Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The solubility product, $K_{\text{sp}}$, quantifies the equilibrium of a sparingly soluble salt like calcium fluoride in a saturated aqueous solution. The dissociation occurs according to the chemical equation $\mathrm{CaF}_2 (s) \rightleftharpoons \mathrm{Ca}^{2+} (aq) + 2\mathrm{F}^{-} (aq)$, where defining the molar solubility as $s$ implies an ion concentration product of K_{\text{sp}} = \[\mathrm{Ca}^{2+}]$$[\mathrm{F}^{-}]^2 = 4s^3$.

The given solubility of $2.34 \times 10^{-3} \text{ g}$ per $100 \text{ mL}$ must be converted into molar units of moles per liter. Scaling the mass to $1000 \text{ mL}$ results in $2.34 \times 10^{-2} \text{ g/L}$, and dividing by the molar mass of $78 \text{ g/mol}$ provides a molar solubility of $s = 3 \times 10^{-4} \text{ mol/L}$. Applying this to the equilibrium expression gives $K_{\text{sp}} = 4 \times (3 \times 10^{-4})^3$, resulting in $108 \times 10^{-12} \text{ (mol/L)}^3$. Adjusting the power of ten to match the requested $10^{-8}$ format yields $0.0108 \times 10^{-8} \text{ (mol/L)}^3$, which simplifies to 0 when rounded to the nearest whole number.

Potassium permanganate acts as a strong oxidizing agent, making it necessary to select an acidic medium that does not interfere with the titration process. Hydrochloric acid is inappropriate for this application because the permanganate ions readily oxidize chloride ions into chlorine gas. This chemical interaction is represented by the reaction $2\mathrm{KMnO}_4 + 16\mathrm{HCl} \rightarrow 2\mathrm{MnCl}_2 + 2\mathrm{KCl} + 8\mathrm{H}_2\mathrm{O} + 5\mathrm{Cl}_2$, which consumes the permanganate reagent and leads to inaccurate results.

Butyric acid functions as a weak acid in aqueous solution, undergoing partial dissociation defined by its acid dissociation constant. Calculating the pH involves the relationship between the acid dissociation constant and the molar concentration of the solution, relying on the assumption that the degree of dissociation remains minimal. Converting the dissociation constant $K_{a} = 2 \times 10^{-5}$ into its logarithmic form provides the value of 4.7 for $\mathrm{p}K_{a}$, as this is derived from $5 - \log 2$ and the value of $\log 2$ is 0.30.

The pH is subsequently calculated using the expression $\mathrm{pH} = \frac{1}{2} \mathrm{p}K_{a} - \frac{1}{2} \log C$. Substituting the concentration 0.2 and the previously determined $\mathrm{p}K_{a}$ value into the formula, the expression becomes $\mathrm{pH} = \frac{1}{2} (4.7) - \frac{1}{2} \log(0.2)$. Given that $\log(0.2) = -0.70$, the calculation proceeds as $\mathrm{pH} = 2.35 - \frac{1}{2}(-0.70) = 2.35 + 0.35$, which results in 2.7. This final value is expressed as $27 \times 10^{-1}$.

When titrating a weak base like $\mathrm{NH}_4\mathrm{OH}$ with a strong acid such as $\mathrm{HCl}$, the initial pH of the solution resides at a basic level, typically above 7. As the strong acid is incrementally added to the base, the consumption of hydroxide ions causes the pH to drop steadily, tracing a characteristic downward curve. This decline continues throughout the process, reflecting the transition from a basic solution toward the equivalence point, which naturally occurs below pH 7 due to the hydrolysis of the resulting acidic salt.

When a weak base is neutralized by an equivalent amount of a strong acid, the resulting solution contains only the corresponding salt, which undergoes hydrolysis to establish an acidic pH determined by the salt's concentration and the dissociation constant of the base. In this mixture of 20 mL of 0.1 M $\mathrm{NH_4OH}$ and 40 mL of 0.05 M $\mathrm{HCl}$, both reactants contribute 2 mmoles, leading to the formation of 2 mmoles of $\mathrm{NH_4Cl}$ in a total volume of 60 mL. This leaves the salt concentration at $C = 2/60 = 1/30\ \mathrm{M}$. Using the formula $\mathrm{pH} = \frac{1}{2} (\mathrm{pK_w} - \mathrm{pK_b} - \log C)$, and substituting the values $\mathrm{pK_w} = 14$, $\mathrm{pK_b} = 5$, and $\log C = \log(1/30) = -(\log 3 + \log 10) = -1.48$, the pH is $\frac{1}{2} (14 - 5 - (-1.48))$. This calculation results in a pH of 5.24.

