How is H plus from a weak acid affected by a common ion?

Adding a salt or strong acid with an ion common to the weak electrolyte shifts its ionization equilibrium backward by Le Chatelier's principle. For example, adding sodium acetate to acetic acid decreases the acid's alpha and hence its H plus. This is called the common ion effect and is the basis of buffer solutions, salt hydrolysis calculations, and controlled precipitation in analytical chemistry.

Ionic Equilibrium In Solution

Q: What is the difference between strong and weak electrolytes?

Strong electrolytes ionize almost completely in water so their solutions conduct electricity strongly. Examples include HCl, NaOH, and NaCl. Weak electrolytes ionize only partially, setting up an equilibrium between ions and unionized molecules; their solutions conduct poorly. Examples include acetic acid and ammonia solution.

Q: Why does the degree of ionization of a weak electrolyte increase on dilution?

Ostwald's dilution law shows alpha is proportional to the square root of volume, or inversely proportional to the square root of concentration. Dilution reduces the concentration of ions, so by Le Chatelier's principle the equilibrium shifts toward more ionization. At infinite dilution, alpha approaches unity, so even weak electrolytes are essentially fully ionized.

Q: Is water an acid or a base?

Water is amphoteric or amphiprotic: it can act as an acid (donating H plus to give OH minus) or as a base (accepting H plus to give H3O plus). Whether water behaves as an acid or base in a particular reaction depends on the other substance present. When mixed with HCl water is a base; when mixed with NH3 water is an acid.

Q: Why is H3BO3 not considered an Arrhenius acid?

Boric acid does not release H plus directly. Instead it accepts an OH minus from water, and the water becomes deprotonated releasing H plus. Because the acidity arises from H3BO3 acting as a Lewis acid, accepting an electron pair from OH minus, it is best classified as a Lewis acid.

Q: Why does the pH of pure water differ at different temperatures?

The self-ionization of water is endothermic, so Kw increases with temperature. At 25 degrees C, Kw is 10 to the minus 14 and pure water has pH 7. At 80 degrees C, Kw is about 10 to the minus 13, so pure water has neutral pH about 6.5. Water remains neutral (equal H plus and OH minus) at all temperatures; only the numerical neutral pH changes.

ChemistryEquilibriumFor NEET aspirants

Ionic equilibrium is the equilibrium between an undissociated electrolyte and its ions in aqueous solution. Strong electrolytes such as $H C l$ , $N a O H$ , and $N a C l$ dissociate almost completely; weak electrolytes such as $C H_{3} C O O H$ and $N H_{4} O H$ dissociate only partially, and their ionization follows the law of mass action through the Ostwald dilution law. This concept covers electrolytes, degree of ionization $α$ , all four theories of acids and bases (Arrhenius, Bronsted-Lowry, Lewis, Lux-Flood), the self-ionization of water ( $K_{w} = 1 0^{- 14}$ at $25 ° C$ ), the pH scale, and the common ion effect that underpins buffers and salt hydrolysis.

Key Formulas - Quick Reference

Degree of ionization: $α = \frac{moles dissociated}{moles taken initially}$
Ostwald dilution law (weak electrolyte $A B$ ): $K_{a} = \frac{C α ^{2}}{1 - α}$ ; for $α ≪ 1$ : $α = \frac{K _{a}}{C}$
Ionic product of water: $K_{w} = [H^{+}] [O H^{-}] = 1 0^{- 14}$ at $25 ° C$
pH definition: $p H = - lo g [H^{+}]$ ; $pO H = - lo g [O H^{-}]$ ; $p H + pO H = p K_{w} = 14$ at $25 ° C$
pH of a strong monoprotic acid (concentration $> 1 0^{- 6} M$ ): $p H = - lo g C$
pH of a strong base (concentration $> 1 0^{- 6} M$ ): $pO H = - lo g C$ ; $p H = 14 - pO H$
Mixture of strong acids: $[H^{+}] = \frac{N _{1} V _{1} + N _{2} V _{2}}{V _{1} + V _{2}}$
Conjugate acid-base pair: $K_{a} \times K_{b} = K_{w}$ ; $p K_{a} + p K_{b} = p K_{w} = 14$

1. Electrolytes and Non-electrolytes

An electrolyte is a substance whose aqueous solution or molten form conducts electricity by breaking into ions. A non-electrolyte dissolves without producing ions and does not conduct electricity.

