The ionization constant of benzoic acid is 6.46 × 10⁻⁵ and Ksp for silver benzoate is 2.5 × 10⁻¹³. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

The ionization constant of benzoic acid is 6.46 × 10⁻⁵ and Ksp for silver benzoate is 2.5 × 10⁻¹³. How many ti

Exercises · 6.70

Q. The ionization constant of benzoic acid is 6.46 \times 10⁻⁵ and Ksp for silver benzoate is 2.5 \times 10⁻¹³. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

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Concept understanding — Solubility Product Constant

Solubility Product Constant: From Intuition to Precision

Imagine you drop a pinch of salt into a glass of water. The salt crystals disappear — they dissolve. But what if you keep adding salt, spoonful after spoonful? At some point, the water can't hold any more; the extra salt just sits at the bottom, undissolved. That's a saturated solution — the maximum amount of solute has dissolved at that temperature.

Now, here's the key question: even in that saturated solution, is everything static? Not at all. At the microscopic level, salt ions are constantly leaving the solid crystal and entering the solution (dissolving), while other ions in solution bump into the crystal and stick back (precipitating). At saturation, these two processes happen at exactly the same rate. The system is in dynamic equilibrium.

Note

A saturated solution is not "full" in a static sense — it's a busy, balanced dance between dissolving and precipitating.

The Intuition: A Crowded Dance Floor

Think of a dance hall with a capacity limit. The "dancers" are ions (like Na⁺ and Cl⁻ from table salt). The dance floor is the solution. When the floor is empty, dancers easily find space — dissolution is fast. As more dancers arrive, they start bumping into each other and some leave the floor (precipitate). At the maximum capacity, the number of dancers entering equals the number leaving. That equilibrium number of dancers is what we call solubility.

But here's the twist: for many salts, the "dancers" come in different types — say, positive ions and negative ions. The equilibrium isn't just about the total number; it's about the product of their concentrations. Why product? Because the chance of a positive and a negative ion meeting to form a solid depends on both their concentrations. If you double the concentration of positive ions, the chance of a collision doubles. If you double both, it quadruples.

That product — at equilibrium, for a saturated solution — is a constant. That's the solubility product constant, $K_{sp}$ .

The Precise Statement

For a sparingly soluble salt that dissociates in water as:

A_mB_n(s) \rightleftharpoons m A^{n+}(aq) + n B^{m-}(aq)

the solubility product constant is defined as:

K_{sp} = [A^{n+}]^m \cdot [B^{m-}]^n

where the square brackets denote molar concentrations (mol/L) at saturation (equilibrium with the solid).

Watch out

The solid $A_mB_n$ does not appear in the expression. Its concentration is constant (pure solid) and is absorbed into $K_{sp}$ . Never write $[A_mB_n]$ in the $K_{sp}$ expression.

What $K_{sp}$ Tells You

Small $K_{sp}$ (e.g., $10^{-30}$ ): The salt is very insoluble. Only a tiny amount dissolves.
Large $K_{sp}$ (e.g., $10^{-2}$ ): The salt is relatively soluble.
$K_{sp}$ is temperature-dependent — always quote the temperature (usually 25°C).

Important

$K_{sp}$ is an equilibrium constant. It only applies to saturated solutions in contact with undissolved solid. If no solid is present, the solution may be unsaturated ( $Q < K_{sp}$ ) or supersaturated ( $Q > K_{sp}$ ), but $K_{sp}$ itself doesn't change.

A Concrete Example: Silver Chloride

Silver chloride, AgCl, is a classic sparingly soluble salt. Its dissolution:

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

The $K_{sp}$ expression:

K_{sp} = [Ag^+][Cl^-]

At 25°C, $K_{sp} = 1.8 \times 10^{-10}$ . This tiny number means that in a saturated solution, the product of the two ion concentrations is only $1.8 \times 10^{-10}$ .

If you know the solubility of AgCl is $s$ mol/L, then $[Ag^+] = s$ and $[Cl^-] = s$ , so: …

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The solubility of silver benzoate in a buffer of pH 3.19 is about 3.2 times its solubility in pure water. This happens because the buffer protonates benzoate ions, shifting the dissolution equilibrium forward.

