Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The Henderson-Hasselbalch equation $\mathrm{pH} = \mathrm{p}K_a + \log \frac{[\text{salt}]}{[\text{acid}]}$ dictates buffer conditions based on the equilibrium between a weak acid and its conjugate base. Substituting the known parameters into this relationship yields $5.74 = 4.74 + \log \frac{[\text{salt}]}{1.0}$. Subtracting the $\mathrm{p}K_a$ from the $\mathrm{pH}$ leaves $1 = \log \frac{[\text{salt}]}{1.0}$, which simplifies to $\log[\text{salt}] = 1$. Calculating the antilogarithm of 1 confirms that the concentration of sodium acetate is 10 M.

Because the first acid dissociation constant $K_{a_{1}} = 1.7 \times 10^{-2}$ is significantly larger than the second constant $K_{a_{2}} = 6.4 \times 10^{-8}$, the hydronium ion concentration in a sulphurous acid solution is determined predominantly by the first ionization, $\mathrm{H}_{2}\mathrm{SO}_{3} \rightleftharpoons \mathrm{H}^{+} + \mathrm{HSO}_{3}^{-}$. For a $0.588\ \mathrm{M}$ solution, the extent of dissociation is small enough that the equilibrium concentration of $\mathrm{H}_{2}\mathrm{SO}_{3}$ is effectively $0.588\ \mathrm{M}$. By defining the concentration of $\mathrm{H}^{+}$ and $\mathrm{HSO}_{3}^{-}$ as $x$, the equilibrium constant expression K_{a_{1}} = \frac{[\mathrm{H}^{+}]\$[\mathrm{HSO}{3}^{-}]$}{[\mathrm{H}{2}\mathrm{SO}_{3}]}$becomes$1.7 × 10^{-2} = \frac{x^{2}}{0.588}$. Solving for$x$yields$x^{2} \approx 0.01$, meaning$[\mathrm{H}^{+}] \approx 0.1\ \mathrm{M}$. Taking the negative logarithm of this hydronium ion concentration confirms the$\mathrm{pH}$is$1$.

The common ion effect explains why the presence of $\mathrm{Pb}^{2+}$ ions from the complete dissociation of lead nitrate ($\mathrm{Pb(NO_3)_2}$) suppresses the solubility of lead iodide ($\mathrm{PbI}_{2}$). Because lead nitrate is a strong electrolyte, a $0.1 \mathrm{M}$ solution provides a concentration of $0.1 \mathrm{M}$ $\mathrm{Pb}^{2+}$ ions immediately. Since lead iodide is only sparingly soluble, its own contribution of $\mathrm{Pb}^{2+}$ ions to the total concentration is negligible compared to the $0.1 \mathrm{M}$ already present, allowing us to treat the concentration of lead ions in the solubility product expression as effectively equal to $0.1 \mathrm{M}$.

For the equilibrium reaction $\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^{-}(aq)$, the iodide ion concentration is $2S$, where $S$ is the solubility in $\mathrm{mol/L}$. Substituting these values into the solubility product expression gives:
$8.0 \times 10^{-9} = \$[\mathrm{Pb}^{2+}]\$\$[\mathrm{I}^{-}]^2 = (0.1)(2S)^2$
Simplifying this expression, we have:
$8.0 \times 10^{-9} = 0.1 \times 4S^2 = 0.4S^2$
Dividing by $0.4$ yields $S^2 = 2.0 \times 10^{-8}$, and taking the square root gives $S = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4} \mathrm{mol/L}$. Using the provided value of $\sqrt{2} = 1.41$, the solubility $S$ becomes $1.41 \times 10^{-4} \mathrm{mol/L}$, which is equivalent to $141 \times 10^{-6} \mathrm{mol/L}$. Comparing this to the given format, the value of $x$ is 141.

