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.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2026Set D31 markMCQ
Q.A 0.15 mole of pyridinium chloride has been added to 500 cm3 of 0.2M pyridine solution (a base). Assuming there is no change in volume upon mixing, the pH of the resulting solution is (Note: Kb for pyridine is 1.5×10−9)
(A) 5
(B) 6
(C) 7
(D) 8
›Reveal solutionSolution
This is a buffer of pyridine (weak base) and pyridinium chloride (its conjugate acid salt); the Henderson–Hasselbalch equation for bases gives the pOH, from which pH follows.
Step 1 — Moles of each species
Moles of pyridine (base) =0.500 L×0.2 mol/L=0.10 mol. Moles of pyridinium chloride (conjugate acid, the salt) added =0.15 mol. Since the volume is unchanged, this is a base/conjugate-acid buffer.
Q.What volume of 0.2MCH3COOH needs to be added to 100 ml of 0.4MCH3COONa solution to prepare a buffer of pH equal to 4.91 ? ( pKa of CH3COOH is 4.76 )
(A) 282.6 ml
(B) 213.65 ml
(C) 101.41 ml
(D) 141.54 ml
›Reveal solutionSolution
Use the Henderson–Hasselbalch equation to find the required ratio of conjugate base to acid, then solve for the volume of acetic acid solution needed. The answer is 141.54 mL, option (D).
We are making a buffer by mixing acetic acid (CH₃COOH) with its conjugate base (CH₃COO⁻ from sodium acetate). The pH of a buffer is given by the Henderson–Hasselbalch equation:
pH=pKa+log[CH3COOH][CH3COO−]
Here, the concentrations are those after mixing, but since both species are in the same final volume, the ratio of concentrations equals the ratio of moles. So we can work directly with moles.
Step-by-step solution:
Identify given data
Volume of CH₃COONa solution: Vbase=100ml=0.100L
Concentration of CH₃COONa: 0.4M → moles of CH₃COO⁻ initially:
nbase=0.100×0.4=0.040mol
Concentration of CH₃COOH: 0.2M
Target pH = 4.91, pKₐ = 4.76
Apply Henderson–Hasselbalch
4.91=4.76+log[CH3COOH][CH3COO−]
log[CH3COOH][CH3COO−]=4.91−4.76=0.15
[CH3COOH][CH3COO−]=100.15≈1.4125
Interpret ratio in terms of moles
Since the final volume is the same for both, the concentration ratio equals the mole ratio:
nacidnbase=1.4125
The moles of base remain 0.040mol (we add no extra base, only acid). So:
Q.What volume of 0.2M Acetic acid is to be added to 100ml of 0.4M Sodium acetate so that a Buffer solution of pH equal to 4.94 is obtained? (pKa of CH3COOH=4.76)
(A) 132.1 ml
(B) 125.3 ml
(C) 150.2 ml
(D) 110.6 ml
›Reveal solutionSolution
Use the Henderson–Hasselbalch equation to relate pH, pKa, and the ratio of conjugate base to acid. The required volume of 0.2 M acetic acid is found to be 132.1 mL, which corresponds to option (A).
We are mixing a weak acid (acetic acid, CH₃COOH) with its conjugate base (sodium acetate, CH₃COONa) to form a buffer. The pH of a buffer is given by the Henderson–Hasselbalch equation:
pH=pKa+log[acetic acid][acetate]
Here, we know the target pH (4.94), the pKa (4.76), and the concentration and volume of the acetate solution. We need to find the volume of acetic acid solution to add.
Why this works: The equation directly links the pH to the ratio of concentrations of the two species. Since both are in the same final solution, the ratio of concentrations equals the ratio of moles (because volume cancels). So we can work in moles from the start.
Find the required ratio of [acetate] to [acetic acid].
Using the Henderson–Hasselbalch equation:
4.94=4.76+log[acetic acid][acetate]
log[acetic acid][acetate]=4.94−4.76=0.18
[acetic acid][acetate]=100.18≈1.5136
This means in the final buffer, the moles of acetate must be about 1.5136 times the moles of acetic acid.
Calculate the moles of acetate we start with.
Sodium acetate solution: 100 mL of 0.4 M.
moles of acetate=0.100L×0.4mol/L=0.04mol
Let the volume of 0.2 M acetic acid added be V mL (so V/1000 L).
Moles of acetic acid added: