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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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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