The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴ and 4.8 × 10⁻⁹ respectively. Calculate the ionization constants of the corresponding conjugate base.

> [!TLDR] > The conjugate base of an acid is weaker when the acid is stronger. Using $K_w = K_a \times K_b$, we find $K_b(\text{F}^-) = 1.47 \times 10^{-11}$, $K_b(\text{HCOO}^-) = 5.56 \times 10^{-11}$, and $K_b(\text{CN}^-) = 2.08 \times 10^{-6}$. The key to this problem is the relationship between an acid and its conjugate base. In the Brønsted-Lowry theory, when an acid donates a proton ($\text{H}^+$), what remains is its conjugate base. For example: $$\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-$$ Here, $\text{F}^-$ is the conjugate base of HF. The strength of a conjugate base is inversely related to the strength of the acid: a strong acid has a very weak conjugate base, and a weak acid has a comparatively stronger conjugate base. The quantitative link is given by the ion-product constant of water, $K_w$, at 298 K: > [!FORMULA] > $$K_w = K_a \times K_b = 1.0 \times 10^{-14}$$ This holds for any conjugate acid-base pair in water. So to find $K_b$ of the conjugate base, we simply rearrange: $$K_b = \frac{K_w}{K_a}$$ Let’s apply this to each acid. 1. **For HF ($K_a = 6.8 \times 10^{-4}$)** The conjugate base is $\text{F}^-$ (fluoride ion). $$K_b(\text{F}^-) = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}$$ Notice how small this is — HF is a moderately weak acid, so its conjugate base is very weak indeed. 2. **For HCOOH ($K_a = 1.8 \times 10^{-4}$)** The conjugate base is $\text{HCOO}^-$ (formate ion). $$K_b(\text{HCOO}^-) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = 5.56 \times 10^{-11}$$ Formic acid is slightly weaker than HF, so its conjugate base is slightly stronger — but still very weak. 3. **For HCN ($K_a = 4.8 \times 10^{-9}$)** The conjugate base is $\text{CN}^-$ (cyanide ion). …

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴ and 4.8 × 10⁻⁹ respectively. C

Exercises · 6.43

Q. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 \times 10⁻⁴, 1.8 \times 10⁻⁴ and 4.8 \times 10⁻⁹ respectively. Calculate the ionization constants of the corresponding conjugate base.

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Concept understanding — Brønsted Lowry Conjugate Pairs

From Intuition to Precision: Brønsted–Lowry Conjugate Pairs

Imagine you're at a party where people are passing around a single, very special coin. The coin represents a proton ( $\text{H}^+$ ). The game is simple: someone can give the coin to someone else, but only if the other person is willing to take it. You can't just throw the coin; you need a willing receiver.

In chemistry, acids and bases play exactly this game. An acid is the person who gives away the proton. A base is the person who accepts it. But here's the twist: the moment the acid gives away its proton, it transforms into something new — something that can now accept a proton back. That transformed form is called the conjugate base. Similarly, the base, after accepting the proton, becomes something that can donate it back — the conjugate acid.

This is the core of the Brønsted–Lowry theory: every acid-base reaction involves a pair of substances that are linked by the gain or loss of a single proton.

The Precise Statement

\text{Acid} \rightleftharpoons \text{Conjugate Base} + \text{H}^+

A conjugate acid-base pair consists of two species that differ by exactly one proton ( $\text{H}^+$ ). The acid has the proton; the conjugate base does not.

When an acid donates a proton, it becomes its conjugate base.

When a base accepts a proton, it becomes its conjugate acid.

Important

In any Brønsted–Lowry reaction, there are two conjugate pairs: one on the reactant side and one on the product side. They always appear together.

Seeing It in Action

Take the classic reaction between hydrochloric acid and water:

\text{HCl} + \text{H}_2\text{O} \rightarrow \text{Cl}^- + \text{H}_3\text{O}^+

Let's identify the pairs:

Pair 1: $\text{HCl}$ (acid) and $\text{Cl}^-$ (conjugate base). They differ by one $\text{H}^+$ .
Pair 2: $\text{H}_2\text{O}$ (base) and $\text{H}_3\text{O}^+$ (conjugate acid). They also differ by one $\text{H}^+$ .

Notice: water acted as a base here — it accepted the proton from HCl. But water can also act as an acid in other reactions. That's the beauty of the Brønsted–Lowry theory: a substance's role depends on the reaction, not on a fixed label.

Tip

To find the conjugate base of any acid, simply remove one $\text{H}^+$ and reduce the charge by +1.

To find the conjugate acid of any base, add one $\text{H}^+$ and increase the charge by +1.

A Quick Reference Table

Acid	Conjugate Base	Base	Conjugate Acid
$\text{HCl}$	$\text{Cl}^-$	$\text{NH}_3$	$\text{NH}_4^+$
$\text{H}_2\text{SO}_4$	$\text{HSO}_4^-$	$\text{H}_2\text{O}$	$\text{H}_3\text{O}^+$
$\text{NH}_4^+$	$\text{NH}_3$	$\text{OH}^-$	$\text{H}_2\text{O}$
$\text{H}_2\text{O}$	$\text{OH}^-$	$\text{CO}_3^{2-}$	$\text{HCO}_3^-$

Notice how water appears in both columns — it's amphoteric, meaning it can act as either an acid or a base depending on its partner.

Why This Matters …

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The conjugate base of an acid is weaker when the acid is stronger. Using $K_w = K_a \times K_b$ , we find $K_b(\text{F}^-) = 1.47 \times 10^{-11}$ , $K_b(\text{HCOO}^-) = 5.56 \times 10^{-11}$ , and $K_b(\text{CN}^-) = 2.08 \times 10^{-6}$ .

The key to this problem is the relationship between an acid and its conjugate base. In the Brønsted-Lowry theory, when an acid donates a proton ( $\text{H}^+$ ), what remains is its conjugate base. For example:

\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-

Here, $\text{F}^-$ is the conjugate base of HF. The strength of a conjugate base is inversely related to the strength of the acid: a strong acid has a very weak conjugate base, and a weak acid has a comparatively stronger conjugate base.

The quantitative link is given by the ion-product constant of water, $K_w$ , at 298 K:

K_w = K_a \times K_b = 1.0 \times 10^{-14}

This holds for any conjugate acid-base pair in water. So to find $K_b$ of the conjugate base, we simply rearrange:

K_b = \frac{K_w}{K_a}

Let’s apply this to each acid.

For HF ( $K_a = 6.8 \times 10^{-4}$ ) The conjugate base is $\text{F}^-$ (fluoride ion).

K_b(\text{F}^-) = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}

Notice how small this is — HF is a moderately weak acid, so its conjugate base is very weak indeed.

For HCOOH ( $K_a = 1.8 \times 10^{-4}$ ) The conjugate base is $\text{HCOO}^-$ (formate ion).

K_b(\text{HCOO}^-) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} = 5.56 \times 10^{-11}

Formic acid is slightly weaker than HF, so its conjugate base is slightly stronger — but still very weak.

For HCN ( $K_a = 4.8 \times 10^{-9}$ ) The conjugate base is $\text{CN}^-$ (cyanide ion). …

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Q.The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴ and 4.8 × 10⁻⁹ respectively. Calculate the ionization constants of the corresponding conjugate base.

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