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.

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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 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	$s = \sqrt{K_{sp}} = 1.1 \times 10^{-5}$	$1.1 \times 10^{-5}$	$1.1 \times 10^{-5}$

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

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$x$
›Reveal solutionSolution

Gold number is defined for $10\text{ mL}$ sol $+\,1\text{ mL}$ of $10\%$ NaCl; here both sol and NaCl are scaled up $\times 50$ , so mass needed $= 50x$ mg.

The gold number $x$ = milligrams of protective colloid $A$ that just prevents coagulation of $10\text{ mL}$ of standard gold sol on adding $1\text{ mL}$ of $10\%$ NaCl.

In this problem:

$\text{gold sol} = 500\text{ mL} = 50 \times 10\text{ mL}, \qquad 10\%\ \text{NaCl} = 50\text{ mL} = 50 \times 1\text{ mL}$ …
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TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$Q. AB crystallizes in a bcc lattice. If the distance between two oppositely charged ions in the lattice is 335 pm, then the edge length of it (in pm) is (A) 376.8 (B) 366.8 (C) 386.8 (D) 396.8$
›Reveal solutionSolution
In a body-centered cubic (bcc) lattice of a compound AB, the distance between oppositely charged ions is half the body diagonal, so the edge length is $a = \frac{2 \times 335}{\sqrt{3}} \approx 386.8$ pm, matching option (C).

Concept & Intuition

The key is to visualize the geometry of a bcc lattice. In AB crystallizing in bcc, the ions are arranged such that one type (say A) sits at the corners and the other (B) at the body center, or vice versa. The shortest distance between an A and a B ion is along the body diagonal — from a corner to the center of the cube. That distance is exactly half the length of the full body diagonal. The body diagonal of a cube of edge length $a$ is $\sqrt{3}\,a$ . So if the given distance (335 pm) is that half-diagonal, we can solve for $a$ .

Step-by-step solution

Identify the relevant distance

In a bcc lattice, the closest oppositely charged ions are a corner ion and the body-centered ion. The distance between them is half the body diagonal of the cube.

Write the relationship

Body diagonal of a cube = $\sqrt{3}\,a$ .

Half of that = $\frac{\sqrt{3}}{2}\,a$ .

This is given as 335 pm:

$\frac{\sqrt{3}}{2}\,a = 335$

Solve for edge length $a$ Multiply both sides by 2:

$\sqrt{3}\,a = 670$
Divide by $\sqrt{3}$ :
$a = \frac{670}{\sqrt{3}} …$
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$^{2+}$
›Reveal solutionSolution

2Cu²⁺+4I⁻→Cu₂I₂+I₂; 5×10⁻³ mol Cu²⁺ gives 2.5×10⁻³ mol each of I₂ and the precipitate.

Cu²⁺ does not form a stable CuI₂; it oxidises I⁻ and is itself reduced to Cu(I), precipitating as Cu₂I₂: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂. n(Cu²⁺)=0.100×0.05=5×10⁻³ mol is limiting (I⁻ = 1×0.1 = 0.1 mol, large excess). For every 2 mol Cu²⁺, 1 mol I₂ and 1 m …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$\mathrm{Li}^{2+}$
›Reveal solutionSolution
The radius of an electron's orbit in a hydrogen-like atom is directly proportional to the square of the principal quantum number ( $n^2$ ) and inversely proportional to the atomic number ( $Z$ ). For the second orbits of $\mathrm{Li}^{2+}$ and $\mathrm{Be}^{3+}$ , the ratio of their radii is $\boxed{4:3}$ .

The problem asks for the ratio of the radii of the second orbits of two hydrogen-like species: $\mathrm{Li}^{2+}$ and $\mathrm{Be}^{3+}$ . To solve this, we need to understand how the radius of an electron's orbit is determined in such systems.

Concept and Intuition

Bohr's model, while having limitations, accurately describes the energy and radii of orbits for hydrogen and hydrogen-like species (atoms or ions with only one electron). Both $\mathrm{Li}^{2+}$ (Lithium with 3 protons, losing 2 electrons leaves 1 electron) and $\mathrm{Be}^{3+}$ (Beryllium with 4 protons, losing 3 electrons leaves 1 electron) are hydrogen-like species.

According to Bohr's model, the radius of the $n$ -th orbit ( $r_n$ ) for a hydrogen-like atom with atomic number $Z$ is given by a specific formula. This formula shows that the radius depends on two main factors:

The principal quantum number ( $n$ ), which defines the energy level or orbit. Higher $n$ means larger orbits.

The atomic number ( $Z$ ), which represents the number of protons in the nucleus. A higher $Z$ means a stronger attractive force from the nucleus, pulling the electron closer and resulting in smaller orbits.

The relationship is that $r_n$ is directly proportional to $n^2$ and inversely proportional to $Z$ . This proportionality is crucial for calculating ratios, as many constants cancel out.
$n$
Let's apply this understanding to find the required ratio.

Step-by-step Solution

Identify the relevant parameters for $\mathrm{Li}^{2+}$ :

For Lithium (Li), the atomic number $Z = 3$ .

The problem specifies the "second orbit", so the principal quantum number $n = 2$ .

Using the proportionality $r_n \propto \frac{n^2}{Z}$ , the radius of the second orbit for $\mathrm{Li}^{2+}$ can be expressed as: …
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TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$K_c$
(g) + \mathrm{H}_2
(g) \rightleftharpoons \frac{2}{3} \mathrm{NH}_3 (g) $is 50. The$ K_c $value for the reaction$ 2\mathrm{NH}_3
(g) \rightleftharpoons \mathrm{N}_2
(g) + 3\mathrm{H}_2 (g) $at the same temperature is (A)$ 4 \times 10^{-6} $(B)$ 8 \times 10^{-6} $(C)$ 6 \times 10^{-6} $(D)$ 8 \times 10^{-3}$
›Reveal solutionSolution
The key idea is to relate equilibrium constants when the reaction is reversed and scaled. Given $K_c = 50$ for $\frac{1}{3}\mathrm{N}_2 + \mathrm{H}_2 \rightleftharpoons \frac{2}{3}\mathrm{NH}_3$ , the $K_c$ for $2\mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3\mathrm{H}_2$ is $8 \times 10^{-6}$ .

Concept and Intuition

Equilibrium constants depend on how the reaction is written. If you multiply a reaction by a factor, the $K_c$ is raised to that power. If you reverse a reaction, $K_c$ becomes its reciprocal. Here, the target reaction is the reverse of the given reaction, but also multiplied by 3 to clear fractions. So we need to apply both transformations.

Step-by-step solution

Write the given reaction and its $K_c$ Given:

$\frac{1}{3}\mathrm{N}_2(g) + \mathrm{H}_2(g) \rightleftharpoons \frac{2}{3}\mathrm{NH}_3(g) \quad K_{c1} = 50$

Reverse the reaction Reversing gives:

$\frac{2}{3}\mathrm{NH}_3(g) \rightleftharpoons \frac{1}{3}\mathrm{N}_2(g) + \mathrm{H}_2(g)$
For a reversed reaction, $K_c$ becomes the reciprocal:
$K_{c,\text{reverse}} = \frac{1}{K_{c1}} = \frac{1}{50}$

Multiply the reversed reaction by 3 to clear fractions Multiply every coefficient by 3:

$2\mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g) + 3\mathrm{H}_2(g)$
When a reaction is multiplied by a factor $n$ , the equilibrium constant is raised to the $n$ th power. Here $n = 3$ , so: …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$^3$
›Reveal solutionSolution
Schottky defects are thermally-activated vacancies that occur in ionic crystals. At room temperature, the equilibrium concentration follows Arrhenius behavior, yielding approximately $10^6$ defects per cm³ in NaCl.

Understanding Schottky Defects

A Schottky defect occurs when a pair of ions (one cation and one anion) are missing from their lattice sites in an ionic crystal, maintaining electrical neutrality. These defects form spontaneously due to thermal energy and represent an equilibrium between the entropy gain from disorder and the energy cost of creating vacancies.

The key insight is that defect concentration is governed by thermodynamics, not kinetics. Even though creating a vacancy requires energy, at any finite temperature, some defects will exist because they increase the entropy of the crystal.

Calculating Equilibrium Defect Concentration

The number of Schottky defects follows the Arrhenius equation:
$n = N \exp\left(-\frac{E_s}{2k_BT}\right)$
1. Determine the number of lattice sites ( $N$ )

For NaCl, there are approximately $2 \times 10^{22}$ ion pairs per cm³ (this comes from the density and molar volume of NaCl).

2. Identify the formation energy ( $E_s$ )

For NaCl, the Schottky defect formation energy is approximately $E_s \approx 2$ eV (this is an experimentally determined value).

3. Apply room temperature conditions

At room temperature ( $T \approx 300$ K):

$k_BT \approx 0.026$ eV

Therefore: $\frac{E_s}{2k_BT} = \frac{2}{2 \times 0.026} \approx 38$

4. Calculate the defect fraction
$\frac{n}{N} = \exp(-38) \approx 3 \times 10^{-17}$
5. Find the absolute number of defects …
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$\begin{array}{ll} \text{Metal Sulfide} & \text{Solubility product} \\ \text{A) PbS} & \text{I) } 4.0 \times 10^{-53} \\ \text{B) HgS} & \text{II) } 8.0 \times 10^{-28} \\ \text{C) MnS} & \text{III) } 1.6 \times 10^{-24} \\ \text{D) ZnS} & \text{IV) } 2.5 \times 10^{-13} \\ \end{array}$
›Reveal solutionSolution
The solubility product $K_{sp}$ of a metal sulfide depends on the solubility of the sulfide in water, which is linked to the position of the metal in the qualitative analysis scheme. The order of increasing $K_{sp}$ is PbS < HgS < ZnS < MnS, matching A→I, B→II, C→IV, D→III, so the correct option is (B).

The key concept here is solubility product and its relation to the qualitative analysis of cations. In the classical H₂S scheme, metal sulfides precipitate at different pH values because their $K_{sp}$ values vary enormously. The less soluble the sulfide, the smaller its $K_{sp}$ . PbS and HgS are very insoluble (Group II in the H₂S scheme), while MnS is much more soluble (Group IV). ZnS sits in between (Group II but soluble in dilute acid). So the order of increasing $K_{sp}$ (from smallest to largest) is: PbS < HgS < ZnS < MnS.

Now let’s match the given numbers to this order.

Identify the smallest $K_{sp}$ .

The smallest value among the four is $4.0 \times 10^{-53}$ (I). This must belong to the most insoluble sulfide. Between PbS and HgS, HgS is famously the least soluble — its $K_{sp}$ is astronomically small. So HgS → I is a strong candidate. But let’s check: PbS has $K_{sp} \approx 10^{-28}$ range, so indeed HgS gets the $10^{-53}$ value. So B → I.

Next smallest $K_{sp}$ .

The next smallest is $8.0 \times 10^{-28}$ (II). This fits PbS, which is very insoluble but not as extreme as HgS. So A → II.

The two larger $K_{sp}$ values. …
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TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$0.7 : 0.92 : 1.0 : 1.0$
›Reveal solutionSolution
The packing densities are FCC = 0.74, HCP = 0.74, BCC = 0.68, simple cubic = 0.52.

Their ratio in the order FCC : BCC : simple cubic : HCP is 0.74 : 0.68 : 0.52 : 0.74, which simplifies to 1.0 : 0.92 : 0.7 : 1.0.

The correct option is (C).

Concept & Intuition

Packing density (or atomic packing factor, APF) is the fraction of volume in a crystal structure that is actually occupied by atoms. It depends on two things: how many atoms are in a unit cell, and how efficiently they fill space. For hard spheres of equal size, the densest possible packing is 0.74 — achieved by both FCC and HCP. BCC is slightly less dense (0.68), and simple cubic is the loosest (0.52). The question asks for the ratio of these values in the order FCC, BCC, simple cubic, HCP. So we just compute each APF and compare.

Step-by-step reasoning

Simple cubic (SC)

One atom per unit cell (8 corners × 1/8).

Edge length $a = 2r$ (atoms touch along the edge).

Volume of atom = $\frac{4}{3}\pi r^3$ , cell volume = $a^3 = 8r^3$ .

APF = $\frac{1 \times \frac{4}{3}\pi r^3}{8r^3} = \frac{\pi}{6} \approx 0.5236$ .

Body-centered cubic (BCC)

2 atoms per unit cell (1 center + 8 corners × 1/8).

Atoms touch along the body diagonal: $\sqrt{3}a = 4r$ → $a = \frac{4r}{\sqrt{3}}$ .

Cell volume = $a^3 = \frac{64r^3}{3\sqrt{3}}$ .

APF = $\frac{2 \times \frac{4}{3}\pi r^3}{64r^3/(3\sqrt{3})} = \frac{\pi\sqrt{3}}{8} \approx 0.6802$ .

Face-centered cubic (FCC)

4 atoms per unit cell (6 face centers × 1/2 + 8 corners × 1/8).

Atoms touch along the face diagonal: $\sqrt{2}a = 4r$ → $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}\,r$ .

Cell volume = $a^3 = 16\sqrt{2}\,r^3$ .

APF = $\frac{4 \times \frac{4}{3}\pi r^3}{16\sqrt{2}\,r^3} = \frac{\pi}{3\sqrt{2}} \approx 0.7405$ .

Hexagonal close-packed (HCP)

For ideal HCP (c/a ratio = $\sqrt{8/3} \approx 1.633$ ), the packing density is the same as FCC: 0.7405.

Reason: both are close-packed structures; the difference is only in stacking order (ABCABC vs ABAB), not in the fraction of filled space.

Form the ratio …
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TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
$0.5$
›Reveal solutionSolution
Schottky defects reduce the effective number of atoms in a unit cell, thereby decreasing the crystal's density. For potassium in an FCC lattice with $0.1\%$ Schottky defects, the approximate density is $\boxed{2.1 \text{ g/cm}^3}$ .

Concept and Intuition

Crystal Lattice and Unit Cell: Potassium crystallizes in a Face-Centered Cubic (FCC) lattice. In an FCC unit cell, atoms are located at all eight corners and the center of each of the six faces. The effective number of atoms belonging to one FCC unit cell ( $Z$ ) is calculated as $(8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 1 + 3 = 4$ .

Density of a Crystal: The density ( $\rho$ ) of a crystalline solid is determined by the total mass of atoms within a unit cell divided by the volume of the unit cell. The mass of the unit cell is the number of atoms ( $Z$ ) multiplied by the mass of a single atom (Molar mass $M$ divided by Avogadro's number $N_A$ ). The volume of a cubic unit cell is $a^3$ , where $a$ is the unit cell length.

Schottky Defects: A Schottky defect is a type of point defect in a crystal lattice where an atom (or an ion pair in ionic crystals) is missing from its regular lattice site, creating a vacancy.

The presence of these missing atoms means that the total mass within a given volume of the crystal is less than that of a perfect crystal.

Consequently, Schottky defects decrease the overall density of the crystal.

If a crystal has $x\%$ Schottky defects, it implies that $x\%$ of the lattice sites are vacant, and thus, the effective number of atoms contributing to the mass of the unit cell is reduced by $x\%$ .

Step-by-Step Solution

Identify Given Values and Constants:

Lattice type: FCC, so the number of atoms per unit cell for a perfect crystal ( $Z$ ) is $4$ .

Unit cell length ( $a$ ) = $0.5 \text{ nm}$ .

Percentage of Schottky defects = $0.1\%$ .

Molar mass of Potassium ( $M$ ) = $39 \text{ g/mol}$ (standard atomic weight).

Avogadro's number ( $N_A$ ) = $6.022 \times 10^{23} \text{ mol}^{-1}$ .

Convert Unit Cell Length to Centimeters:

The density is required in $\mathrm{g\ cm^{-3}}$ , so we convert the unit cell length from nanometers to centimeters:

$a = 0.5 \text{ nm} = 0.5 \times 10^{-9} \text{ m} = 0.5 \times 10^{-7} \text{ cm}$ .

Calculate the Volume of the Unit Cell:

For a cubic unit cell, the volume ( $V$ ) is $a^3$ :

$V = (0.5 \times 10^{-7} \text{ cm})^3 = 0.125 \times 10^{-21} \text{ cm}^3$ .

Determine the Effective Number of Atoms per Unit Cell ( $Z'$ ) with Schottky Defects:

A $0.1\%$ Schottky defect means that $0.1\%$ of the lattice sites are vacant. This directly translates to a $0.1\%$ reduction in the effective number of atoms present in the unit cell.

Fraction of defects = $0.1\% = \frac{0.1}{100} = 0.001$ .

The effective number of atoms per unit cell ( $Z'$ ) is:

$Z' = Z \times (1 - \text{fraction of defects})$

$Z' = 4 \times (1 - 0.001) = 4 \times 0.999 = 3.996$ . …
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TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
$\mathrm{p}K_b$
›Reveal solutionSolution
When the molar concentrations of a weak base and its conjugate acid are equal in a buffer, the Henderson–Hasselbalch equation for bases gives $\text{pOH} = \text{p}K_b$ of the base. The correct option is (A).

The key here is the Henderson–Hasselbalch equation for a basic buffer. A buffer made from a weak base and its conjugate acid (the salt of that base) resists changes in pH. The equation that governs its pOH is:
$\text{pOH} = \text{p}K_b + \log\frac{[\text{conjugate acid}]}{[\text{base}]}$
This is the direct analogue of the acid-buffer equation $\text{pH} = \text{p}K_a + \log\frac{[\text{conjugate base}]}{[\text{acid}]}$ . The logic is identical: the ratio of the two species determines how far the pOH is from the base's $\text{p}K_b$ .

Now, the problem states that the molar concentrations of the base and its conjugate acid are the same. That means $[\text{base}] = [\text{conjugate acid}]$ , so the ratio $\frac{[\text{conjugate acid}]}{[\text{base}]} = 1$ .

Plug this into the Henderson–Hasselbalch equation for bases:

$\text{pOH} = \text{p}K_b + \log(1)$

Since $\log(1) = 0$ , the equation simplifies immediately to:

$\text{pOH} = \text{p}K_b$
That’s the entire reasoning — no further calculation needed. The pOH of the buffer equals the $\text{p}K_b$ of the weak base when the two components are at equal concentration. …
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TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
$Q. If the density of a 2 M solution of ethylene glycol in water is 1.11 g/ml, the molality (in 'm') of the solution is approximately (A) 1.92 (B) 1.57 (C) 2.05 (D) 2.15$
›Reveal solutionSolution
Take 1 L of solution: mass $=1110$ g, solute $=124$ g, water $=986$ g; molality $=2/0.986 \approx 2.05\ m$ .

Take exactly $1\ \text{L}$ of the $2\ \text{M}$ solution.

Moles of ethylene glycol (C $_2$ H $_6$ O $_2$ , $M = 62\ \text{g/mol}$ ) $= 2\ \text{mol}$ .

Mass of solute $= 2 \times 62 = 124\ \text{g}$ .

Mass of solution $= 1000\ \text{mL} \times 1.11\ \text{g/mL} = 1110\ \text{g}$ . …
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$4 : 3$
›Reveal solutionSolution
The highest wavelength in the Lyman series corresponds to the smallest energy transition ( $n=2$ to $n=1$ ), while the lowest wavelength corresponds to the largest energy transition ( $n=\infty$ to $n=1$ ). The ratio of these wavelengths is $\boxed{4:3}$ .

Concept and Intuition

When an electron in a hydrogen atom (or a hydrogen-like ion) jumps from a higher energy level to a lower one, it emits a photon. The energy of this photon corresponds to the energy difference between the two levels, and this energy determines the photon's wavelength.

The energy levels in a hydrogen atom are quantized, given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ , where $n$ is the principal quantum number ( $n=1, 2, 3, \dots$ ).

The energy of the emitted photon is $\Delta E = E_{n_i} - E_{n_f}$ , where $n_i$ is the initial (higher) energy level and $n_f$ is the final (lower) energy level.

This energy is also related to the wavelength $\lambda$ by the equation $\Delta E = \frac{hc}{\lambda}$ , where $h$ is Planck's constant and $c$ is the speed of light.

From this, we see that $\lambda = \frac{hc}{\Delta E}$ . This means:

A larger energy difference ( $\Delta E$ ) results in a shorter wavelength ( $\lambda$ ).

A smaller energy difference ( $\Delta E$ ) results in a longer wavelength ( $\lambda$ ).

The Lyman series specifically refers to transitions where electrons fall to the ground state, meaning the final energy level is $n_f = 1$ . The initial energy level $n_i$ can be $2, 3, 4, \dots, \infty$ .

To find the highest wavelength ( $\lambda_{max}$ ) in the Lyman series, we need the smallest possible energy difference. This occurs for the transition from $n_i = 2$ to $n_f = 1$ .

To find the lowest wavelength ( $\lambda_{min}$ ) in the Lyman series, we need the largest possible energy difference. This occurs for the transition from $n_i = \infty$ (the ionization limit) to $n_f = 1$ .

We can use the Rydberg formula, which directly relates the wavelength of emitted light to the principal quantum numbers of the initial and final states.
$\lambda$
Step-by-step Derivation

Identify the Lyman series parameters: For the Lyman series, electrons transition to the ground state. Therefore, the final principal quantum number is $n_f = 1$ . The Rydberg formula for the Lyman series becomes:

$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_i^2} \right) = R \left( 1 - \frac{1}{n_i^2} \right)$

Calculate the highest wavelength ( $\lambda_{max}$ ): The highest wavelength corresponds to the smallest energy transition. For the Lyman series ( $n_f=1$ ), the smallest energy transition occurs when the electron falls from the very next higher level, which is $n_i = 2$ . Substitute $n_i = 2$ into the Rydberg formula: $\frac{1}{\lambda_{max}} = R \left( 1 - \frac{1}{2^2} \right)$ …
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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​

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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, $Hg_2I_2$

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