A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution

NCERT Exemplar · Q31

Q. A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4 in water is 8 \times 10^-4 mol dm^-3. Calculate its solubility in 0.01 mol dm^-3 of H2SO4.

Sikkim CbseShort· 3mImportance★★★★★est

85% · 132/155 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

In pure water BaSO₄ dissolves to $8 \times 10^{-4}$ mol dm⁻³; in 0.01 M H₂SO₄ the common sulfate ion suppresses dissociation, reducing solubility to $6.4 \times 10^{-5}$ mol dm⁻³.

When a sparingly soluble salt like barium sulfate sits in equilibrium with its ions, the product of those ion concentrations is fixed at a constant called the solubility product, $K_{\text{sp}}$ . Adding a solution that already contains one of those ions—here, sulfate from sulfuric acid—shifts the equilibrium backward (Le Chatelier), forcing less BaSO₄ to dissolve. This is the common-ion effect.

The key is to first find $K_{\text{sp}}$ from the pure-water solubility, then use that same $K_{\text{sp}}$ to find the new solubility when sulfate is already present.

Step-by-step solution

1. Write the dissolution equilibrium and $K_{\text{sp}}$ expression.

Barium sulfate dissociates as

\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq).

The solubility product is

K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}].

2. Calculate $K_{\text{sp}}$ from the solubility in pure water.

In pure water, if the solubility is $s = 8 \times 10^{-4}$ mol dm⁻³, then at equilibrium

[\text{Ba}^{2+}] = s = 8 \times 10^{-4} \, \text{mol dm}^{-3}, \quad [\text{SO}_4^{2-}] = s = 8 \times 10^{-4} \, \text{mol dm}^{-3}.

Hence

K_{\text{sp}} = (8 \times 10^{-4})(8 \times 10^{-4}) = 64 \times 10^{-8} = 6.4 \times 10^{-7}.

K_{\text{sp}}(\text{BaSO}_4) = 6.4 \times 10^{-7}

3. Set up the equilibrium in 0.01 M H₂SO₄.

Sulfuric acid is a strong acid and fully dissociates (both protons in dilute solution):

\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}.

So the initial sulfate concentration is $[\text{SO}_4^{2-}]_0 = 0.01$ mol dm⁻³.

Let the solubility of BaSO₄ in this solution be $s'$ . Then at equilibrium:

[\text{Ba}^{2+}] = s', \quad [\text{SO}_4^{2-}] = 0.01 + s'.

4. Apply the solubility-product condition.

Since $K_{\text{sp}}$ is unchanged,

K_{\text{sp}} = s' \cdot (0.01 + s') = 6.4 \times 10^{-7}.

5. Make the approximation $s' \ll 0.01$ .

Because the common ion suppresses solubility, we expect $s'$ to be much smaller than 0.01. Assume $0.01 + s' \approx 0.01$ : …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ30 — pH of a solution of a strong acid is 5.0. What will be the pH of the s…→Next questionQ32 — pH of 0.08 mol dm^-3 HOCl solution is 2.85. Calculate its ionisation c…

Q.A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4 in water is 8 × 10^-4 mol dm^-3. Calculate its solubility in 0.01 mol dm^-3 of H2SO4.

Step-by-step solution

🎓Unlock everything free for 14 days

Unlock everything free for 14 days