atom, more $O$ atoms: more electron withdrawal by $O$ , more stable conjugate base, stronger acid. $H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}$ .
Same number of $O$ atoms, more electronegative central atom: stronger acid. $H O I < H O B r < H O C l$ .
Across a period, oxoacids increase in strength: $H_{3} P O_{4} < H_{2} S O_{4} < H C l O_{4}$ .

Solved Example 4

Which is the strongest acid:

H C l O_{4}, H B r O_{4}, H I O_{4}

, or

H N O_{3}

Solution:

Among $H C l O_{4}, H B r O_{4}, H I O_{4}$ , acidic strength decreases down the group as central-atom electronegativity decreases. So $H C l O_{4} > H B r O_{4} > H I O_{4}$ . Also, $H C l O_{4}$ has oxidation state $+ 7$ on Cl vs $+ 5$ on N in $H N O_{3}$ ; the higher oxidation state (with more O atoms) makes $H C l O_{4}$ the strongest. Answer: $H C l O_{4}$ .

8. Salt Hydrolysis - Overview

Salt hydrolysis is the reaction of the ions of a dissolved salt with water to produce either

H^{+}

O H^{-}

ions, making the solution acidic or basic (or leaving it neutral).

Four categories of salts, based on the strength of their parent acid and base:

Salt type	Example	Which ion hydrolyzes?	Solution
Strong acid + strong base	$N a C l, K N O_{3}$	Neither	Neutral ( $p H = 7$ )
Weak acid + strong base	$C H_{3} C O O N a, K C N$	Anion (basic hydrolysis)	Basic ( $p H > 7$ )
Strong acid + weak base	$N H_{4} C l, A l C l_{3}$	Cation (acidic hydrolysis)	Acidic ( $p H < 7$ )
Weak acid + weak base	$C H_{3} C O O N H_{4}$	Both	Depends on $K_{a}$ vs $K_{b}$

9. Hydrolysis of a Salt of Weak Acid + Strong Base

Example: sodium acetate. $C H_{3} C O O N a \to C H_{3} C O O^{-} + N a^{+}$ (complete). The acetate ion (conjugate base of a weak acid) hydrolyzes:

$C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$

The hydrolysis constant $K_{h}$ is derived by multiplying and dividing by $[H^{+}]$ :

$K_{h} = \frac{[ C H _{3} C O O H ] [ O H ^{-} ]}{[ C H _{3} C O O ^{-} ]} = \frac{K _{w}}{K _{a}}$

Setting up ICE for $C$ molar salt with degree of hydrolysis $h$ :

$K_{h} = \frac{C h \cdot C h}{C ( 1 - h )} \approx C h^{2} \Rightarrow h = \frac{K _{h}}{C} = \frac{K _{w}}{K _{a} \cdot C}$

$[O H^{-}] = C h = K_{w} \cdot C / K_{a}$ ; $[H^{+}] = K_{w} / [O H^{-}] = K_{w} \cdot K_{a} / C$ ; taking logs:

$p H = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$

Solved Example 5

Calculate the degree of hydrolysis and pH of a solution formed by dissolving

0.1

mol of

C H_{3} C O O N a

in water to give

100 L

of solution.

K_{a} (C H_{3} C O O H) = 2 \times 1 0^{- 5}

Solution:

$C = 0.1/100 = 1 0^{- 3} M$ . $K_{h} = K_{w} / K_{a} = 1 0^{- 14} / (2 \times 1 0^{- 5}) = 5 \times 1 0^{- 10}$ .

$h = K_{h} / C = 5 \times 1 0^{- 10} /1 0^{- 3} = 5 \times 1 0^{- 7} \approx 7.07 \times 1 0^{- 4}$ or $\approx 0.07%$ .

$p H = \frac{1}{2} (14 + 4.7 - 3) = \frac{1}{2} (15.7) \approx 7.85$ .

10. Hydrolysis of a Salt of Strong Acid + Weak Base

Example: ammonium chloride. $N H_{4} C l \to N H_{4}^{+} + C l^{-}$ (complete). The ammonium ion (conjugate acid of the weak base $N H_{3}$ ) hydrolyzes:

$N H_{4}^{+} + H_{2} O ⇌ N H_{4} O H + H^{+}$

Symmetric derivation gives:

$K_{h} = \frac{K _{w}}{K _{b}}, h = \frac{K _{w}}{K _{b} \cdot C}, [H^{+}] = \frac{K _{w} \cdot C}{K _{b}}$

$p H = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$

11. Hydrolysis of a Salt of Weak Acid + Weak Base

Example: ammonium acetate. Both ions hydrolyze:

$C H_{3} C O O^{-} + N H_{4}^{+} + H_{2} O ⇌ C H_{3} C O O H + N H_{4} O H$

Multiplying and dividing by $[H^{+}] [O H^{-}]$ :