Strong electrolytes

These ionize almost completely in water ( $α \approx 1$ , or $100%$ ). Examples include:

Strong acids: $H C l, H B r, H I, H C l O_{4}, H N O_{3}, H_{2} S O_{4}$
Strong bases: $N a O H, K O H, R b O H, C s O H, C a (O H)_{2}, B a (O H)_{2}$
All soluble salts: $N a C l, K N O_{3}, N H_{4} C l$ , etc.

Because ionization is essentially complete, we write these with a single arrow: $H C l + a q \to H^{+} (a q) + C l^{-} (a q)$ .

Weak electrolytes

These ionize only partially ( $α ≪ 1$ , typically less than $10%$ at ordinary concentrations). An equilibrium exists between the unionized molecule and the ions:

Weak acids: $C H_{3} C O O H, H C N, H F, H_{2} C O_{3}, H_{3} P O_{4}, H C O O H$
Weak bases: $N H_{4} O H (N H_{3} \cdot H_{2} O), Z n (O H)_{2}, F e (O H)_{3}$ , most organic amines
Sparingly soluble salts: $A g C l, B a S O_{4}, P b C l_{2}$

Weak electrolytes are written with the double arrow: $C H_{3} C O O H + a q ⇌ C H_{3} C O O^{-} (a q) + H^{+} (a q)$ .

Non-electrolytes

Substances such as sugar (sucrose), urea, glucose, ethanol, and diethyl ether dissolve as molecules without dissociation. They do not conduct electricity in solution.

2. Degree of Ionization ( $α$ )

The degree of ionization is the fraction of one mole of an electrolyte that has dissociated into ions under the given conditions:

α = \frac{moles dissociated}{moles taken initially}

Percentage ionization $= α \times 100$ . The value of $α$ depends on several factors:

Nature of the electrolyte: strong electrolytes ionize completely; weak electrolytes only partially.
Nature of the solvent: solvents with high dielectric constant (like water) favour ionization by weakening the electrostatic attraction between ions.
Dilution: for weak electrolytes, $α$ increases with dilution and approaches unity at infinite dilution (Ostwald's dilution law).
Temperature: increasing temperature generally increases $α$ because ionization is usually endothermic.
Presence of a common ion: adding a salt with an ion common to the weak electrolyte suppresses its ionization (common ion effect).

3. Ostwald's Dilution Law

For a weak binary electrolyte $A B$ dissolved in water at concentration $C$ mol/L, if $α$ is the degree of ionization:

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

Applying the law of mass action:

$K_{e q} = \frac{[ A ^{+} ] [ B ^{-} ]}{[ A B ]} = \frac{( C α ) ( C α )}{C ( 1 - α )} = \frac{C α ^{2}}{1 - α}$

For weak electrolytes with $α ≪ 1$ , we approximate $1 - α \approx 1$ :

$K_{e q} \approx C α^{2} \Rightarrow α = \frac{K _{e q}}{C}$

Validity of approximation: The formula

α = K / C

is safe when

α < 0.1

(or

10%

). If

α

turns out larger, solve the quadratic

C α^{2} + K α - K = 0

exactly.

Since $C = 1/ V$ (where $V$ is the dilution - volume in litres containing one mole), Ostwald's law can also be written:

$α = K_{e q} \cdot V$

So $α$ is directly proportional to $V$ : the degree of ionization of a weak electrolyte is directly proportional to the square root of the dilution.

4. Common Ion Effect

Common ion effect: the ionization of a weak electrolyte is suppressed when a strong electrolyte containing an ion common to it is added to the solution.

Consider $C H_{3} C O O H ⇌ C H_{3} C O O^{-} + H^{+}$ . If we add sodium acetate $C H_{3} C O O N a$ (which ionizes completely to give $C H_{3} C O O^{-}$ and $N a^{+}$ ), the increased $[C H_{3} C O O^{-}]$ shifts the acetic acid equilibrium to the left (by Le Chatelier's principle), decreasing $[H^{+}]$ and hence $α$ .

Solved Example 1

Explain why acetic acid is less acidic in sodium acetate solution than in sodium chloride solution.

Solution:

Sodium acetate provides the common $C H_{3} C O O^{-}$ ion, which suppresses the ionization of $C H_{3} C O O H$ via the common ion effect, reducing $[H^{+}]$ . Sodium chloride does not share any ion with $C H_{3} C O O H$ and hence does not suppress its ionization.

5. Theories of Acids and Bases

(a) Arrhenius theory (1887)

Arrhenius acid: a substance that produces

H^{+}

ions when dissolved in water.
Arrhenius base: a substance that produces

O H^{-}

ions when dissolved in water.

Examples: $H C l (a q) \to H^{+} (a q) + C l^{-} (a q)$ ; $N a O H (a q) \to N a^{+} (a q) + O H^{-} (a q)$ .

Acids are classified by basicity (number of $H^{+}$ per molecule):

Monobasic (monoprotic): $H C l, H N O_{3}, H C l O_{4}$
Dibasic (diprotic): $H_{2} S O_{4}, H_{2} C O_{3}$
Tribasic (triprotic): $H_{3} P O_{4}$

Bases are similarly classified by acidity (number of $O H^{-}$ ): mono-acidic ( $N a O H$ ), di-acidic ( $C a (O H)_{2}$ ), tri-acidic ( $F e (O H)_{3}$ ).

Two important caveats: (i)

H_{3} B O_{3}

(boric acid) is not an Arrhenius acid - it does not release

H^{+}

directly. Instead it reacts with water:

H_{3} B O_{3} + H_{2} O ⇌ [B (O H)_{4}]^{-} + H^{+}

. It is a Lewis acid. (ii)

H^{+}

never exists free in solution; it is hydrated as

H_{3} O^{+}

(hydronium),

H_{5} O_{2}^{+}

, or

H_{7} O_{3}^{+}

Limitations of Arrhenius theory:

Limited to aqueous solutions only.
Cannot explain the acidic or basic nature of substances with no $H^{+}$ or $O H^{-}$ (e.g., $C O_{2}, S O_{2}$ are acidic; $N H_{3}, C a O$ are basic).
Cannot explain non-aqueous acid-base reactions like $N H_{3} (g) + H C l (g) \to N H_{4} C l (s)$ .

(b) Bronsted-Lowry theory (1923) - Proton concept

Bronsted acid: a proton (

H^{+}

) donor.
Bronsted base: a proton acceptor.

Every acid has a conjugate base (what remains after donating $H^{+}$ ), and every base has a conjugate acid (what forms after accepting $H^{+}$ ). A conjugate acid-base pair differs by exactly one proton.

Acid	Conjugate base	Base	Conjugate acid
$H C l$	$C l^{-}$	$N H_{3}$	$N H_{4}^{+}$
$H_{2} S O_{4}$	$H S O_{4}^{-}$	$H_{2} O$	$H_{3} O^{+}$
$H S O_{4}^{-}$	$S O_{4}^{2 -}$	$O H^{-}$	$H_{2} O$
$H_{2} O$	$O H^{-}$	$R N H_{2}$	$R N H_{3}^{+}$

Rules:

A strong acid has a weak conjugate base, and vice versa.
An acid-base reaction proceeds spontaneously from stronger acid + stronger base to weaker acid + weaker base.

Amphoteric (amphiprotic) species can act as either an acid or a base depending on the reaction. Examples:

H_{2} O, H C O_{3}^{-}, H S O_{4}^{-}, H S^{-}, H_{2} P O_{4}^{-}, H P O_{4}^{2 -}

For water: as an acid, $H_{2} O \to H^{+} + O H^{-}$ ; as a base, $H_{2} O + H^{+} \to H_{3} O^{+}$ .

(c) Lewis theory (1923) - Electronic concept

Lewis acid: an electron pair acceptor.
Lewis base: an electron pair donor.

Lewis acids include:

Electron-deficient molecules: $B F_{3}, A l C l_{3}, B C l_{3}$ (incomplete octet on central atom).
Simple cations: $H^{+}, F e^{3 +}, A g^{+}, C u^{2 +}$ .
Molecules with vacant $d$ -orbitals: $S n C l_{4}, S i F_{4}, P F_{5}$ .
Molecules with polar multiple bonds: $C O_{2}, S O_{2}, S O_{3}$ .