Why solubility changes in a buffer

Silver benzoate (AgC₆H₅COO) is a sparingly soluble salt. In pure water, it dissolves according to:

\text{AgC}_6\text{H}_5\text{COO(s)} \rightleftharpoons \text{Ag}^+(\text{aq}) + \text{C}_6\text{H}_5\text{COO}^-(\text{aq})

The solubility product is $K_{sp} = [\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-] = 2.5 \times 10^{-13}$ .

Now, benzoate ion ( $\text{C}_6\text{H}_5\text{COO}^-$ ) is the conjugate base of benzoic acid ( $\text{C}_6\text{H}_5\text{COOH}$ ), which has $K_a = 6.46 \times 10^{-5}$ . In an acidic buffer, the benzoate ions get protonated:

\text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \rightleftharpoons \text{C}_6\text{H}_5\text{COOH}

This removes benzoate ions from solution, so more silver benzoate must dissolve to restore equilibrium — solubility increases.

The key insight: in pure water, the only equilibrium is the $K_{sp}$ ; in the buffer, we have two coupled equilibria — dissolution and protonation. The total solubility $S$ equals the concentration of silver ions, which now equals the sum of all benzoate-containing species: $[\text{C}_6\text{H}_5\text{COO}^-] + [\text{C}_6\text{H}_5\text{COOH}]$ .

K_{sp}' = K_{sp} \left(1 + \frac{[\text{H}^+]}{K_a}\right)

Let's derive this step by step.

Step-by-step solution

1. Find the solubility in pure water

In pure water, let $S_0$ be the solubility. Then $[\text{Ag}^+] = S_0$ and $[\text{C}_6\text{H}_5\text{COO}^-] = S_0$ .

K_{sp} = S_0 \times S_0 = S_0^2

S_0 = \sqrt{K_{sp}} = \sqrt{2.5 \times 10^{-13}} = \sqrt{25 \times 10^{-14}} = 5 \times 10^{-7} \ \text{M}

So in pure water, solubility is $5.0 \times 10^{-7}$ mol/L.

2. Determine $[\text{H}^+]$ in the buffer

The buffer pH is 3.19.

[\text{H}^+] = 10^{-3.19} = 10^{-3} \times 10^{-0.19}

Now $10^{-0.19} = 10^{-0.20 + 0.01} \approx 0.631 \times 1.023 \approx 0.646$ (or use calculator: $10^{-0.19} = 0.6457$ ). So:

[\text{H}^+] \approx 6.46 \times 10^{-4} \ \text{M}

Note

Notice that $[\text{H}^+] = 6.46 \times 10^{-4}$ is exactly $10 \times K_a$ (since $K_a = 6.46 \times 10^{-5}$ ). This is a neat coincidence from the given numbers — the pH was chosen to make the arithmetic clean.

3. Set up the buffer solubility

Let $S$ be the solubility in the buffer. Then $[\text{Ag}^+] = S$ .

The total benzoate species in solution is also $S$ (mass balance: every dissolved silver comes with one benzoate group, whether as ion or acid):

[\text{C}_6\text{H}_5\text{COO}^-] + [\text{C}_6\text{H}_5\text{COOH}] = S

The acid-base equilibrium gives:

K_a = \frac{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}

So $[\text{C}_6\text{H}_5\text{COOH}] = \dfrac{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}{K_a}$ .

Substitute into the mass balance:

[\text{C}_6\text{H}_5\text{COO}^-] + \frac{[\text{H}^+]}{K_a} [\text{C}_6\text{H}_5\text{COO}^-] = S

$[\text{C}_6\text{H}_5\text{COO}^-] \left(1 + \frac{[\text{H}^+]}{K_a}\right) = S$ …

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Q.The ionization constant of benzoic acid is 6.46 × 10⁻⁵ and Ksp for silver benzoate is 2.5 × 10⁻¹³. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Why solubility changes in a buffer

Step-by-step solution

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