The common ion effect explains why the solubility of a sparingly soluble salt decreases when a common ion is introduced into the solution, as this shifts the dissolution equilibrium back toward the solid form. To determine the solubility in an acidic environment, we first calculate the solubility product constant, $K_{\text{sp}}$, for $\mathrm{CdSO}_4$ by squaring its solubility in pure water:
$K_{\text{sp}} = (8.0 \times 10^{-4})^2 = 64 \times 10^{-8}$

When $\mathrm{CdSO}_4$ is added to $0.01 \, \mathrm{M}$ $\mathrm{H}_2\mathrm{SO}_4$, the strong acid dissociates completely, providing an initial concentration of $0.01 \, \mathrm{M}$ for the $\mathrm{SO}_4^{2-}$ ions. If we denote the new solubility of $\mathrm{CdSO}_4$ as $x$, the concentration of $\mathrm{Cd}^{2+}$ ions becomes $x$, while the total concentration of $\mathrm{SO}_4^{2-}$ ions becomes $x + 0.01$. Given that the solubility is very low, we can simplify the expression by approximating $x + 0.01 \approx 0.01$. Using the equilibrium expression K_{\text{sp}} = \[\mathrm{Cd}^{2+}]$[\mathrm{SO}_4^{2-}], we can write: $$64 \times 10^{-8} = x(0.01)$$ Solving this equation results in x = 64 \times 10^{-6} , \mathrm{mol} , \mathrm{L}^{-1}$.

The selection of an appropriate acid-base indicator relies on the indicator's pH range encompassing the pH at the equivalence point of the titration. For the titration of a strong acid and a weak base, the resulting salt hydrolyzes to create an acidic medium, and because methyl orange changes color within the pH range of $3.2 - 4.4$, it effectively signals the equivalence point, making statement I true. Conversely, titrating a weak acid like acetic acid with a strong base like $\mathrm{NaOH}$ results in a basic environment at the equivalence point, which aligns perfectly with the pH range of $8.2 - 10.0$ for phenolphthalein, rendering the claim that it is unsuitable in statement II incorrect.

Oxalic acid acts as a primary standard, meaning it is non-hygroscopic and can be weighed accurately to prepare a standard solution for titrating against a secondary standard like aqueous $NaOH$. In contrast, concentrated $H_2SO_4$ is hygroscopic and unstable in concentration, preventing its use as a primary standard for direct and accurate determination. Consequently, titrating aqueous $NaOH$ from a burette into a conical flask containing a known amount of aqueous oxalic acid minimizes errors associated with solution preparation and chemical stability.

The speed at which a chemical reaction occurs is governed by kinetic parameters, whereas the extent to which a reaction proceeds to completion is determined by thermodynamic parameters. The observation that Cu(II) ions and sulphide ions react rapidly to form a solid precipitate is a kinetic phenomenon, relating to the activation energy barrier and collision dynamics of the species involved. Conversely, the equilibrium constant for the reaction ${Cu}^{2+}({aq})+ {S}^{2-}( {aq})⇌ {CuS}( {s})$ is indeed very high, which is mathematically expressed as the inverse of the solubility product, $K_{eq} = 1/K_{sp}$. Because $K_{sp}$ for copper sulphide is exceptionally low, the equilibrium constant is extremely high, indicating that the reaction is thermodynamically favored to reach near-complete conversion, but this thermodynamic favorability does not account for the kinetic speed at which the reaction proceeds.

The solubility product constant characterizes the equilibrium between a sparingly soluble salt and its dissolved ions, providing a direct link between ion concentrations and the solubility $S$. For the dissociation process $\mathrm{AB}_2(\mathrm{s}) \rightleftharpoons \mathrm{A}^{2+}(\mathrm{aq}) + 2\mathrm{B}^-(\mathrm{aq})$, the ionic concentrations in a saturated solution are $[\mathrm{A}^{2+}] = S$ and $[\mathrm{B}^-] = 2S$, which leads to the solubility product expression:
$K_{sp} = \$[\mathrm{A}^{2+}]\$\$[\mathrm{B}^-]^2 = 4S^3$
Substituting $K_{sp} = 3.20 \times 10^{-11}$ into the equation gives $S^3 = 8.0 \times 10^{-12}$, and solving for $S$ yields a solubility of $2.00 \times 10^{-4}$ mol L⁻¹.

