A 0.02M sol ution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Exercises · 6.62

Q. A 0.02M sol ution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

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Concept understanding — Buffer Solution pH

What is a Buffer Solution?

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:

\text{pH} = \text{p}K_a + \log_{10} \left( \frac{[A^-]}{[HA]} \right)

Where:

$\text{p}K_a = -\log_{10} K_a$ (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 $\text{pH} = \text{p}K_a$ . 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 \rightleftharpoons H^+ + A^-

The acid dissociation constant is:

K_a = \frac{[H^+][A^-]}{[HA]}

Take negative logs of both sides:

-\log K_a = -\log [H^+] - \log \frac{[A^-]}{[HA]}

Which gives:

\text{p}K_a = \text{pH} - \log \frac{[A^-]}{[HA]}

Rearrange:

\text{pH} = \text{p}K_a + \log \frac{[A^-]}{[HA]}

That's it. The derivation is just algebra on the definition of $K_a$ .

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 ( $\text{p}K_a = 4.76$ ) with 0.1 M sodium acetate. What's the pH?

$\text{pH} = 4.76 + \log \frac{0.1}{0.1} = 4.76 + \log 1 = 4.76$ …

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This is a problem about finding the base dissociation constant ( $K_b$ ) of pyridine from the pH of its conjugate acid's salt solution. The key is to recognize that pyridinium hydrochloride ( $\text{C}_5\text{H}_5\text{NH}^+\text{Cl}^-$ ) is the salt of a weak base (pyridine) and a strong acid (HCl). Its solution is acidic due to hydrolysis of the pyridinium ion. Using the given pH and concentration, we find $K_a$ for the conjugate acid, then $K_b$ for pyridine via $K_w = K_a \cdot K_b$ . The final answer is $K_b = 1.5 \times 10^{-9}$ .

Concept and Intuition

Pyridinium hydrochloride is a salt formed from pyridine (a weak base) and hydrochloric acid (a strong acid). When dissolved in water, it dissociates completely into $\text{C}_5\text{H}_5\text{NH}^+$ (the conjugate acid of pyridine) and $\text{Cl}^-$ (which is neutral). The pyridinium ion then undergoes hydrolysis — it reacts with water to regenerate some pyridine and produce $\text{H}_3\text{O}^+$ , making the solution acidic.

The pH tells us the concentration of $\text{H}^+$ ions at equilibrium. From that, we can work backwards: first find the equilibrium constant for the hydrolysis reaction (which is $K_a$ of the conjugate acid), then use the relationship between $K_a$ and $K_b$ through the ion-product of water.

K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ at } 25^\circ\text{C}

Step-by-step solution

1. Write the hydrolysis reaction and the expression for $K_a$ .

The pyridinium ion ( $\text{PyH}^+$ ) acts as a weak acid:

\text{PyH}^+ + \text{H}_2\text{O} \rightleftharpoons \text{Py} + \text{H}_3\text{O}^+

The acid dissociation constant for this equilibrium is:

K_a = \frac{[\text{Py}][\text{H}_3\text{O}^+]}{[\text{PyH}^+]}

2. Find $[\text{H}^+]$ from the given pH.

\text{pH} = 3.44 \quad \Rightarrow \quad [\text{H}^+] = 10^{-3.44}

Calculate this:

10^{-3.44} = 10^{-4} \times 10^{0.56} = 10^{-4} \times 3.63 \approx 3.63 \times 10^{-4} \text{ M}

Tip

To compute $10^{0.56}$ quickly: $\log 3.63 \approx 0.56$ , so $10^{0.56} \approx 3.63$ . In exams, you can often leave it as $10^{-3.44}$ and simplify later.

3. Set up an ICE table for the hydrolysis.

Initial concentration of $\text{PyH}^+$ is $0.02 \text{ M}$ . Let $x$ be the amount that hydrolyzes.

Species	Initial (M)	Change (M)	Equilibrium (M)
$\text{PyH}^+$	$0.02$	$-x$	$0.02 - x$
$\text{Py}$	$0$	$+x$	$x$
$\text{H}_3\text{O}^+$	$0$ (approx)	$+x$	$x$

From step 2, we know $x = [\text{H}^+] = 3.63 \times 10^{-4} \text{ M}$ .

4. Check whether $x$ is negligible compared to initial concentration.

\frac{x}{0.02} = \frac{3.63 \times 10^{-4}}{0.02} = 0.01815 \approx 1.8\%

Since this is less than 5%, the approximation $0.02 - x \approx 0.02$ is acceptable for most exam purposes. We'll use it.

