Q.Determine the degree of ionization and pH of a 0.05M of ammonia solution. The ionization constant of ammonia can be taken from Table 6.7. Also, calculate the ionization constant of the conjugate acid of ammonia.
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Imagine you're making lemonade. If you add a few drops of lemon juice to a glass of water, the pH drops sharply — it becomes very acidic. But if you add the same few drops to a glass of already acidic lemonade, the pH barely changes. Why? Because lemonade contains a buffer — a mixture that resists pH change when small amounts of acid or base are added.
A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid). It "soaks up" added H⁺ or OH⁻ ions without letting the pH swing wildly.
Note
The key is that both components must be present in significant amounts. A weak acid alone won't buffer — you need its conjugate base partner too.
The Intuition: A Chemical Sponge
Think of a buffer as a two-way sponge:
If you add acid (H⁺): The conjugate base in the buffer grabs the extra H⁺, turning into the weak acid. The H⁺ is "absorbed" — pH barely drops.
If you add base (OH⁻): The weak acid donates an H⁺ to neutralise the OH⁻, turning into the conjugate base. The OH⁻ is "absorbed" — pH barely rises.
The buffer works best when the amounts of weak acid and conjugate base are roughly equal. That's when the sponge is most "spongy" — it can absorb shocks in either direction.
The Precise Statement: The Henderson–Hasselbalch Equation
For a buffer made from a weak acid HA and its conjugate base A−, the pH is given by:
pH=pKa+log10([HA][A−])
Where:
pKa=−log10Ka (a measure of the weak acid's strength — lower pKa = stronger acid)
[A−] = concentration of the conjugate base
[HA] = concentration of the weak acid
This equation tells you exactly how the pH depends on the ratio of base to acid, not their absolute amounts.
Tip
When [A−]=[HA], the ratio is 1, log(1)=0, so pH=pKa. This is the buffer's optimal pH — it resists change most strongly here.
Why This Works: A Quick Derivation
Start from the weak acid equilibrium:
HA⇌H++A−
The acid dissociation constant is:
Ka=[HA][H+][A−]
Take negative logs of both sides:
−logKa=−log[H+]−log[HA][A−]
Which gives:
pKa=pH−log[HA][A−]
Rearrange:
pH=pKa+log[HA][A−]
That's it. The derivation is just algebra on the definition of Ka.
Watch out
The Henderson–Hasselbalch equation assumes that the concentrations [HA] and [A−] are the initial concentrations you mixed. It works well when both are much larger than [H+] or [OH−] from dissociation — which is true for a properly made buffer.
Example: Making an Acetate Buffer
You mix 0.1 M acetic acid (pKa=4.76) with 0.1 M sodium acetate. What's the pH?
Concept: Buffer Solution pH — but here it's a weak base (ammonia) in water, so we use the base dissociation constant Kb and the relation [OH−]=Kb⋅C for a weak base.
Step 1 — Find Kb and Ka of conjugate acid
From Table 6.7, Kb for NH3 = 1.77×10−5.
For the conjugate acid NH4+,
Ka=KbKw=1.77×10−51.0×10−14=5.65×10−10.
Step 2 — Degree of ionization (α)
For a weak base, α=CKb=0.051.77×10−5=3.54×10−4=0.0188 (or 1.88%).
For a weak base like ammonia, the degree of ionization (α) is found from Kb=Cα2/(1−α), and pH follows from [OH−]=Cα. Using Kb=1.77×10−5 for 0.05 M NH₃, we get α≈0.0188, pH ≈10.95, and Ka for NH₄⁺ is 5.65×10−10.
Why This Approach Works
Ammonia in water is a classic weak base — it doesn't fully ionize. Instead, it establishes an equilibrium:
NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)
The ionization constant Kb tells us how far this reaction goes. From Table 6.7 (NCERT), Kb for ammonia is 1.77×10−5 at 25°C.
The degree of ionization α is the fraction of ammonia molecules that have accepted a proton. For a weak base, α is small, so we can often simplify calculations — but we'll check that assumption.
The conjugate acid of ammonia is the ammonium ion, NH₄⁺. For any conjugate acid-base pair, Ka×Kb=Kw, where Kw=1.0×10−14 at 25°C. This lets us find Ka for NH₄⁺ directly.
Step-by-Step Solution
1. Set up the equilibrium table
Let initial concentration of NH₃ be C=0.05 M. If α is the degree of ionization:
Species
Initial (M)
Change (M)
Equilibrium (M)
NH₃
C
−Cα
C(1−α)
NH₄⁺
0
+Cα
Cα
OH⁻
0
+Cα
Cα
2. Write the Kb expression
Kb=[NH3][NH4+][OH−]=C(1−α)(Cα)(Cα)=1−αCα2
Substitute known values:
1.77×10−5=1−α0.05⋅α2
3. Solve for α
This is a quadratic in α. Multiply through:
1.77×10−5(1−α)=0.05α2
1.77×10−5−1.77×10−5α=0.05α2
Rearrange:
0.05α2+1.77×10−5α−1.77×10−5=0
Using the quadratic formula α=2a−b±b2−4ac with a=0.05, b=1.77×10−5, c=−1.77×10−5:
The negative root gives a negative α (impossible), so take the positive root:
α=0.1−1.77×10−5+3.13×10−10+3.54×10−6
α=0.1−1.77×10−5+3.5403×10−6
α=0.1−1.77×10−5+1.8816×10−3
α=0.11.8639×10−3=0.01864
Tip
Since α≈0.019 is much less than 0.05, we could have used the approximation 1−α≈1, giving α=Kb/C=1.77×10−5/0.05=3.54×10−4=0.0188. The exact value (0.01864) is very close — the approximation works well here.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2025Set eng-2025-04284 marksMCQ
Q.4 g of NaOH were dissolved in 1 litre of a solution containing 1 mole of CH3COOH and 1 mole of CH3COONa. The [H+] in the resultant solution is
(Given: Ka (CH3COOH)=1.1×10−5)
(A) 1.47×10−5 M
(B) 2×10−5 M
(C) 2.5×10−5 M
(D) 1.5×10−5 M
(E) 0.9×10−5 M
›Reveal solutionSolution
Buffer: [H+]=Ka[salt][acid].
Moles of NaOH =4/40=0.1 mol. NaOH neutralises acetic acid: