What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Cal

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Q. What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

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Concept understanding — Weak Acid Ionization

Weak Acid Ionization: From Intuition to Precision

Imagine you drop a spoonful of sugar into a glass of water. Some sugar dissolves, but a lot just sits at the bottom. Now imagine you drop a spoonful of salt — it all dissolves completely. Acids behave the same way. Some acids, like hydrochloric acid (HCl), dissolve completely in water — every single molecule breaks apart. Others, like acetic acid (vinegar), only partially break apart. Most of the acid molecules stay intact, and only a few actually ionize.

That's the core idea: weak acids are shy about giving away their hydrogen ion. They don't fully commit.

The Precise Statement

A weak acid ( $HA$ ) in water establishes an equilibrium between the intact acid molecule and its ions:

HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}

The double arrow ( $\rightleftharpoons$ ) is the key. It tells you the reaction happens in both directions simultaneously. Some $HA$ molecules break apart to form $H_3O^+$ and $A^-$ , while some $H_3O^+$ and $A^-$ recombine back into $HA$ . At equilibrium, both processes happen at the same rate — so the concentrations stop changing.

Important

For a weak acid, most of the acid remains as $HA$ at equilibrium. Only a tiny fraction exists as ions. This is the opposite of a strong acid, where the forward reaction goes to completion (single arrow: $\rightarrow$ ).

The Quantitative Measure: $K_a$

Every weak acid has a number that tells you exactly how "shy" it is — the acid dissociation constant, $K_a$ :

K_a = \frac{[H_3O^+][A^-]}{[HA]}

K_a = \frac{[H_3O^+][A^-]}{[HA]}

The smaller the $K_a$ , the weaker the acid. For acetic acid (vinegar), $K_a \approx 1.8 \times 10^{-5}$ . That tiny number means the numerator (ions) is very small compared to the denominator (intact acid). For a strong acid like HCl, $K_a$ is effectively infinite — the denominator is essentially zero because all the acid has ionized.

A Concrete Example

Suppose you dissolve 0.10 mol of acetic acid ( $CH_3COOH$ ) in 1 L of water. At equilibrium, you'll find:

$[CH_3COOH] \approx 0.0998$ M (almost all of it is still intact)
$[H_3O^+] \approx 0.0013$ M (only about 1.3% has ionized)
$[CH_3COO^-] \approx 0.0013$ M

Note

|---------|-------------|------------|-----------------|

| $CH_3COOH$ | 0.10 | $-x$ | $0.10 - x$ |

| $H_3O^+$ | 0 | $+x$ | $x$ |

| $CH_3COO^-$ | 0 | $+x$ | $x$ |

Plugging into $K_a = \frac{x^2}{0.10 - x} = 1.8 \times 10^{-5}$ and solving gives $x \approx 0.0013$ M.

Why This Matters …

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Aniline is a weak base, so we use the weak base ionization equilibrium. For 0.001 M aniline with $K_b = 4.27 \times 10^{-10}$ , the degree of ionization $\alpha \approx 6.5 \times 10^{-4}$ , the pH is about 7.82, and the ionization constant of its conjugate acid (anilinium ion) is $K_a \approx 2.34 \times 10^{-5}$ .

The Concept: Weak Base Ionization

Aniline ( $C_6H_5NH_2$ ) is a weak base. It does not fully dissociate in water; instead, it establishes an equilibrium:

C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^-

The equilibrium constant for this reaction is the base ionization constant, $K_b$ . From Table 6.7 (standard NCERT data), the $K_b$ of aniline is $4.27 \times 10^{-10}$ at 298 K.

Because $K_b$ is very small, the ionization is tiny. This lets us use the approximation that the equilibrium concentration of aniline is nearly the same as its initial concentration — a classic simplification for weak bases.

Watch out

A common mistake is to treat aniline as a strong base or to forget that the $OH^-$ from water itself (at $10^{-7}$ M) can become significant when the base is extremely weak and dilute. Here, the calculated $[OH^-]$ is about $6.5 \times 10^{-7}$ M — comparable to pure water's $10^{-7}$ M. We must check whether the water autoionization contribution matters.

Step-by-Step Solution

1. Write the equilibrium expression.

For the reaction:

C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^-

The equilibrium constant is:

K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} = 4.27 \times 10^{-10}

Let the initial concentration of aniline be $c = 0.001$ M. If $\alpha$ is the degree of ionization, then at equilibrium:

$[C_6H_5NH_2] = c(1 - \alpha) \approx c$ (since $\alpha$ is tiny)
$[C_6H_5NH_3^+] = c\alpha$
$[OH^-] = c\alpha$ (from the base, plus a tiny bit from water)

2. Calculate the degree of ionization $\alpha$ using the approximation.

Substitute into the $K_b$ expression:

K_b = \frac{(c\alpha)(c\alpha)}{c} = c\alpha^2

So:

\alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{4.27 \times 10^{-10}}{0.001}} = \sqrt{4.27 \times 10^{-7}}

\alpha = \sqrt{4.27} \times 10^{-3.5} \approx 2.066 \times 10^{-3.5}

Since $10^{-3.5} = 10^{-4} \times 10^{0.5} = 3.162 \times 10^{-4}$ , we get:

\alpha \approx 2.066 \times 3.162 \times 10^{-4} \approx 6.53 \times 10^{-4}

Tip

A quick check: $\alpha = 6.5 \times 10^{-4}$ means only about 0.065% of aniline molecules ionize. This justifies the approximation $1 - \alpha \approx 1$ .

3. Find $[OH^-]$ and then $pOH$ , then $pH$ .

From the base ionization:

[OH^-]_{\text{base}} = c\alpha = 0.001 \times 6.53 \times 10^{-4} = 6.53 \times 10^{-7} \text{ M}

Now, pure water contributes $[OH^-]_{\text{water}} = 1.0 \times 10^{-7}$ M. The total $[OH^-]$ is not simply the sum because the common ion effect suppresses water autoionization. However, when the base contribution is comparable to $10^{-7}$ , we must solve the exact equation.

Let $[OH^-] = x$ . The charge balance gives:

[C_6H_5NH_3^+] + [H^+] = [OH^-]

But $[C_6H_5NH_3^+] = c\alpha = \frac{K_b c}{[OH^-]}$ (from the $K_b$ expression rearranged). And $[H^+] = \frac{K_w}{[OH^-]}$ . So:

\frac{K_b c}{x} + \frac{K_w}{x} = x

Multiply through by $x$ :

K_b c + K_w = x^2

Thus:

$x = \sqrt{K_b c + K_w}$ …

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Q.What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

The Concept: Weak Base Ionization

Step-by-Step Solution

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