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.
TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
Q.A certain buffer solution contains [X−] and [HX] in the ratio of 1:10.
If Kb of X− is 10−10, then the pH value of the buffer is
(A) 9
(B) 5
(C) 3
(D) 11
›Reveal solutionSolution
This is a buffer of a weak acid HX and its conjugate base X⁻. Given the ratio [X⁻]:[HX] = 1:10 and Kb of X⁻ = 10⁻¹⁰, we find Ka of HX = 10⁻⁴, then use the Henderson–Hasselbalch equation to get pH = 3. The correct option is (C).
Concept & Intuition
A buffer resists pH change when it contains a weak acid and its conjugate base. Here we have HX (weak acid) and X⁻ (its conjugate base). The key is that we are given the base dissociation constantKb of X⁻, not the acid dissociation constant Ka of HX. But for a conjugate pair, Ka×Kb=Kw=10−14 at 25°C. So we can find Ka for HX, then use the Henderson–Hasselbalch equation:
pH=pKa+log[acid][base]
where “base” is X⁻ and “acid” is HX.
Step-by-step
Find Ka of HX from Kb of X⁻
For the conjugate pair:
Ka(HX)×Kb(X−)=Kw=10−14
Given Kb=10−10:
Ka=10−1010−14=10−4
Find pKa
pKa=−log(10−4)=4
Apply the Henderson–Hasselbalch equation
The ratio is [X−]:[HX]=1:10, so:
Q.What is the pH of 10−2 M aqueous solution of aniline at 298 K? (Kb (aniline) =4×10−10; log5=0.7, log2=0.3; log4=0.6)
(A) 7.3
(B) 7.4
(C) 9.4
(D) 8.3
›Reveal solutionSolution
Aniline is a weak base. We calculate the hydroxide ion concentration using its Kb value, then find pOH, and finally convert to pH. The pH of the solution is 8.3.
Aniline (C6H5NH2) is an organic compound that acts as a weak base. When dissolved in water, it undergoes partial ionization, accepting a proton from water to form its conjugate acid (anilinium ion, C6H5NH3+) and hydroxide ions (OH−). The presence of these hydroxide ions makes the solution basic.
To find the pH of a weak base solution, we first need to determine the equilibrium concentration of hydroxide ions, [OH−]. This is done using the base ionization constant, Kb, which describes the extent of ionization. Once [OH−] is known, we can calculate the pOH of the solution using the formula pOH=−log[OH−]. Finally, we use the relationship pH+pOH=14 (at 298 K) to find the pH.
Here's how to solve the problem step-by-step:
Write the ionization equilibrium for aniline in water.
Aniline accepts a proton from water, forming its conjugate acid and hydroxide ions:
C6H5NH2(aq)+H2O(l)⇌C6H5NH3+(aq)+OH−(aq)
Set up an ICE (Initial, Change, Equilibrium) table for concentrations.
Let C be the initial concentration of aniline, which is 10−2 M. Let x be the concentration of aniline that ionizes at equilibrium.
Species
Initial (M)
Change (M)
Equilibrium (M)
C6H5NH2
10−2
−x
10−2−x
C6H5NH3+
0
+x
x
OH−
0
+x
x
Write the expression for the base ionization constant (Kb) and substitute equilibrium concentrations.
For a weak base B ionizing as B+H2O⇌BH++OH−, the base ionization constant is given by:
Kb=[B][BH+][OH−]
For aniline:
Kb=[C6H5NH2][C6H5NH3+][OH−]
Substituting the equilibrium concentrations from the ICE table:
Kb=10−2−x(x)(x)=10−2−xx2
We are given $K_b (\text{aniline}) = 4 \times 10^{-10}$.
4×10−10=10−2−xx2
Solve for x, which represents [OH−] at equilibrium.
Since Kb is very small (4×10−10) compared to the initial concentration of aniline (10−2 M), we can assume that x is much smaller than 10−2. This allows us to simplify the denominator: 10−2−x≈10−2.
To verify this approximation, we check the ratio C/Kb:
C/Kb=10−2/(4×10−10)=0.25×108=2.5×107.
Since 2.5×107≫500, the approximation is valid. …
Q.Acetylsalicylic acid has pKa value 3.5. The pH of gastric juice in human stomach is 2–3 and the pH in the small intestine is approximately 7.4. Then acetylsalicylic acid will be
(A) Unionized in the stomach and ionized in the small intestine
(B) Unionized in the small intestine and in the stomach
(C) Completely get ionized in both small intestine and stomach
(D) Ionized in the stomach and almost unionized in the small intestine
›Reveal solutionSolution
The key idea is the Henderson–Hasselbalch principle: a weak acid is mostly unionized when pH < pKa and mostly ionized when pH > pKa. Since aspirin’s pKa = 3.5, in the stomach (pH ≈ 2–3) it is unionized, and in the small intestine (pH ≈ 7.4) it is ionized. The correct option is (A).
Concept & Intuition
Acetylsalicylic acid (aspirin) is a weak acid. Its behavior in different pH environments is governed by the Henderson–Hasselbalch equation, which relates the ratio of ionized (conjugate base) to unionized (acid) forms to the pH and pKa.
When pH < pKa, the environment is more acidic than the acid’s own strength, so the acid stays mostly in its neutral (unionized) form.
When pH > pKa, the environment is more basic, so the acid loses a proton and becomes ionized (charged).
The stomach is highly acidic (pH 2–3), well below pKa 3.5, so aspirin remains unionized. The small intestine is nearly neutral (pH 7.4), far above pKa, so aspirin becomes ionized. Unionized molecules can cross cell membranes easily; ionized ones cannot — this is why aspirin is absorbed in the stomach but not in the small intestine.
Step-by-step reasoning
Recall the Henderson–Hasselbalch equation for a weak acid HA:
pH=pKa+log([HA][A−])
Rearranging:
[HA][A−]=10pH−pKa
Here, [HA] is the unionized form, [A−] is the ionized form.
Apply to the stomach (pH ≈ 2–3, pKa = 3.5):
The difference pH−pKa is negative (e.g., 2 – 3.5 = –1.5).
[HA][A−]=10−1.5≈0.032
This means the ionized form is only about 3% of the unionized form — the drug is overwhelmingly unionized in the stomach.
Apply to the small intestine (pH ≈ 7.4, pKa = 3.5):
Now pH−pKa=7.4−3.5=3.9, a large positive number.
[HA][A−]=103.9≈7943
The ionized form is nearly 8000 times more abundant than the unionized form — the drug is almost completely ionized in the small intestine.