$K_{h} = \frac{K _{w}}{K _{a} \cdot K _{b}}$

Setting up ICE with degree of hydrolysis $h$ (assumed equal for both ions):

$K_{h} = \frac{C h \cdot C h}{C ( 1 - h ) \cdot C ( 1 - h )} = (\frac{h}{1 - h})^{2} \Rightarrow \frac{h}{1 - h} = K_{h} = K_{w} / K_{a} K_{b}$

Notice $K_{h}$ has no $C$ dependence - the degree of hydrolysis is independent of concentration for this salt type. From $[H^{+}] = K_{a} \cdot \frac{h}{1 - h}$ :

$p H = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$

Since

p H

depends only on

p K_{a}

p K_{b}

, three cases arise:

$K_{a} = K_{b}$ : $p H = 7$ (neutral). E.g., $C H_{3} C O O N H_{4}$ has $p K_{a} \approx p K_{b} \approx 4.74$ , so it is essentially neutral.
$K_{a} > K_{b}$ : $p H < 7$ (acidic).
$K_{a} < K_{b}$ : $p H > 7$ (basic).

Solved Example 6

Calculate the pH and degree of hydrolysis of

1 0^{- 2} M N H_{4} C N

. Given

K_{a} (H C N) = 5 \times 1 0^{- 10}

and

K_{b} (N H_{3}) = 2 \times 1 0^{- 5}

Solution:

$p K_{a} = 10 - lo g 5 = 9.30$ ; $p K_{b} = 5 - lo g 2 = 4.70$ .

$p H = \frac{1}{2} (14 + 9.30 - 4.70) = \frac{1}{2} (18.60) = 9.30$ (basic, since $C N^{-}$ is a much stronger base than $N H_{4}^{+}$ is an acid).

$\frac{h}{1 - h} = K_{w} / (K_{a} K_{b}) = 1 0^{- 14} / (5 \times 1 0^{- 10} \times 2 \times 1 0^{- 5}) = 1 = 1$ .

So $h = 1 - h \Rightarrow h = 0.5$ or $50%$ . The salt is half-hydrolyzed.

12. Hydrolysis of Polyvalent Ions JEE Advanced

For salts like $N a_{3} P O_{4}, K_{2} C O_{3}, Z n S O_{4}$ , or $F e C l_{3}$ , the polyvalent ion hydrolyzes in steps:

$P O_{4}^{3 -} + H_{2} O ⇌ H P O_{4}^{2 -} + O H^{-}; K_{h_{1}}$

$H P O_{4}^{2 -} + H_{2} O ⇌ H_{2} P O_{4}^{-} + O H^{-}; K_{h_{2}}$

$H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} P O_{4} + O H^{-}; K_{h_{3}}$

The hydrolysis constants relate to the acid dissociation constants of the parent acid via:

$K_{a_{1}} \times K_{h_{3}} = K_{w}, K_{a_{2}} \times K_{h_{2}} = K_{w}, K_{a_{3}} \times K_{h_{1}} = K_{w}$

Because $K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}$ , we have $K_{h_{1}} ≫ K_{h_{2}} ≫ K_{h_{3}}$ . The first hydrolysis step dominates:

$p H \approx \frac{1}{2} (p K_{w} + p K_{a_{3}} + lo g C)$

13. Hydrolysis of Amphiprotic Anions

Anions like $H C O_{3}^{-}, H S^{-}, H_{2} P O_{4}^{-}, H P O_{4}^{2 -}$ can both accept and donate a proton, so they undergo both ionization and hydrolysis:

$H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -} (acts as acid, K_{a_{2}})$

$H C O_{3}^{-} + H_{2} O ⇌ H_{2} C O_{3} + O H^{-} (acts as base)$

The pH of such a solution is remarkably simple - independent of concentration:

$p H (H C O_{3}^{-}) = \frac{1}{2} (p K_{a_{1}} + p K_{a_{2}})$

Similarly for $H_{2} P O_{4}^{-}$ : $p H = \frac{1}{2} (p K_{a_{1}} + p K_{a_{2}})$ (of phosphoric acid). For $H P O_{4}^{2 -}$ : $p H = \frac{1}{2} (p K_{a_{2}} + p K_{a_{3}})$ .

14. Ostwald Theory of Indicators

An acid-base indicator is a weak organic acid ( $H I n$ ) or base whose ionized and unionized forms have distinctly different colours. For an acidic indicator like phenolphthalein:

$H I n ⇌ H^{+} + I n^{-}$

$colourless pink$

Applying the law of mass action: $K_{I n} = \frac{[ H ^{+} ] [ I n ^{-} ]}{[ H I n ]}$ . Taking logs and rearranging:

$p H = p K_{I n} + lo g \frac{[ I n ^{-} ]}{[ H I n ]}$

The eye distinguishes the ionized colour when $[I n^{-}] / [H I n] \geq 10$ (about $91%$ ionized), corresponding to $p H \geq p K_{I n} + 1$ . It distinguishes the unionized colour when $[I n^{-}] / [H I n] \leq 1/10$ , corresponding to $p H \leq p K_{I n} - 1$ .

So an indicator's useful pH range is $p K_{I n} \pm 1$ .

Indicator	$p K_{I n}$	pH range	Acid form	Base form
Methyl orange	$3.7$	$3.1 - 4.4$	Red	Yellow
Methyl red	$5.0$	$4.4 - 6.2$	Red	Yellow
Litmus	$6.5$	$5.5 - 7.5$	Red	Blue
Phenolphthalein	$9.4$	$8.3 - 10.0$	Colourless	Pink

At the equivalence point of a titration, the indicator's colour changes distinctly if its

p K_{I n}

matches the pH at equivalence. For a strong acid vs strong base titration (equivalence at

p H = 7

), both methyl orange and phenolphthalein work. For a weak acid vs strong base (equivalence at

p H > 7

), only phenolphthalein works. For a weak base vs strong acid (equivalence at

p H < 7

), only methyl orange works.

Common Mistakes to Avoid

Watch out

Using $α = K_{a} / C$ when $α > 0.1$ . The approximation $1 - α \approx 1$ fails for dilute or moderately weak electrolytes. Always sanity-check $α$ and fall back to the exact quadratic when needed.
Forgetting that $K_{a} \times K_{b} = K_{w}$ applies only to conjugate pairs. $K_{a}$ of $C H_{3} C O O H$ times $K_{b}$ of $C H_{3} C O O^{-}$ equals $K_{w}$ - but $K_{a}$ of $C H_{3} C O O H$ times $K_{b}$ of $N H_{3}$ has no such simple relation.
Using $K_{a_{1}}$ for later steps of a polyprotic acid. Each step has its own constant, and $K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}$ ; do not mix them up.
Assuming $[H^{+}]$ from all steps of a polyprotic acid add up. The first step dominates so severely that the later steps are effectively suppressed. $[H^{+}] \approx K_{a_{1}} C$ works.
Applying $p H = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ blindly. This formula is for a salt of a weak acid and a strong base only. Choose the correct hydrolysis formula from the salt type.
Assuming $C H_{3} C O O N H_{4}$ solution has $p H = 7$ because both hydrolyze. It happens to be nearly neutral only because $p K_{a} (C H_{3} C O O H) \approx p K_{b} (N H_{3}) \approx 4.74$ . For $N H_{4} C N$ , $p H$ is around $9.3$ because $C N^{-}$ is a much stronger base than $N H_{4}^{+}$ is an acid.
Choosing the wrong indicator for a titration. The indicator's $p K_{I n}$ must lie within the steep-rise region of the titration curve, close to the equivalence pH.
Confusing basicity of an acid with basicity of a solution. $H_{3} P O_{4}$ is tribasic (three acidic $H^{+}$ ), but a $H_{3} P O_{4}$ solution is acidic, not basic. The two "basicities" mean different things.

Frequently Asked Questions

How is $p K_{a}$ related to acid strength?

$p K_{a} = - lo g K_{a}$ . Smaller $p K_{a}$ (large $K_{a}$ ) means a stronger acid. Strong acids like $H C l$ have $p K_{a}$ around $- 6$ ; weak acids like acetic acid have $p K_{a}$ around $4.74$ ; very weak acids like $H C N$ have $p K_{a}$ around $9.3$ . A difference of one unit in $p K_{a}$ corresponds to a factor of $10$ in $K_{a}$ .

Why does the second ionization of a polyprotic acid always have a smaller $K_{a}$ ?

Removing a positive $H^{+}$ from a neutral $H_{3} P O_{4}$ molecule is easier than removing another $H^{+}$ from the negatively charged $H_{2} P O_{4}^{-}$ ion, and much harder from the doubly negative $H P O_{4}^{2 -}$ . Each successive removal fights an increasing negative charge that holds the remaining protons more tightly, so $K_{a_{1}} ≫ K_{a_{2}} ≫ K_{a_{3}}$ , typically by factors of $1 0^{4}$ to $1 0^{6}$ .

Why is $H C l O_{4}$ a stronger acid than $H C l O$ ?

In $H C l O_{4}$ , chlorine is bonded to four oxygens (three double-bonded, one bonded to $H$ ), all of which withdraw electron density from the $O - H$ bond and stabilize the resulting $C l O_{4}^{-}$ anion by delocalizing negative charge over four oxygens. In $H C l O$ , chlorine is bonded to just one oxygen, so much less electron withdrawal and less charge delocalization. This makes $H C l O_{4}$ much stronger.

