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.
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.