In which of the following solvents is silver chloride most soluble? (i) 0.1 mol dm^-3 AgNO3 solution (ii) 0.1

NCERT Exemplar · Q12

Q. In which of the following solvents is silver chloride most soluble?

(i) 0.1 mol dm^-3 AgNO3 solution

(ii) 0.1 mol dm^-3 HCl solution

(iii) H2O

(iv) Aqueous ammonia

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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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Silver chloride is most soluble in aqueous ammonia because ammonia forms a stable complex ion with silver ions, effectively removing them from solution and shifting the dissolution equilibrium. The correct option is (iv).

The solubility of an ionic compound like silver chloride ( $AgCl$ ) in a solvent depends on the extent to which its dissolution equilibrium can be shifted. Silver chloride is a sparingly soluble salt, meaning only a small amount dissolves in water. Its dissolution can be represented by the equilibrium:

AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

The solubility product constant ( $K_{sp}$ ) for this equilibrium is given by $K_{sp} = [Ag^+][Cl^-]$ . A higher concentration of $Ag^+$ or $Cl^-$ ions in the solution (due to other sources) will shift this equilibrium to the left, decreasing the solubility of $AgCl$ . Conversely, if $Ag^+$ or $Cl^-$ ions are removed from the solution, the equilibrium will shift to the right, increasing the solubility.

Let's analyze each solvent option:

Solubility in H $_2$ O (option (iii))

In pure water, $AgCl$ dissolves to a small extent, establishing the equilibrium:

$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

If $s$ is the molar solubility of $AgCl$ in water, then $[Ag^+] = s$ and $[Cl^-] = s$ .

So, $K_{sp} = s^2$ , and $s = \sqrt{K_{sp}}$ . This serves as our baseline for comparison. For $AgCl$ , $K_{sp}$ is approximately $1.8 \times 10^{-10}$ at $25^\circ C$ , so $s \approx 1.34 \times 10^{-5}$ mol dm $^{-3}$ .
Solubility in 0.1 mol dm $^{-3}$ AgNO $_3$ solution (option (i))

Silver nitrate ( $AgNO_3$ ) is a strong electrolyte and dissociates completely in water:

$AgNO_3(aq) \rightarrow Ag^+(aq) + NO_3^-(aq)$

This solution introduces a significant concentration of $Ag^+$ ions ( $0.1$ mol dm $^{-3}$ ) into the solvent. According to Le Chatelier's principle, the presence of a common ion ( $Ag^+$ ) will shift the dissolution equilibrium of $AgCl$ to the left:

$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

This reduces the solubility of $AgCl$ compared to its solubility in pure water. If $s'$ is the solubility in $AgNO_3$ , then $[Ag^+] \approx 0.1$ M (since $s'$ will be very small) and $[Cl^-] = s'$ .

$K_{sp} = [Ag^+][Cl^-] \approx (0.1)(s')$ .

$s' = K_{sp}/0.1$ . This value will be much smaller than $\sqrt{K_{sp}}$ .
Solubility in 0.1 mol dm $^{-3}$ HCl solution (option (ii))

Hydrochloric acid ( $HCl$ ) is a strong acid and dissociates completely in water:

$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$

This solution introduces a significant concentration of $Cl^-$ ions ( $0.1$ mol dm $^{-3}$ ) into the solvent. Similar to the $AgNO_3$ case, the presence of a common ion ( $Cl^-$ ) will shift the dissolution equilibrium of $AgCl$ to the left:

$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$

This also reduces the solubility of $AgCl$ compared to its solubility in pure water. If $s''$ is the solubility in $HCl$ , then $[Ag^+] = s''$ and $[Cl^-] \approx 0.1$ M.

$K_{sp} = [Ag^+][Cl^-] \approx (s'')(0.1)$ .

$s'' = K_{sp}/0.1$ . This value will also be much smaller than $\sqrt{K_{sp}}$ . …

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Q.In which of the following solvents is silver chloride most soluble?

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