Why is $C H_{3} C O O N a$ solution basic?

Sodium acetate dissociates completely into $N a^{+}$ and $C H_{3} C O O^{-}$ . $N a^{+}$ is the cation of a strong base ( $N a O H$ ) and does not hydrolyze. But $C H_{3} C O O^{-}$ is the conjugate base of the weak acid $C H_{3} C O O H$ ; it is basic enough to accept a proton from water, producing $O H^{-}$ : $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$ . The excess $O H^{-}$ makes the solution basic.

Why does the pH of $C H_{3} C O O N H_{4}$ not depend on concentration?

Both ions hydrolyze. The overall hydrolysis constant $K_{h} = K_{w} / (K_{a} K_{b})$ leads to the ratio $h / (1 - h) = K_{h}$ , which has no $C$ term. The final pH formula $p H = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ is independent of concentration. Physically, dilution shifts both hydrolysis equilibria equally, so their competing effects cancel out at first order.

Why is $N a H C O_{3}$ solution slightly basic?

$H C O_{3}^{-}$ is amphiprotic: it can donate $H^{+}$ (giving $C O_{3}^{2 -}$ , using $K_{a_{2}}$ of $H_{2} C O_{3}$ ) or accept $H^{+}$ (giving $H_{2} C O_{3}$ , using $K_{b}$ related to $K_{a_{1}}$ ). Its pH is $\frac{1}{2} (p K_{a_{1}} + p K_{a_{2}}) \approx \frac{1}{2} (6.4 + 10.3) \approx 8.35$ , which is slightly basic. The formula reflects a balance between the anion's acid and base tendencies.

What is the useful pH range of an indicator?

An indicator changes colour distinctly when the ratio of its two forms shifts by a factor of $10$ from equal, corresponding to a pH range of $p K_{I n} \pm 1$ . Within this range the colour transitions from mostly one form to mostly the other. Outside the range, one form dominates and no colour change is visible.

Why is phenolphthalein a suitable indicator for weak acid vs strong base titration but not for weak base vs strong acid?

Phenolphthalein has $p K_{I n} \approx 9.4$ , so its useful range is about $8.3 - 10.0$ . In a weak-acid vs strong-base titration, the equivalence point pH lies in the basic range ( $> 7$ ), which falls within phenolphthalein's transition zone. In weak base vs strong acid titration, equivalence pH is in the acidic range ( $< 7$ ), where phenolphthalein remains colourless throughout - it cannot signal the endpoint. For that titration, methyl orange or methyl red is used.

Previous year questions on Ionization Of Acids And Bases

13 questions from past papers, each with a step-by-step solution.

Show all 13 questions

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1. Dissociation Constants Ka​ and Kb​

2. pKa​ and pKb​

3. Relation Between Ka​ and Kb​ for a Conjugate Pair

4. pH of a Weak Monoprotic Acid

5. pH of a Mixture of Two Weak Acids

6. Polyprotic Acids JEE Advanced

7. Factors Affecting Acid Strength

(a) H-A bond strength

(b) Electronegativity

(c) Stability of the conjugate base

(d) Oxoacid trends

8. Salt Hydrolysis - Overview

9. Hydrolysis of a Salt of Weak Acid + Strong Base

10. Hydrolysis of a Salt of Strong Acid + Weak Base

11. Hydrolysis of a Salt of Weak Acid + Weak Base

12. Hydrolysis of Polyvalent Ions JEE Advanced

13. Hydrolysis of Amphiprotic Anions

14. Ostwald Theory of Indicators

Common Mistakes to Avoid

Frequently Asked Questions

How is pKa​ related to acid strength?

Why does the second ionization of a polyprotic acid always have a smaller Ka​?

Why is HClO4​ a stronger acid than HClO?

Why is CH3​COONa solution basic?

Why does the pH of CH3​COONH4​ not depend on concentration?

Why is NaHCO3​ solution slightly basic?

What is the useful pH range of an indicator?

Why is phenolphthalein a suitable indicator for weak acid vs strong base titration but not for weak base vs strong acid?

Previous year questions on Ionization Of Acids And Bases

Ready to master Equilibrium?

1. Dissociation Constants $K_{a}$ and $K_{b}$

2. $p K_{a}$ and $p K_{b}$

3. Relation Between $K_{a}$ and $K_{b}$ for a Conjugate Pair

How is $p K_{a}$ related to acid strength?

Why does the second ionization of a polyprotic acid always have a smaller $K_{a}$ ?

Why is $H C l O_{4}$ a stronger acid than $H C l O$ ?

Why is $C H_{3} C O O N a$ solution basic?

Why does the pH of $C H_{3} C O O N H_{4}$ not depend on concentration?

Why is $N a H C O_{3}$ solution slightly basic?