5. Substitute into the $K_a$ expression. …

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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
$[\mathrm{X}^-]$
›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 constant $K_b$ of X⁻, not the acid dissociation constant $K_a$ of HX. But for a conjugate pair, $K_a \times K_b = K_w = 10^{-14}$ at 25°C. So we can find $K_a$ for HX, then use the Henderson–Hasselbalch equation:
$\text{pH} = \text{p}K_a + \log\frac{[\text{base}]}{[\text{acid}]}$
where “base” is X⁻ and “acid” is HX.

Step-by-step

Find $K_a$ of HX from $K_b$ of X⁻ For the conjugate pair:

$K_a(\text{HX}) \times K_b(\text{X}^-) = K_w = 10^{-14}$
Given $K_b = 10^{-10}$ :
$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$

Find $\text{p}K_a$

$\text{p}K_a = -\log(10^{-4}) = 4$

Apply the Henderson–Hasselbalch equation The ratio is $[\text{X}^-] : [\text{HX}] = 1 : 10$ , so:

$\frac{[\text{X}^-]}{[\text{HX}]} = \frac{1}{10} = 0.1$
Then:

$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{X}^-]}{[\text{HX}]}\right) = 4 + \log(0.1)$ …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ

10^{-2}

›Reveal solution

Aniline is a weak base. We calculate the hydroxide ion concentration using its $K_b$ value, then find pOH, and finally convert to pH. The pH of the solution is $\boxed{8.3}$ .

Aniline ( $\text{C}_6\text{H}_5\text{NH}_2$ ) 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, $\text{C}_6\text{H}_5\text{NH}_3^+$ ) and hydroxide ions ( $\text{OH}^-$ ). 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, $[\text{OH}^-]$ . This is done using the base ionization constant, $K_b$ , which describes the extent of ionization. Once $[\text{OH}^-]$ is known, we can calculate the pOH of the solution using the formula $\text{pOH} = -\log[\text{OH}^-]$ . Finally, we use the relationship $\text{pH} + \text{pOH} = 14$ (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:

\text{C}_6\text{H}_5\text{NH}_2 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_3^+ (aq) + \text{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)
$\text{C}_6\text{H}_5\text{NH}_2$	$10^{-2}$	$-x$	$10^{-2} - x$
$\text{C}_6\text{H}_5\text{NH}_3^+$	$0$	$+x$	$x$
$\text{OH}^-$	$0$	$+x$	$x$

Write the expression for the base ionization constant ( $K_b$ ) and substitute equilibrium concentrations.
$\text{B}$

K_b = \frac{[\text{C}_6\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_6\text{H}_5\text{NH}_2]}

Substituting the equilibrium concentrations from the ICE table:

K_b = \frac{(x)(x)}{10^{-2} - x} = \frac{x^2}{10^{-2} - x}

We are given $K_b (\text{aniline}) = 4 \times 10^{-10}$.

4 \times 10^{-10} = \frac{x^2}{10^{-2} - x}

Solve for $x$ , which represents $[\text{OH}^-]$ at equilibrium. Since $K_b$ is very small ( $4 \times 10^{-10}$ ) 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: $10^{-2} - x \approx 10^{-2}$ . To verify this approximation, we check the ratio $C/K_b$ : $C/K_b = 10^{-2} / (4 \times 10^{-10}) = 0.25 \times 10^8 = 2.5 \times 10^7$ . Since $2.5 \times 10^7 \gg 500$ , the approximation is valid. …

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TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$pK_a$
›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:

$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$
Rearranging:
$\frac{[\text{A}^-]}{[\text{HA}]} = 10^{\text{pH} - \text{p}K_a}$
Here, $[\text{HA}]$ is the unionized form, $[\text{A}^-]$ is the ionized form.

Apply to the stomach (pH ≈ 2–3, pKa = 3.5): The difference $\text{pH} - \text{p}K_a$ is negative (e.g., 2 – 3.5 = –1.5).

$\frac{[\text{A}^-]}{[\text{HA}]} = 10^{-1.5} \approx 0.032$
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 $\text{pH} - \text{p}K_a = 7.4 - 3.5 = 3.9$ , a large positive number.

$\frac{[\text{A}^-]}{[\text{HA}]} = 10^{3.9} \approx 7943$
The ionized form is nearly 8000 times more abundant than the unionized form — the drug is almost completely ionized in the small intestine.

Interpret the biological consequence: …
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Q.A 0.02M sol ution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Concept and Intuition

Step-by-step solution

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