Ostwald's dilution law (weak acid) - α = √(Ka - C)

What is Ostwald's Dilution Law?

WHY does this matter? Weak acids (acetic acid, HF, carbonic acid) don't fully ionize. Their behavior depends on both Ka (intrinsic strength) and C (concentration). Ostwald's law lets us predict pH, buffer capacity, and titration curves.

Derivation from First Principles

Consider a weak monoprotic acid HA: $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

Step 1: Set up the equilibrium table

	HA	H⁺	A⁻
Initial	C	0	0
Change	-Cα	+Cα	+Cα
Equilibrium	C(1-α)	Cα	Cα

WHY Cα? If C moles/L start, and fraction α dissociates, then Cα moles/L dissociate.

Step 2: Write the Ka expression

Step 3: Apply the weak acid approximation

For weak acids, α << 1, so (1 - α) ≈ 1. This simplification is valid when α < 0.05 (5% ionization).

WHY does this work? If α = 0.01, then 1 - 0.01 = 0.99 ≈ 1 (error < 1%). We're essentially saying "so little dissociates that the denominator stays ~C".

$K_a \approx C\alpha^2$

Step 4: Solve for α

Figure — Ostwald's dilution law (weak acid) - α = √(Ka - C)

Step 5: Derive [H⁺] and pH

Since [H⁺] = Cα: $[\text{[H}^+\text{]} = C\alpha = C \cdot \sqrt{\frac{K_a}{C}} = \sqrt{K_a \cdot C}$

$\text{pH} = -\log[\text{H}^+] = -\frac{1}{2}\log(K_a \cdot C) = \frac{1}{2}(\text{pK}_a - \log C)$

WHY the ½ factor? The square root in α becomes a ½ when you take logs.

Worked Examples

Common Mistakes & How to Fix Them

Active Recall Flashcards

#flashcards/chemistry

What is the degree of ionization α?

The fraction of weak electrolyte molecules that dissociate into ions at equilibrium: α = moles dissociated / total moles initially.

State Ostwald's dilution law for a weak acid

α = √(Ka/C), where α is degree of ionization, Ka is acid dissociation constant, C is initial concentration.

How does α change with dilution (qualitative)?

α increases with dilution because α ∝ 1/√C. Lower concentration shifts equilibrium toward ions.

How does [H⁺] change with dilution for a weak acid?

[H⁺] = √(KaC) decreases with dilution because it's proportional to √C, not C.

When is the approximation (1-α) ≈ 1 valid?

When α < 0.05 (5% ionization). For higher α, must use the full quadratic Ka = Cα²/(1-α).

Derive [H⁺] for a weak acid from Ostwald's law

[H⁺] = Cα = C√(Ka/C) = √(KaC). pH = ½(pKa - log C).

Why does Ka remain constant while α changes with dilution?

Ka is a thermodynamic constant (depends only on temperature). α adjusts to maintain Ka = Cα²/(1-α) as C changes.

For 0.1 M acetic acid (Ka=1.8×10⁻⁵), estimate α

α = √(1.8×10⁻⁵/0.1) = √(1.8×10⁻⁴) ≈ 0.013 or 1.3%.

Deep Understanding

Recall Feynman Explanation (to a 12-year-old)

Imagine you have a jar of 100 "acid robots." These robots are shy—only a few dare to split apart into "H⁺ bots" and "A⁻ bots." In a crowded jar (high concentration), only 1 or 2 split because there's no room.

Now pour the robots into a bigger jar with more water (dilution). Suddenly there's space! Maybe 4 robots feel brave enough to split. The percentage that split (α) went up. But here's the trick: you still have fewer total split robots in the whole jar, because you started with the same 100 but spread them out more.

Ostwald's law is the math that predicts exactly how many split: it's like √(bravery ÷ crowding). More crowding (C) → fewer splits. More bravery (Ka) → more splits. The √ comes from the fact that when one robot splits, it makes two pieces, so the math squares.

Connections:

Le Chatelier's Principle: dilution shifts equilibrium toward ions (more particles)
Henderson-Hasselbalch Equation: at half-neutralization, pH = pKa (when α → 0.5 region, buffer action starts)
Common Ion Effect: adding A⁻ suppresses α by shifting equilibrium left
Polyprotic Acids: successive Ka values decrease, so Ostwald's law applies to each step
Conductivity of Solutions: conductance ∝ Cα, so weak acids have non-linear conductance vs C
Buffer Capacity: depends on α; optimal at α ≈ 0.5 (equal amounts HA and A⁻)
pH Calculation Methods: Ostwald provides the approximate [H⁺] = √(KaC) shortcut

Summary: Ostwald's dilution law reveals that weak acids are a balancing act between Ka (intrinsic strength) and C (concentration). The √(Ka/C) relationship emerges from the quadratic equilibrium expression when ionization is small. Dilution increases % ionization but decreases total [H⁺]—a subtle distinction that trips up many students. Master this, and you understand buffer behavior, titration curves, and the entire weak electrolyte landscape.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Ostwald ka yeh law weak acids ke liye bahut zaroori hai. Socho agar tumhare pas ek weak acid hai jaise acetic acid (sirka), toh woh puri tarah se ions mein nahi totta—sirf thoda sa totta hai. Degree of ionization (α) yeh bata hai kitna fraction tota. Jab tum acid ko dilute karte ho (zyada pani milate ho), toh ek ajeb chez hoti hai: percentage ionization badhta hai, lekin total hydrogen ions ki sankhya kam ho jati hai!

Formula hai α = √(Ka/C). Iska matlab, agar concentration C kam karoge (dilution), toh α badhega kyunki square root ke neeche C hai. Par [H⁺] = Cα = √(Ka×C) hota hai, jo √C ke saath badhta hai, na ki C ke saath. Isliye dilution karne se solution kam acidic (higher pH) hota hai, chahe α badh jaye. Yeh counter-intuitive lagta hai par thermodynamics ka basic rule hai—equilibrium constant Ka fixed rehta hai, toh α aur C ko balance karna padta hai.

Iska use pH calculations mein, buffer design mein, aur titration curves samajhne mein hota hai. Agar tumhe kisi weak acid ka pH nikalna hai, direct formula use karo: pH = ½(pKa - log C). Par dhyan rakho, yeh approximation tab tak valid hai jab α < 5% ho. Agar zyada ionization ho raha hai, toh quadratic equation solve karna padega. Yeh law chemistry mein weak electrolytes ka backbone hai—samajh gaye toh equilibrium ka adha chapter clear!