Ka for CH3COOH is 1.8 × 10^-5 and Kb for NH4OH is 1.8 × 10^-5. The pH of ammonium acetate will be (i) 7.005 (i

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Q. Ka for CH3COOH is 1.8 \times 10^-5 and Kb for NH4OH is 1.8 \times 10^-5. The pH of ammonium acetate will be

(i) 7.005

(ii) 4.75

(iii) 7.0

(iv) Between 6 and 7

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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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For a salt of a weak acid and a weak base with equal $K_a$ and $K_b$ , the pH is given by $\frac{1}{2} pK_w = 7.0$ , independent of concentration. The answer is 7.0.

Ammonium acetate is a classic example of a salt formed from a weak acid (acetic acid, $CH_3COOH$ ) and a weak base (ammonium hydroxide, $NH_4OH$ ). When such a salt dissolves in water, both the cation ( $NH_4^+$ ) and the anion ( $CH_3COO^-$ ) hydrolyze — that is, they react with water to re-establish the acid-base equilibria. The key insight is that the pH of the solution depends on the relative strengths of the conjugate acid and base, not on the salt concentration.

Here, both $K_a$ and $K_b$ are given as $1.8 \times 10^{-5}$ . This equality is the crucial fact: the acid and base are exactly matched in strength. Let’s see why that leads to a neutral pH.

Write the hydrolysis reactions. The acetate ion ( $CH_3COO^-$ ) is the conjugate base of acetic acid. It reacts with water:

CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-

The ammonium ion ( $NH_4^+$ ) is the conjugate acid of ammonia. It reacts with water:

NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+

Notice that one reaction produces $OH^-$ (making the solution basic) and the other produces $H^+$ (making it acidic). The net pH depends on which effect dominates.

Recall the formula for pH of a salt of weak acid + weak base. For a salt like $NH_4CH_3COO$ , the hydrogen ion concentration is given by:

[H^+] = \sqrt{\frac{K_w \cdot K_a}{K_b}}

This formula is derived by combining the equilibrium expressions for the two hydrolysis reactions and the autoionization of water. It assumes that the salt concentration is large enough that the approximations hold, but the result is independent of concentration.

[H^+] = \sqrt{\frac{K_w \cdot K_a}{K_b}}

Plug in the values. Here $K_a = 1.8 \times 10^{-5}$ , $K_b = 1.8 \times 10^{-5}$ , and $K_w = 1.0 \times 10^{-14}$ at 25°C. …

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Q.Ka for CH3COOH is 1.8 × 10^-5 and Kb for NH4OH is 1.8 × 10^-5. The pH of ammonium acetate will be

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