The solubility product constant describes the equilibrium state where the rate of dissolution equals the rate of precipitation for a sparingly soluble salt, providing the basis to calculate ion concentrations in a saturated solution. The dissociation of chromium hydroxide follows the equation $\mathrm{Cr}(\mathrm{OH})_3 \rightleftharpoons \mathrm{Cr}^{3+} + 3\mathrm{OH}^{-}$, which dictates that the solubility product constant is K_{sp} = \[\mathrm{Cr}^{3+}]$$[\mathrm{OH}^{-}]^3 = 6.0 × 10^{-31}$.

Because each unit of chromium hydroxide yields three hydroxide ions for every single chromium ion, the concentration relationship is established as [\mathrm{Cr}^{3+}] = \frac{[\mathrm{OH}^{-}]\}{3}$. Substituting this into the$K_{sp}$expression results \in$K_{sp} = \frac{[\mathrm{OH}^{-}]^4}{3}$. Rearranging this equation to solve for the fourth power of the hydroxide concentration gives$[\mathrm{OH}^{-}]^4 = 3 × K_{sp}$, which equals$3 × 6.0 × 10^{-31} = 18 × 10^{-31}$. Consequently, the hydroxide ion concentration is$(18 × 10^{-31})^{1/4}$.

The equilibrium of a saturated salt solution is determined by the stoichiometry of its dissociation, where the concentrations of the constituent ions dictate the solubility product constant. In this system, the concentration of the anionic species Y reaches $2 \times 10^{-3}\,\mathrm{M}$ while the cationic species X reaches $1 \times 10^{-3}\,\mathrm{M}$, confirming a 1:2 stoichiometric ratio consistent with the formula $\mathrm{XY}_2$. The dissociation equilibrium is represented as:
$\mathrm{XY}_2(s) \rightleftharpoons \mathrm{X}^{2+}(aq) + 2\mathrm{Y}^-(aq)$
The solubility product is calculated using the expression K_{sp} = \[\mathrm{X}^{2+}]$$[\mathrm{Y}^-]^2$. By substituting the measured concentrations into this expression, the calculation$(1 × 10^{-3}) × (2 × 10^{-3})^2$yields a final solubility product of$4 × 10^{-9},\mathrm{M}^3$.

The ionization of water, $\mathrm{H_2O}(l) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq)$, is an endothermic process characterized by $\Delta H > 0$, which proves the reason provided is false. As temperature increases, the equilibrium shifts forward to accommodate the added heat, resulting in a higher concentration of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ ions. This increase in hydrogen ion concentration causes the pH to decrease, which demonstrates that the assertion is also incorrect.

The $\mathrm{pOH}$ of an aqueous solution is inversely related to its hydroxide ion concentration, meaning that acidic solutions exhibit higher $\mathrm{pOH}$ values, while basic solutions show lower $\mathrm{pOH}$ values. A solution of $0.01\,\mathrm{M}\,\mathrm{HCl}$ is a strong acid that provides a high concentration of $\mathrm{H}^{+}$ ions, resulting in a low $\mathrm{pH}$ of $2$ and a correspondingly high $\mathrm{pOH}$ of $12$. In contrast, $0.01\,\mathrm{M}\,\mathrm{NaCl}$ is a neutral salt solution with a fixed $\mathrm{pOH}$ of $7$, whereas $0.01\,\mathrm{M}\,\mathrm{CH}_3\mathrm{COONa}$ undergoes anionic hydrolysis to yield a basic solution with a $\mathrm{pOH}$ less than $7$. Finally, $0.01\,\mathrm{M}\,\mathrm{NaOH}$ is a strong base that fully dissociates to provide hydroxide ions, resulting in a minimal $\mathrm{pOH}$ of $2$. Comparing these values leads to the decreasing order of $\mathrm{pOH}$ as $(A) > (D) > (C) > (B)$.

Determining whether a precipitate forms when mixing two salt solutions requires comparing the ion product, $Q$, to the solubility product constant, $K_{sp}$. If $Q$ exceeds $K_{sp}$, the concentrations of the ions in the mixture are high enough that the solution is supersaturated, and a precipitate must form.