Lewis bases include:

Molecules with lone pairs: $N H_{3}, H_{2} O, R - O H, R - N H_{2}, P H_{3}$ .
All anions: $O H^{-}, C l^{-}, C N^{-}, F^{-}$ .

Neutralization in Lewis terms involves donation of an electron pair from base to acid, forming a coordinate bond:

$B F_{3} + : N H_{3} \to F_{3} B \leftarrow N H_{3}$

Comparison of concepts: Lewis

\supset

Bronsted-Lowry

\supset

Arrhenius. Every Arrhenius acid is a Bronsted acid (donates

H^{+}

); every Bronsted acid is a Lewis acid (

H^{+}

accepts an electron pair). But not every Lewis acid is a Bronsted acid:

B F_{3}

has no proton to donate.

Solved Example 2

In which of the following reactions does

N H_{3}

act as an acid? (A)

N H_{3} (a q) + H^{+} \to N H_{4}^{+}

(B)

N H_{3} (l) + N a (s) \to N a N H_{2} + \frac{1}{2} H_{2}

(C)

N H_{3} (g) + H C l (g) \to N H_{4} C l (s)

Solution:

In reaction (B), $N H_{3}$ loses a proton to become $N H_{2}^{-}$ (amide ion), so it is acting as a Bronsted acid. In (A) and (C), $N H_{3}$ accepts a proton, acting as a base. Answer: (B).

6. Levelling Effect of Water JEE Advanced

All strong acids ( $H C l, H B r, H I, H N O_{3}, H C l O_{4}$ ) appear equally strong in water because water is basic enough to fully accept a proton from any of them. The strongest acid that can exist in water is $H_{3} O^{+}$ ; any acid stronger than $H_{3} O^{+}$ is completely converted to $H_{3} O^{+}$ . This is the levelling effect of water on acids.

Similarly, all strong bases ( $N a O H, K O H, C a (O H)_{2}$ ) are levelled to $O H^{-}$ in water. To distinguish among strong acids or strong bases, we must use a non-aqueous solvent - for example, glacial acetic acid can differentiate strengths of $H C l O_{4}, H C l, H N O_{3}, H_{2} S O_{4}$ .

7. Properties of Water and Self-Ionization

Amphoteric nature

Water can act as either an acid or a base. It behaves as an acid toward $N H_{3}$ and as a base toward $H C l$ . This makes it an ideal solvent for a wide range of acid-base reactions.

Molar concentration of water

Density of water $= 1 g / c m^{3} = 1000 g / L$ . Molar mass $= 18 g / m o l$ .

$[H_{2} O] = \frac{1000}{18} \approx 55.55 M$

Ionic product of water ( $K_{w}$ )

Water self-ionizes to a tiny extent:

$2 H_{2} O (l) ⇌ H_{3} O^{+} (a q) + O H^{-} (a q)$

or more simply: $H_{2} O ⇌ H^{+} + O H^{-}$ .

Since $[H_{2} O]$ is essentially constant, we absorb it into the constant:

$K_{w} = [H^{+}] [O H^{-}] = 1 0^{- 14} M^{2} at 25 ° C$

Degree of dissociation and $K_{a}$ of water

At $25 ° C$ : $[H^{+}] = [O H^{-}] = 1 0^{- 7} M$ . So the degree of dissociation of water:

$α = \frac{1 0 ^{- 7}}{55.55} \approx 1.8 \times 1 0^{- 9} = 1.8 \times 1 0^{- 7} %$

Absolute $K_{a}$ (or $K_{b}$ ) of water:

$K_{a} (H_{2} O) = \frac{[ H ^{+} ] [ O H ^{-} ]}{[ H _{2} O ]} = \frac{1 0 ^{- 14}}{55.55} \approx 1.8 \times 1 0^{- 16}$

Therefore $p K_{a} (H_{2} O) = p K_{b} (H_{2} O) = 15.74$ .

Temperature dependence of $K_{w}$ : ionization of water is endothermic, so

K_{w}

increases with temperature. At

100 ° C

K_{w} \approx 1 0^{- 12}

; at

80 ° C

K_{w} \approx 1 0^{- 13}

. Consequently, the neutral pH shifts: at

80 ° C

pure water is neutral at

p H = 6.5

, not

7

. This is a common JEE Advanced trap.