Calculating the acidity of a mixture involves determining the total concentration of hydrogen ions present in the final solution. Since both hydrochloric acid and sulfuric acid are strong acids that fully dissociate in water, we first determine the total quantity of hydrogen ions provided by each source. Hydrochloric acid is monoprotic, yielding $0.01 \times 200 = 2 \text{ mmol}$ of $\text{H}^+$, while sulfuric acid is diprotic and releases two moles of hydrogen ions per mole of solute, contributing $2 \times 0.01 \times 400 = 8 \text{ mmol}$. Summing these provides a total of $10 \text{ mmol}$ of $\text{H}^+$ within a total final volume of $600 \text{ mL}$.

The resulting molar concentration of hydrogen ions is $[\text{H}^+] = \frac{10 \text{ mmol}}{600 \text{ mL}} = \frac{1}{60} \text{ M}$, which simplifies to $\frac{5}{3} \times 10^{-2} \text{ M}$. The $\text{pH}$ is derived from the negative logarithm of this concentration:
$\text{pH} = -\log \left( \frac{5}{3} \times 10^{-2} \right)$
Applying logarithmic properties, this expression expands to $-(\log 5 - \log 3 - 2)$. Using the provided values of $\log 5 = 0.70$ and $\log 3 = 0.48$, the calculation becomes $-(0.70 - 0.48 - 2)$, which simplifies to $-(-1.78)$ and results in a final $\text{pH}$ value of $1.78$.

The solubility of a sparingly soluble salt like $\mathrm{AgCl}$ is governed by the equilibrium existing between its solid form and dissolved ions: $\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{Cl}^-(aq)$. When a solution contains common ions like $\mathrm{Ag}^+$ from 0.01 M $\mathrm{AgNO}_3$ or $\mathrm{Cl}^-$ from 0.01 M KCl and 0.01 M HCl, Le Chatelier's principle dictates that the equilibrium shifts to the left, decreasing the solubility of the salt. Consequently, deionized water provides the most favorable environment for dissolution because it lacks these common ions that would otherwise suppress the solubility of $\mathrm{AgCl}$.

The Henderson-Hasselbalch equation defines the relationship between the pH of a buffer, the acid dissociation constant, and the concentrations of the conjugate base and weak acid, expressed as \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COO}^{-}]\}{[\mathrm{CH}3\mathrm{CH}2\mathrm{COOH}]}$. Given that the required pH is$4$and the$K{\mathrm{a}}$for propanoic acid is$1.3 × 10^{-5}$, we first determine the$\mathrm{p}K{\mathrm{a}}$by calculating$-\log(1.3 × 10^{-5})$, which yields approximately$4.886$.

Substituting these known values into the equation results in 4 = 4.886 + \log \frac{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COO}^{-}]\}{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}]}$. Subtracting$4.886$from both sides isolates the logarithmic term, giving$\log \frac{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COO}^{-}]$}{[\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}]} = -0.886$. Calculating the antilogarithm of$-0.886$provides the necessary concentration ratio, which is$10^{-0.886}$or approximately$0.13$.

To prepare a basic buffer solution, we rely on the Henderson-Hasselbalch equation, which establishes a relationship between the hydroxide ion concentration and the ratio of the conjugate acid salt to the weak base. Since the target is a buffer with a pH of 8.26, the first essential step is determining the corresponding pOH value using the relation $\mathrm{pH} + \mathrm{pOH} = 14$, which results in a pOH of 5.74.