Calculate the final ion concentrations by accounting for the dilution into the total 400 mL volume. The concentration of $\mathrm{Pb}^{2+}$ is $\frac{300 \times 0.134}{400} = 0.1005$ M, while the concentration of $\mathrm{Cl}^{-}$ is $\frac{100 \times 0.4}{400} = 0.1$ M. Applying these values to the expression Q = \[\mathrm{Pb}^{2+}]$$[\mathrm{Cl}^{-}]^2$results \in$0.1005 × (0.1)^2 = 1.005 × 10^{-3}$. Since this reaction quotient of$1.005 × 10^{-3}$is significantly larger than the given$K_{sp}$of$1.6 × 10^{-5}$, the system remains \in the state where$Q > K_{sp}$.

The solubility of a gas in a liquid is proportional to its partial pressure as defined by Henry's Law, providing a direct relationship between the gas above the solution and its concentration within it. Since $44\ \mathrm{g}$ of $\mathrm{CO}_2$ represents $1\ \mathrm{mol}$, dissolving this amount into $1\ \mathrm{kg}$ of water-which has a volume of $1\ \mathrm{L}$ given the density of $1\ \mathrm{g\ mL}^{-1}$-results in a $1\ \mathrm{M}$ solution at a partial pressure of $30\ \mathrm{bar}$. Scaling this relationship to the operational pressure of $3\ \mathrm{bar}$ indicates that the concentration of dissolved $\mathrm{CO}_2$ in the soft drink is $0.1\ \mathrm{M}$.

Carbonic acid acts as a weak acid in water, dissociating into $\mathrm{H}^+$ and $\mathrm{HCO}_3^-$ with a first dissociation constant of $4.0 \times 10^{-7}$. Using the standard approximation for a weak acid where the hydrogen ion concentration $[H^+]$ equals the square root of the product of the dissociation constant $K_a$ and the molar concentration $C$, we find:
$[H^+] = \sqrt{4.0 \times 10^{-7} \times 0.1} = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4}\ \mathrm{M}$
The $\mathrm{pH}$ is the negative logarithm of the hydrogen ion concentration, calculated as $-\log(2.0 \times 10^{-4})$, which simplifies to $4 - \log 2$. Using the value $\log 2 = 0.3$, the $\mathrm{pH}$ equates to $3.7$, which is expressed as $37 \times 10^{-1}$.

When a strong acid and a strong base are mixed, they undergo a neutralization reaction where hydroxide ions react with hydrogen ions to form water, and the pH of the mixture depends on the concentration of the excess reactant remaining in the total volume.

Determine the molar concentration of the stock solutions. Sodium hydroxide has a molar mass of 40 g/mol, so 4 g per 100 L equals 0.1 mol per 100 L, which is a concentration of $10^{-3} \, \text{M}$. Sulfuric acid ($\mathrm{H_2SO_4}$) has a molar mass of 98 g/mol, so 9.8 g per 100 L is 0.1 mol per 100 L, which is also a concentration of $10^{-3} \, \text{M}$. Because sulfuric acid is diprotic, it provides two moles of hydrogen ions per mole of the acid.

Calculate the moles contributed by the volumes being mixed. Using 40 L of the sodium hydroxide solution provides $40 \times 10^{-3} = 0.04$ moles of hydroxide ions. Combining this with 10 L of the sulfuric acid solution introduces $10 \times 10^{-3} \times 2 = 0.02$ moles of hydrogen ions. The neutralization reaction consumes 0.02 moles of hydroxide ions, leaving a net of $0.04 - 0.02 = 0.02$ moles of hydroxide ions in the total volume.

The final concentration of hydroxide ions is determined by dividing these remaining moles by the total volume of 50 L, resulting in $[\text{OH}^-] = 0.02 / 50 = 4 \times 10^{-4} \, \text{M}$. The pOH is calculated as $-\log(4 \times 10^{-4}) = 4 - 0.60 = 3.40$. Finally, the pH of the resultant solution is determined by the equation $14 - 3.40 = 10.60$.

Determining the final pH of this mixture relies on applying the Henderson-Hasselbalch equation, which requires identifying the concentrations of the weak acid and its conjugate base after all neutralization reactions are complete. Because the system contains a strong acid, a weak acid, and a strong base, the hydroxide ions will neutralize the stronger acid species sequentially until consumed.

First, determining the molar amounts shows the 500 mL solution contains 50 millimoles of acetic acid-based on the 3-gram mass-and 25 millimoles of HCl. Using a 20 mL portion of this solution captures 2 millimoles of acetic acid and 1 millimole of HCl. Adding 0.5 mL of 5 M NaOH provides 2.5 millimoles of hydroxide, which neutralizes the 1 millimole of HCl completely and then reacts with 1.5 millimoles of the acetic acid to form 1.5 millimoles of acetate salt. This process leaves behind 0.5 millimoles of unreacted acetic acid, allowing the use of $\mathrm{pH} = \mathrm{p}K_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$ with the ratio of salt to acid. Substituting the values gives $\mathrm{pH} = 4.75 + \log \frac{1.5}{0.5}$, which simplifies to $4.75 + \log 3$. Incorporating the given $\log 3 = 0.4771$, the final pH is approximately 5.23.

The neutralization of a strong acid by a strong base is defined by the rapid depletion of $\mathrm{H}^+$ ions near the equivalence point, which produces a characteristic sigmoidal titration curve. Initially, the pH is very low, but as 0.1 M $\mathrm{NaOH}$ is added to 100 mL of 0.1 M $\mathrm{HCl}$, the pH increases gradually before undergoing a sharp, vertical spike as the solution reaches the equivalence point. This abrupt transition occurs because the concentration of $\mathrm{H}^+$ ions falls drastically as they are consumed, causing the solution to pivot rapidly from acidic to basic.

An acidic buffer is created by mixing a weak acid with its conjugate base, which is typically supplied by a salt formed from that weak acid and a strong base. Combining 100 mL of 0.1 M $\mathrm{HCl}$ with 200 mL of 0.1 M $\mathrm{CH_3COONa}$ provides 10 millimoles of the strong acid $\mathrm{HCl}$ and 20 millimoles of the salt $\mathrm{CH_3COONa}$. The chemical reaction proceeds according to the equation:

$\mathrm{HCl} + \mathrm{CH_3COONa} \rightarrow \mathrm{CH_3COOH} + \mathrm{NaCl}$

Because $\mathrm{HCl}$ acts as the limiting reagent, all 10 millimoles are consumed to produce 10 millimoles of the weak acid $\mathrm{CH_3COOH}$, leaving 10 millimoles of the $\mathrm{CH_3COONa}$ salt unreacted. This final mixture of a weak acid and its conjugate base effectively forms an acidic buffer solution.

Mixing $\mathrm{H_2SO_4}$ with $\mathrm{NH_4OH}$ triggers an acid-base neutralization that leaves a mixture of the weak base $\mathrm{NH_4OH}$ and its conjugate acid $\mathrm{NH_4}^+$, effectively creating a basic buffer. Calculating the reactants yields 4 mmol of $\mathrm{H}^+$ from 20 mL of 0.1 $\mathrm{M}$ $\mathrm{H_2SO_4}$ and 6 mmol of $\mathrm{NH_4OH}$ from 30 mL of 0.2 $\mathrm{M}$ $\mathrm{NH_4OH}$. The neutralization reaction consumes all 4 mmol of $\mathrm{H}^+$, leaving 2 mmol of unreacted $\mathrm{NH_4OH}$ and forming 4 mmol of $\mathrm{NH_4}^+$. Applying the Henderson-Hasselbalch equation gives pOH = pK_b + \log \frac{[\mathrm{NH_4^+}]\}{[\mathrm{NH_4OH}]} = 4.7 + \log(\frac{4}{2}) = 5.0$, which leads to a final pH of$14 - 5.0 = 9.0$.

The electrical conductivity of weak acid solutions depends directly on the extent of ionization, which is dictated by the acid dissociation constant, $K_a$. Because each solution is prepared at the same concentration of 0.1 M, the substance with the highest $K_a$ value generates the most ions and consequently shows the greatest conductivity. With the established order of acid strength being Formic acid > Benzoic acid > Acetic acid, the conductivity follows this same decreasing sequence.