8. The pH Scale

Sorenson (1909) introduced the logarithmic pH scale to conveniently express $H^{+}$ concentrations that range over many orders of magnitude:

$p H = - lo g_{10} [H^{+}] and pO H = - lo g_{10} [O H^{-}]$

From $K_{w} = [H^{+}] [O H^{-}] = 1 0^{- 14}$ , taking negative logarithms:

$p H + pO H = 14 at 25 ° C$

Figure 1: The pH scale at

25 ° C

ranges from

0

(strongly acidic) to

14

(strongly basic), with pure water neutral at

p H = 7

Interpretation at $25 ° C$

$p H < 7$ (or $pO H > 7$ ): acidic solution
$p H = 7 = pO H$ : neutral solution
$p H > 7$ (or $pO H < 7$ ): basic solution

Limitations of the pH scale

The pH scale can extend below $0$ or above $14$ for very concentrated solutions (e.g., $p H$ of $10 M H C l$ is about $- 1$ ).
The pH range shifts with temperature because $K_{w}$ is temperature-dependent.
Strictly, $p H = - lo g (a_{H^{+}})$ where $a_{H^{+}}$ is the activity (concentration $\times$ activity coefficient). At high concentrations, molarity differs from activity.
A $1 0^{- 8} M H C l$ solution does not have $p H = 8$ : at such low acid concentration, the $H^{+}$ from water contributes significantly.

9. pH of Strong Acids and Strong Bases

Strong acid, concentration $> 1 0^{- 6} M$

Water's contribution to $[H^{+}]$ is negligible; the acid provides essentially all of it.

$[H^{+}] = C_{a c i d} \Rightarrow p H = - lo g C_{a c i d}$

Solved Example 3

Calculate the pH of (i)

0.01 N C a (O H)_{2}

and (ii)

1 0^{- 3} M H C l

Solution:

(i) $C a (O H)_{2} \to C a^{2 +} + 2 O H^{-}$ . Since normality accounts for the two $O H^{-}$ groups, $[O H^{-}] = 0.01 N = 0.01 M$ .

$pO H = - lo g (0.01) = 2 \Rightarrow p H = 14 - 2 = 12$ .

(ii) $[H^{+}] = 1 0^{- 3} M \Rightarrow p H = 3$ .

Strong acid, concentration $< 1 0^{- 6} M$

Now $H^{+}$ from water is comparable to that from acid, and must be included. Suppose we take $c$ M of $H C l$ ( $c < 1 0^{- 6}$ ). Let $x$ be the $H^{+}$ contributed by water at equilibrium (so $[O H^{-}] = x$ also comes from water).

$[H^{+}]_{total} = c + x, [O H^{-}] = x$

Applying $K_{w}$ : $(c + x) \cdot x = 1 0^{- 14}$ .

Solved Example 4

Calculate the pH of

1 0^{- 8} M H C l

Solution:

Let $x$ = $[H^{+}]$ from water. Then $[H^{+}]_{total} = 1 0^{- 8} + x$ and $[O H^{-}] = x$ .

$(1 0^{- 8} + x) x = 1 0^{- 14} \Rightarrow x^{2} + 1 0^{- 8} x - 1 0^{- 14} = 0$

Solving the quadratic: $x = \frac{- 1 0 ^{- 8} + 1 0 ^{- 16} + 4 \times 1 0 ^{- 14}}{2} \approx 0.95 \times 1 0^{- 7}$ .

$[H^{+}]_{total} = 1 0^{- 8} + 0.95 \times 1 0^{- 7} \approx 1.05 \times 1 0^{- 7} M$ .

$p H = - lo g (1.05 \times 1 0^{- 7}) \approx 6.98$

Note that the pH is close to $7$ but slightly less, as expected for an acidic solution. Naively writing $p H = - lo g (1 0^{- 8}) = 8$ would be wrong (an acid cannot give a basic pH).

Strong base, concentration $< 1 0^{- 6} M$

Symmetric to the strong acid case. Calculate $[O H^{-}]$ first including water's contribution, then find $[H^{+}] = K_{w} / [O H^{-}]$ .