With the pOH known, we utilize the equation $\mathrm{pOH} = \mathrm{pK}_b + \log \frac{[\text{salt}]}{[\text{base}]}$ to find the required concentration of ammonium chloride. Substituting the known values leads to the expression $5.74 = 4.74 + \log \frac{[\mathrm{NH}_4\mathrm{Cl}]}{0.2}$, which simplifies to $1 = \log \frac{[\mathrm{NH}_4\mathrm{Cl}]}{0.2}$. Taking the antilogarithm on both sides yields the concentration ratio, resulting in the concentration of ammonium chloride being 2 M. Given that the volume of the solution is one litre, we require 2 moles of the salt. Finally, converting this molar quantity to grams using the molar mass of 53.5 g/mol leads to a mass of 107 g to complete the buffer preparation.

When bismuth sulphide reaches equilibrium in a saturated solution, it dissociates according to the reaction $\mathrm{Bi}_2\mathrm{S}_3 \rightleftharpoons 2\mathrm{Bi}^{3+} + 3\mathrm{S}^{2-}$. Representing the molar solubility as $s$, the concentration of the bismuth ions becomes $2s$ and the concentration of the sulphide ions becomes $3s$. The solubility product constant is defined as the product of these concentrations raised to their stoichiometric powers, leading to the expression $K_{\text{sp}} = (2s)^2(3s)^3$.

Expanding this expression results in $K_{\text{sp}} = 4s^2 \times 27s^3$, which simplifies to $108s^5$. Setting this equal to the known value of $1.08 \times 10^{-73}$ produces the equation $108s^5 = 1.08 \times 10^{-73}$. Dividing both sides by 108 gives $s^5 = 1.0 \times 10^{-75}$, and taking the fifth root of this value yields a solubility of $1.0 \times 10^{-15} \, \text{mol/L}$.

When strong acid and base solutions are mixed, the neutralization reaction dictates that the species present in excess determines the final acidity of the mixture. Calculating the milliequivalents for each reactant gives 50 milliequivalents of $\mathrm{H}^+$ from the acid and 30 milliequivalents of $\mathrm{OH}^-$ from the base. After neutralization occurs, an excess of 20 milliequivalents of $\mathrm{H}^+$ remains within the total volume of 80 mL. The concentration of these remaining hydrogen ions is \left[ \mathrm{H}^+ \right]\ = 20 / 80 = 0.25$M, which is equivalent to$2.5 × 10^{-1}$M. Applying the definition of$\mathrm{pH}$leads to the calculation$\mathrm{pH} = -\log \left[ \mathrm{H}^+ \right]$ = -\log(2.5 \times 10^{-1}) = 1 - \log 2.5$. Substituting the provided value gives$\mathrm{pH} = 1 - 0.3979 = 0.6021$. Expressing this result in the required format of$x \times 10^{-4}$indicates that$x = 6021$.

A Lewis base functions by donating a lone pair of electrons, which necessitates that the central atom possesses at least one non-bonding electron pair. In the molecules $\mathrm{NF}_3$, $\mathrm{SF}_4$, and $\mathrm{ClF}_3$, the central atoms retain lone pairs, enabling them to act as donors. Conversely, the central phosphorus atom in $\mathrm{PCl}_5$ has no lone pairs and utilizes its empty $d$-orbitals to accept electron pairs, characterizing it as a Lewis acid rather than a base.

Ammonium phosphate is a salt resulting from the neutralization of a weak acid, phosphoric acid, and a weak base, ammonium hydroxide. The pH of such a salt solution depends entirely on the ionization constants of the parent acid and base and the water dissociation constant, making the solution's concentration irrelevant to the final pH value. This relationship is expressed by the formula $\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_w + \mathrm{p}K_a - \mathrm{p}K_b)$. Substituting the given values where $\mathrm{p}K_w = 14$, $\mathrm{p}K_a = 5.23$, and $\mathrm{p}K_b = 4.75$ results in the calculation $\mathrm{pH} = \frac{1}{2}(14 + 5.23 - 4.75)$. Solving this gives $\mathrm{pH} = \frac{1}{2}(14.48)$, which equals $7.24$, a value approximately equal to $7$.

Solubility equilibrium defines the behavior of sparingly soluble salts such as calcium hydroxide, $\mathrm{Ca(OH)_2}$, in water. Upon dissociation, this compound yields ions following the reaction $\mathrm{Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-}$. By assigning the variable $S$ to represent the molar solubility, the equilibrium concentration of the calcium cation is $S$, whereas the concentration of the hydroxide anion becomes $2S$.

The solubility product constant, $K_{sp}$, is calculated as the product of these ionic concentrations elevated to their stoichiometric powers, expressed as
$K_{sp} = \$[\mathrm{Ca}^{2+}]\$\$[\mathrm{OH}^-]^2$
Replacing the ionic concentrations with their corresponding solubility terms transforms the equation into
$K_{sp} = (S)(2S)^2 = 4S^3$
Given that the known constant $K_{sp} = 5.5 \times 10^{-6}$, the solubility can be isolated using
$S = \sqrt[3]{\frac{K_{sp}}{4}}$
Determining the cube root of the resulting quotient yields a final solubility of $1.11 \times 10^{-2}$ mol/L.

The dissociation of the salt $\mathrm{A}_{3} \mathrm{B}_{2}$ in an aqueous solution produces three $\mathrm{A}^{2+}$ ions and two $\mathrm{B}^{3-}$ ions for every formula unit of the salt. If we denote the molar solubility of the salt as $S$ in mol/L, the equilibrium concentrations of the resulting ions are $[\mathrm{A}^{2+}] = 3S$ and $[\mathrm{B}^{3-}] = 2S$. The solubility product constant is calculated using the concentrations of these ions raised to their respective stoichiometric coefficients:
$K_{\mathrm{sp}} = \$[\mathrm{A}^{2+}]^3 \$[\mathrm{B}^{3-}]^2 = (3S)^3 (2S)^2$
Evaluating this expression gives $K_{\mathrm{sp}} = 27S^3 \cdot 4S^2$, which simplifies to $108S^5$. Since molar solubility $S$ relates to the solubility in g/L ($x$) and molar mass $M$ in g/mol through the relationship $S = \frac{x}{M}$, we substitute this into our result to obtain $K_{\mathrm{sp}} = 108\left(\frac{x}{M}\right)^5$. Comparing this directly to the provided expression $K_{\mathrm{sp}} = a\left(\frac{x}{M}\right)^5$ confirms that the value of $a$ is 108.

The solubility product constant, $K_{sp}$, provides a quantitative link between the ion concentrations in a saturated solution and the molar solubility of a salt, with the specific algebraic form dependent on the stoichiometry of dissociation. For the salt $A_{2}X$, the equilibrium is $A_{2}X(s) \rightleftharpoons 2A^{+}(aq) + X^{2-}(aq)$, where the concentration of ions is related to the molar solubility $S_1$ by $[A^{+}] = 2S_1$ and $[X^{2-}] = S_1$. Substituting these values into the $K_{sp}$ expression yields the following equation:
$K_{sp} = \$[A^{+}]^{2}\$[X^{2-}]\$ = (2S_1)^{2}(S_1) = 4S_1^{3}$
Setting this equal to $4.0 \times 10^{-12}$ allows us to solve for $S_1$, where $4S_1^{3} = 4.0 \times 10^{-12}$, leading to $S_1 = 10^{-4} \text{ M}$.

Similarly, the salt $MX$ dissociates as $MX(s) \rightleftharpoons M^{+}(aq) + X^{-}(aq)$, where both ion concentrations equal the molar solubility $S_2$. The solubility product expression simplifies to:
$K_{sp} = \$[M^{+}]\$\$[X^{-}]\$ = S_2^{2}$
Using the same $K_{sp}$ value of $4.0 \times 10^{-12}$, we find $S_2 = \sqrt{4.0 \times 10^{-12}} = 2.0 \times 10^{-6} \text{ M}$. Calculating the ratio of these solubilities, $\frac{S_1}{S_2}$, results in $\frac{10^{-4}}{2.0 \times 10^{-6}} = 50$.

When $\mathrm{Zn}(\mathrm{OH})_2$ dissolves in a solution containing $\mathrm{NaOH}$, the common ion effect suppresses its solubility, meaning we can treat the total hydroxide concentration as essentially equal to that provided by the $0.1\mathrm{M}$ $\mathrm{NaOH}$ solution. For the dissociation equilibrium $\mathrm{Zn}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Zn}^{2+} + 2\mathrm{OH}^{-}$, the solubility product expression is defined as K_{sp} = \[\mathrm{Zn}^{2+}]$$[\mathrm{OH}^-]^2$. Substituting the molar solubility$S$for$[\mathrm{Zn}^{2+}]$and$0.1\mathrm{M}$for$[\mathrm{OH}^-]$, we arrive at:

$2 \times 10^{-20} = S \times (0.1)^2$

Solving for $S$ gives $S = \frac{2 \times 10^{-20}}{0.01} = 2 \times 10^{-18}\mathrm{M}$, resulting in a value of $2$ for $x$.

Precipitation of a sparingly soluble salt begins when the ionic product of its constituent ions reaches the solubility product constant. When a solution contains multiple anions, the species that requires the lowest concentration of the common cation to satisfy this threshold will precipitate first.

For $\mathrm{AgCl}$, the equilibrium constant expression is represented as:
$\mathrm{K}_{\mathrm{sp}} = \$[\mathrm{Ag}^+]\$[\mathrm{Cl}^-]$
Substituting the known values, the concentration of silver ions required is:
$[\mathrm{Ag}^+] = \frac{1.7 \times 10^{-10}}{0.1} = 1.7 \times 10^{-9} \mathrm{M}$
Similarly, for $\mathrm{Ag}_2\mathrm{CrO}_4$, the condition for precipitation is defined by:
$\mathrm{K}_{\mathrm{sp}} = \$[\mathrm{Ag}^+]^2[\mathrm{CrO}_4^{2-}]$
Solving for the silver ion concentration leads to:
$[\mathrm{Ag}^+] = \sqrt{\frac{1.9 \times 10^{-12}}{0.001}} = 4.3 \times 10^{-5} \mathrm{M}$
Comparing these values shows that $\mathrm{AgCl}$ requires a significantly lower concentration of $\mathrm{Ag}^+$ to initiate precipitation compared to $\mathrm{Ag}_2\mathrm{CrO}_4$. Consequently, $\mathrm{AgCl}$ will precipitate first.

The solubility of silver cyanide, $\mathrm{AgCN}$, in acidic conditions is enhanced because the cyanide ions produced react with hydrogen ions to form the weak acid hydrocyanic acid. This interaction pulls the dissolution equilibrium forward, allowing significantly more solid to dissolve than would occur in pure water. We determine the overall equilibrium constant for this process by combining the solubility product for $\mathrm{AgCN}$ with the inverse of the acid dissociation constant for $\mathrm{HCN}$.

The reaction is $\mathrm{AgCN} + \mathrm{H}^{+} \rightleftharpoons \mathrm{Ag}^{+} + \mathrm{HCN}$, and the equilibrium constant for this process is given by the expression:
$K = \frac{K_{\mathrm{sp}}}{K_{\mathrm{a}}}$
If we let the solubility be $x$, then the concentration of silver ions, $[\mathrm{Ag}^{+}]$, is $x$ and the concentration of undissociated hydrocyanic acid, $[\mathrm{HCN}]$, is also $x$. At a pH of 3, the concentration of hydrogen ions is $10^{-3}$. The expression for the equilibrium constant becomes:
$K = \frac{[\mathrm{Ag}^{+}]\$[\mathrm{HCN}]}{[\mathrm{H}^{+}]\$} = \frac{x^2}{10^{-3}}$
Equating the two expressions for $K$ gives:
$\frac{x^2}{10^{-3}} = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-10}}$
Solving this equation for $x^2$ leads to:
$x^2 = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-10}} \times 10^{-3}$
Calculating the final value results in $x = 1.9 \times 10^{-5}$.

The boiling of water containing temporary magnesium hardness results in the formation of insoluble magnesium hydroxide rather than magnesium carbonate. This chemical outcome occurs because the $K_{sp}$ of $Mg(OH)_2$ is greater than that of $MgCO_3$, driving the precipitation of the hydroxide instead. Since the assertion incorrectly identifies the product formed while the provided reason correctly describes the relative solubility products, the assertion is false while the reason remains true.