Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous

Exercises · 6.67

Q. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.

CBSENCERTSubjective· 3mImportance★★★★★est

61% · 95/155 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 — 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: …

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

The key idea is the Solubility Product Constant ( $K_{sp}$ ): for a sparingly soluble salt $A_xB_y$ , the product of ion concentrations (raised to stoichiometric coefficients) at saturation is constant at a given temperature.

General method (for a salt $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$ ):

Let molar solubility = $s$ mol/L.
Then $[A^{y+}] = xs$ , $[B^{x-}] = ys$ .
$K_{sp} = (xs)^x (ys)^y = x^x y^y s^{x+y}$ .
Solve for $s$ , then compute individual ion molarities.

Using standard $K_{sp}$ values at 298 K (from Table 6.9, NCERT):

Salt	$K_{sp}$	$x$	$y$	$s$ (mol/L)	$[cation]$	$[anion]$
$\text{Ag}_2\text{CrO}_4$	$1.1 \times 10^{-12}$	2	1	$s = \sqrt[3]{\frac{K_{sp}}{4}} = 6.5 \times 10^{-5}$	$1.3 \times 10^{-4}$	$6.5 \times 10^{-5}$
$\text{BaCrO}_4$	$1.2 \times 10^{-10}$	1	1	$s = \sqrt{K_{sp}} = 1.1 \times 10^{-5}$	$1.1 \times 10^{-5}$	$1.1 \times 10^{-5}$

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

The solubility of a sparingly soluble salt is found by relating its $K_{sp}$ expression to the stoichiometric concentrations of its ions. For each salt, we set up the dissolution equilibrium, let $s$ be the molar solubility, substitute into the $K_{sp}$ formula, and solve for $s$ . The individual ion molarities then follow from the stoichiometric coefficients. The results are tabulated below.

The key idea is that the solubility product constant $K_{sp}$ is the equilibrium constant for the dissolution of a sparingly soluble salt. It is the product of the concentrations of the ions, each raised to the power of its stoichiometric coefficient in the balanced equation. For a salt $A_xB_y$ that dissolves as:

A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)

the $K_{sp}$ expression is:

K_{sp} = [A^{y+}]^x [B^{x-}]^y

If we let the molar solubility be $s$ mol/L (the number of moles of salt that dissolve per litre of solution), then from the stoichiometry:

[A^{y+}] = x s \quad \text{and} \quad [B^{x-}] = y s

Substituting into the $K_{sp}$ expression gives:

K_{sp} = (x s)^x (y s)^y = x^x y^y s^{x+y}

We then solve for $s$ . The molarities of the individual ions are then $x s$ and $y s$ respectively.

Now, we apply this to each salt. The $K_{sp}$ values at 298 K are taken from Table 6.9 (standard NCERT data). Let's work through each one.

1. Silver chromate, $Ag_2CrO_4$

Dissociation: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$

Here, $x = 2$ , $y = 1$ . Let solubility = $s$ mol/L.

Then $[Ag^+] = 2s$ , $[CrO_4^{2-}] = s$ .

K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2s)^2 (s) = 4s^3

From Table 6.9, $K_{sp}(Ag_2CrO_4) = 1.1 \times 10^{-12}$ .

4s^3 = 1.1 \times 10^{-12} \implies s^3 = \frac{1.1 \times 10^{-12}}{4} = 2.75 \times 10^{-13}

s = \sqrt[3]{2.75 \times 10^{-13}} = \sqrt[3]{275 \times 10^{-15}} = \sqrt[3]{275} \times 10^{-5}

Since $\sqrt[3]{275} \approx 6.5$ (because $6.5^3 = 274.6$ ), we get:

s \approx 6.5 \times 10^{-5} \text{ mol/L}

Ion molarities: $[Ag^+] = 2s = 1.3 \times 10^{-4}$ M, $[CrO_4^{2-}] = s = 6.5 \times 10^{-5}$ M.

Watch out

A common mistake is to forget the coefficient 2 on $Ag^+$ when squaring. The $K_{sp}$ is $(2s)^2(s) = 4s^3$ , not $s^3$ . Always write the full expression from the balanced equation.

2. Barium chromate, $BaCrO_4$

Dissociation: $BaCrO_4(s) \rightleftharpoons Ba^{2+}(aq) + CrO_4^{2-}(aq)$

Here, $x = 1$ , $y = 1$ . Let solubility = $s$ mol/L.

Then $[Ba^{2+}] = s$ , $[CrO_4^{2-}] = s$ .

K_{sp} = [Ba^{2+}][CrO_4^{2-}] = s \cdot s = s^2

From Table 6.9, $K_{sp}(BaCrO_4) = 1.2 \times 10^{-10}$ .

s^2 = 1.2 \times 10^{-10} \implies s = \sqrt{1.2 \times 10^{-10}} = \sqrt{1.2} \times 10^{-5}

Since $\sqrt{1.2} \approx 1.095$ , we get:

s \approx 1.1 \times 10^{-5} \text{ mol/L}

Ion molarities: $[Ba^{2+}] = s = 1.1 \times 10^{-5}$ M, $[CrO_4^{2-}] = s = 1.1 \times 10^{-5}$ M.

Tip

For a 1:1 salt like $BaCrO_4$ , the solubility is simply $\sqrt{K_{sp}}$ . This is the simplest case.

3. Ferric hydroxide, $Fe(OH)_3$

Dissociation: $Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq)$

Here, $x = 1$ , $y = 3$ . Let solubility = $s$ mol/L.

Then $[Fe^{3+}] = s$ , $[OH^-] = 3s$ .

K_{sp} = [Fe^{3+}][OH^-]^3 = (s)(3s)^3 = s \cdot 27 s^3 = 27 s^4

From Table 6.9, $K_{sp}(Fe(OH)_3) = 1.0 \times 10^{-38}$ .

27 s^4 = 1.0 \times 10^{-38} \implies s^4 = \frac{1.0 \times 10^{-38}}{27} \approx 3.70 \times 10^{-40}

s = \sqrt[4]{3.70 \times 10^{-40}} = \sqrt[4]{3.70} \times 10^{-10}

Since $\sqrt[4]{3.70} \approx 1.39$ (because $1.4^4 = 3.84$ , close enough), we get:

s \approx 1.39 \times 10^{-10} \text{ mol/L}

Ion molarities: $[Fe^{3+}] = s = 1.39 \times 10^{-10}$ M, $[OH^-] = 3s = 4.17 \times 10^{-10}$ M.

Note

The exponent on $s$ is $x+y = 1+3 = 4$ , so we take the fourth root. The very small $K_{sp}$ reflects the extreme insolubility of $Fe(OH)_3$ .

4. Lead chloride, $PbCl_2$

Dissociation: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$

Here, $x = 1$ , $y = 2$ . Let solubility = $s$ mol/L.

Then $[Pb^{2+}] = s$ , $[Cl^-] = 2s$ .

K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = s \cdot 4 s^2 = 4 s^3

From Table 6.9, $K_{sp}(PbCl_2) = 1.6 \times 10^{-5}$ .

4 s^3 = 1.6 \times 10^{-5} \implies s^3 = \frac{1.6 \times 10^{-5}}{4} = 4.0 \times 10^{-6}

s = \sqrt[3]{4.0 \times 10^{-6}} = \sqrt[3]{4.0} \times 10^{-2}

Since $\sqrt[3]{4.0} \approx 1.587$ , we get:

s \approx 1.59 \times 10^{-2} \text{ mol/L}

Ion molarities: $[Pb^{2+}] = s = 1.59 \times 10^{-2}$ M, $[Cl^-] = 2s = 3.18 \times 10^{-2}$ M.

Watch out

$PbCl_2$ has a relatively high $K_{sp}$ compared to the others, so its solubility is in the $10^{-2}$ M range — it is not "insoluble" in the strict sense, but sparingly soluble. Always check the magnitude.

5. Mercurous iodide, $Hg_2I_2$

Dissociation: $Hg_2I_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2I^-(aq)$ …

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

←Previous question · EXERCISES6.66 — Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)2 +…→Next question · EXERCISES6.68 — The solubilit y product constant of Ag 2CrO4 and AgBr are 1.1 × 10⁻¹² …

Q.Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.

1. Silver chromate, Ag2CrO4Ag_2CrO_4Ag2​CrO4​

2. Barium chromate, BaCrO4BaCrO_4BaCrO4​

3. Ferric hydroxide, Fe(OH)3Fe(OH)_3Fe(OH)3​

4. Lead chloride, PbCl2PbCl_2PbCl2​

5. Mercurous iodide, Hg2I2Hg_2I_2Hg2​I2​

🎓Unlock everything free for 14 days

1. Silver chromate, $Ag_2CrO_4$

2. Barium chromate, $BaCrO_4$

3. Ferric hydroxide, $Fe(OH)_3$

4. Lead chloride, $PbCl_2$

5. Mercurous iodide, $Hg_2I_2$

Unlock everything free for 14 days