Solved Example 5

Calculate the pH of

1 0^{- 7} M N a O H

Solution:

$[O H^{-}]$ from $N a O H = 1 0^{- 7}$ . Let water contribute $x$ each of $H^{+}$ and $O H^{-}$ . Then $[O H^{-}]_{total} = 1 0^{- 7} + x$ , $[H^{+}] = x$ .

$(1 0^{- 7} + x) x = 1 0^{- 14} \Rightarrow x^{2} + 1 0^{- 7} x - 1 0^{- 14} = 0$

$x = \frac{- 1 0 ^{- 7} + 1 0 ^{- 14} + 4 \times 1 0 ^{- 14}}{2} = \frac{1 0 ^{- 7} ( 5 - 1 )}{2} \approx 0.618 \times 1 0^{- 7} M$ .

$p H = - lo g (0.618 \times 1 0^{- 7}) \approx 7.21$ .

10. pH of a Mixture of Strong Acids or Strong Bases

Since the acids are strong and none share ions in a way that suppresses the other, they contribute $H^{+}$ ions independently. Using milliequivalent balance:

$[H^{+}] = \frac{N _{1} V _{1} + N _{2} V _{2}}{V _{1} + V _{2}}$

Solved Example 6

Calculate the pH of a mixture of

400 m L

\frac{1}{200} M H_{2} S O_{4}

400 m L

\frac{1}{100} M H C l

, and

200 m L

of pure water.

Solution:

Milliequivalents of $H^{+}$ from $H_{2} S O_{4} = 400 \times \frac{1}{200} \times 2 = 4$ .

Milliequivalents of $H^{+}$ from $H C l = 400 \times \frac{1}{100} = 4$ .

Total milliequivalents of $H^{+} = 4 + 4 = 8$ . Total volume $= 400 + 400 + 200 = 1000 m L$ .

$[H^{+}] = \frac{8}{1000} = 8 \times 1 0^{- 3} M$ .

$p H = - lo g (8 \times 1 0^{- 3}) = 3 - 3 lo g 2 \approx 3 - 0.903 \approx 2.10$

11. pH of a Mixture of a Strong Acid and a Strong Base

Neutralization occurs to completion. The solution's pH depends on which reagent is in excess:

If milliequivalents of acid $>$ milliequivalents of base: solution is acidic. $[H^{+}] = \frac{N _{1} V _{1} - N _{2} V _{2}}{V _{1} + V _{2}}$ .
If milliequivalents of base $>$ milliequivalents of acid: solution is basic. $[O H^{-}] = \frac{N _{2} V _{2} - N _{1} V _{1}}{V _{1} + V _{2}}$ .
If milliequivalents are equal: neutral solution; $p H = 7$ at $25 ° C$ (for salt of strong acid + strong base).

Solved Example 7

Calculate the pH of a mixture of

400 m L

\frac{1}{200} M B a (O H)_{2}

400 m L

\frac{1}{50} M H C l

, and

200 m L

of water.

Solution:

Milliequivalents of $O H^{-}$ from $B a (O H)_{2} = 400 \times \frac{1}{200} \times 2 = 4$ .

Milliequivalents of $H^{+}$ from $H C l = 400 \times \frac{1}{50} = 8$ .

Net milliequivalents of $H^{+} = 8 - 4 = 4$ . Total volume $= 1000 m L$ .

$[H^{+}] = \frac{4}{1000} = 4 \times 1 0^{- 3} M$ . $p H = - lo g (4 \times 1 0^{- 3}) = 3 - 2 lo g 2 \approx 2.40$ .

12. Common Ion Effect - Quantitative Treatment

Consider a weak acid $H A$ (concentration $C_{2}$ ) with a strong acid $H A^{'}$ (concentration $C_{1}$ ) providing a common $H^{+}$ . Let $α$ be the degree of ionization of the weak acid in the presence of the strong acid.

$H A H^{+} A^{-} initial C_{2} C_{1} 0 eqbm C_{2} (1 - α) C_{1} + C_{2} α C_{2} α$

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

Since a strong acid usually dominates, $[H^{+}]_{total} \approx C_{1}$ , and $α_{weak acid}$ is severely suppressed compared to the value it would have alone.

Solved Example 8

Calculate the pH of a solution containing

1 0^{- 1} M H C l

and

1 0^{- 3} M C H_{3} C O O H

. Also calculate the concentration of

C H_{3} C O O^{-}

. Given

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

Solution:

The strong acid dominates: $[H^{+}] \approx 1 0^{- 1} M \Rightarrow p H \approx 1$ .

For acetic acid: $K_{a} = \frac{[ H ^{+} ] [ C H _{3} C O O ^{-} ]}{[ C H _{3} C O O H ]} \approx \frac{( 1 0 ^{- 1} ) \cdot C _{2} α}{C _{2}}$

$2 \times 1 0^{- 5} = \frac{( 1 0 ^{- 1} ) \cdot α}{1} \Rightarrow α = 2 \times 1 0^{- 4}$ .

$[C H_{3} C O O^{-}] = C_{2} α = 1 0^{- 3} \times 2 \times 1 0^{- 4} = 2 \times 1 0^{- 7} M$ .

Note how the strong acid has cut the weak acid's ionization dramatically compared to $α \approx K_{a} / C = 2 \times 1 0^{- 5} /1 0^{- 3} \approx 0.14$ that acetic acid would show alone.

Common Mistakes to Avoid

Watch out

Applying $p H = - lo g C$ blindly for dilute acids. At concentrations below $1 0^{- 6} M$ , water's own $H^{+}$ contribution matters. Setting up the quadratic $(c + x) x = 1 0^{- 14}$ is necessary.
Claiming $p H = 8$ for $1 0^{- 8} M H C l$ . An acid solution cannot have $p H > 7$ . The correct pH is about $6.98$ .
Forgetting the factor of $2$ for $C a (O H)_{2}$ or $H_{2} S O_{4}$ . Molarity times basicity (or acidity) gives normality; use normality when computing $[H^{+}]$ or $[O H^{-}]$ from strong diprotic species.
Applying Ostwald's law with $α > 0.1$ . The formula $α = K / C$ is only valid when $1 - α \approx 1$ . For larger $α$ , solve the full quadratic.
Assuming $p H + pO H = 14$ at all temperatures. This holds only at $25 ° C$ . At higher temperatures $K_{w}$ is larger, so the sum $p K_{w}$ is less than $14$ .
Confusing $H_{3} B O_{3}$ with an Arrhenius acid. Boric acid does not release its own $H^{+}$ ; instead it pulls $O H^{-}$ from water, and the leftover $H^{+}$ makes the solution acidic. It is technically a Lewis acid.
Calling all Bronsted bases Arrhenius bases. $N H_{3}$ is a Bronsted base (accepts $H^{+}$ ) but not strictly an Arrhenius base (it does not directly release $O H^{-}$ ).

Frequently Asked Questions

What is the difference between strong and weak electrolytes?

Strong electrolytes ionize almost completely in water ( $α \approx 1$ ), so their solutions conduct electricity strongly. Examples include $H C l$ , $N a O H$ , and $N a C l$ . Weak electrolytes ionize only partially ( $α ≪ 1$ ), setting up an equilibrium between ions and unionized molecules; their solutions conduct poorly. Examples include acetic acid and ammonia solution.

Why does the degree of ionization of a weak electrolyte increase on dilution?

Ostwald's dilution law shows $α \propto V$ , or equivalently $α \propto 1/ C$ . Dilution reduces the concentration of ions, so by Le Chatelier's principle the equilibrium shifts toward more ionization to restore ion concentration. At infinite dilution, $α$ approaches unity - even weak electrolytes are essentially fully ionized.

Why do all Bronsted acid-base reactions proceed from stronger to weaker?

A stronger acid has a greater tendency to donate protons; a stronger base has a greater tendency to accept them. When they react, the products (weaker acid and weaker base) hold on to their protons and electrons more firmly, so the equilibrium lies to the product side. This is analogous to spontaneous processes moving from higher to lower energy.

Is water an acid or a base?

Water is amphoteric (amphiprotic): it can act as an acid (donating $H^{+}$ to give $O H^{-}$ ) or as a base (accepting $H^{+}$ to give $H_{3} O^{+}$ ). Whether water behaves as an acid or base in a particular reaction depends on the other substance present. When mixed with $H C l$ , water is a base; when mixed with $N H_{3}$ , water is an acid.

Why is $H_{3} B O_{3}$ not considered an Arrhenius acid?

Boric acid does not release $H^{+}$ directly. Instead, it accepts an $O H^{-}$ from water: $H_{3} B O_{3} + H_{2} O ⇌ [B (O H)_{4}]^{-} + H^{+}$ . It is the water that becomes deprotonated, releasing $H^{+}$ . Because the acidity arises from $H_{3} B O_{3}$ acting as a Lewis acid (accepting an electron pair from $O H^{-}$ ), it is best classified as a Lewis acid.

Why does the pH of pure water differ at different temperatures?

The self-ionization of water is endothermic, so $K_{w}$ increases with temperature. At $25 ° C$ , $K_{w} = 1 0^{- 14}$ and pure water has $p H = 7$ . At $80 ° C$ , $K_{w} \approx 1 0^{- 13}$ , so pure water has $[H^{+}] = 1 0^{- 13} \approx 3.16 \times 1 0^{- 7}$ and neutral $p H \approx 6.5$ . Water remains neutral (equal $H^{+}$ and $O H^{-}$ ) at all temperatures - only the numerical value of neutral pH changes.

What is the levelling effect of water?

Water is basic enough to fully accept a proton from any strong acid, so all strong acids ( $H C l, H B r, H I, H N O_{3}, H C l O_{4}$ ) appear equally strong when dissolved in water - they are all "levelled" to $H_{3} O^{+}$ . Similarly, strong bases are all levelled to $O H^{-}$ . To differentiate intrinsic strengths among strong acids or bases, we must use a solvent less basic than water (e.g., glacial acetic acid).

How is $[H^{+}]$ from a weak acid affected by a common ion?

Adding a salt or strong acid with an ion common to the weak electrolyte shifts its ionization equilibrium backward (Le Chatelier). For example, adding sodium acetate to acetic acid decreases the acid's $α$ and hence its $[H^{+}]$ . This is called the common ion effect and is the basis of buffer solutions, salt hydrolysis calculations, and controlled precipitation in analytical chemistry.

Previous year questions on Ionic Equilibrium In Solution

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

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1. Electrolytes and Non-electrolytes

Strong electrolytes

Weak electrolytes

Non-electrolytes

2. Degree of Ionization (α)

3. Ostwald's Dilution Law

4. Common Ion Effect

5. Theories of Acids and Bases

(a) Arrhenius theory (1887)

(b) Bronsted-Lowry theory (1923) - Proton concept

(c) Lewis theory (1923) - Electronic concept

6. Levelling Effect of Water JEE Advanced

7. Properties of Water and Self-Ionization

Amphoteric nature

Molar concentration of water

Ionic product of water (Kw​)

Degree of dissociation and Ka​ of water

8. The pH Scale

Interpretation at 25°C

Limitations of the pH scale

9. pH of Strong Acids and Strong Bases

Strong acid, concentration >10−6M

Strong acid, concentration <10−6M

Strong base, concentration <10−6M

10. pH of a Mixture of Strong Acids or Strong Bases

11. pH of a Mixture of a Strong Acid and a Strong Base

12. Common Ion Effect - Quantitative Treatment

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between strong and weak electrolytes?

Why does the degree of ionization of a weak electrolyte increase on dilution?

Why do all Bronsted acid-base reactions proceed from stronger to weaker?

Is water an acid or a base?

Why is H3​BO3​ not considered an Arrhenius acid?

Why does the pH of pure water differ at different temperatures?

What is the levelling effect of water?

How is [H+] from a weak acid affected by a common ion?

Previous year questions on Ionic Equilibrium In Solution

Ready to master Equilibrium?

2. Degree of Ionization ( $α$ )

Ionic product of water ( $K_{w}$ )

Degree of dissociation and $K_{a}$ of water

Interpretation at $25 ° C$

Strong acid, concentration $> 1 0^{- 6} M$

Strong acid, concentration $< 1 0^{- 6} M$

Strong base, concentration $< 1 0^{- 6} M$

Why is $H_{3} B O_{3}$ not considered an Arrhenius acid?

How is $[H^{+}]$ from a weak acid affected by a common ion?