Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

The stoichiometry of the reaction relies on identifying the limiting reagent, which is determined by comparing the available reactants. Converting 2 g of sodium sulphate into moles using the molar mass of 143 g mol⁻¹ gives 14 mmol of sodium sulphate, which is completely consumed by the excess 100 mmol of calcium hydroxide according to the balanced reaction $\mathrm{Ca(OH)_2} + \mathrm{Na_2SO_4} \rightarrow \mathrm{CaSO_4} + 2\mathrm{NaOH}$. Because the molar ratio of sodium sulphate to calcium sulphate is one-to-one, 14 mmol of calcium sulphate is formed, and the amount of sodium hydroxide produced is 28 mmol.

Calculating the mass of the precipitate involves multiplying the 14 mmol of calcium sulphate by its molar mass of 136 g mol⁻¹, resulting in 1.9 g. The concentration of hydroxide ions is found by taking the 28 mmol of hydroxide produced and dividing it by the total volume of 0.1 L, which gives a final concentration of 0.28 mol L⁻¹.

Ammonium chloride is a salt derived from a strong acid and a weak base, which causes the solution to become acidic because the ammonium ion undergoes hydrolysis in water. The concentration of hydrogen ions produced can be calculated using the relation
$[H^{+}] = \sqrt{\frac{K_w \cdot c}{K_b}}$
where the water dissociation constant $K_w = 10^{-14}$, the concentration $c = 0.02$, and the base dissociation constant $K_b = 10^{-5}$. Substituting these values into the expression yields
$[H^{+}] = \sqrt{\frac{10^{-14} \cdot 2 \times 10^{-2}}{10^{-5}}} = \sqrt{2 \times 10^{-11}} = \sqrt{20} \times 10^{-6} \approx 4.47 \times 10^{-6} \text{ M}$
The pH is found by taking the negative logarithm of this concentration, which corresponds to $pH = -\log(4.47 \times 10^{-6})$. This simplifies to $6 - \log(\sqrt{20})$, and applying the property $\log(\sqrt{20}) = \frac{1}{2}(\log 2 + \log 10) = \frac{1}{2}(0.301 + 1) = 0.6505$ leads to a final pH of approximately $6 - 0.65 = 5.35$.

The solubility product constant, $K_{SP}$, defines the equilibrium state for a saturated solution where the product of ion concentrations, raised to their respective stoichiometric powers, equals the solubility product. Upon dissociation, one unit of $\mathrm{Zr}_3(\mathrm{PO}_4)_4$ breaks down into three $\mathrm{Zr}^{+4}$ ions and four $\mathrm{PO}_4^{-3}$ ions.

Defining molar solubility as $S$, the equilibrium concentrations are $3S$ for the zirconium ions and $4S$ for the phosphate ions. Substituting these values into the equilibrium expression gives:
$K_{SP} = (3S)^3(4S)^4$
Evaluating the exponents results in $27S^3$ and $256S^4$, which multiply to give:
$K_{SP} = 6912S^7$
Rearranging this equation to solve for $S$ provides the final relationship:
$S = \left( \frac{K_{SP}}{6912} \right)^{1/7}$

The solubility of a sparingly soluble salt like $\mathrm{Cd(OH)}_2$ is dictated by its solubility product constant, K_{sp} = \[\mathrm{Cd}^{2+}]$$[\mathrm{OH}^-]^2$, which remains invariant even as the concentration of common ions fluctuates \in different environments. Using the established properties of this substance, the solubility product is determined to be$24.9 × 10^{-15}$.

In a buffer solution with a $\mathrm{pH}$ of 12, the hydroxide ion concentration is fixed at $10^{-2} \ \mathrm{M}$ because the $\mathrm{pOH}$ is 2. Substituting this known hydroxide concentration into the equilibrium expression gives $24.9 \times 10^{-15} = s' \times (10^{-2})^2$, where $s'$ represents the new solubility. Solving for $s'$ results in $24.9 \times 10^{-11} \ \mathrm{M}$, which simplifies to $2.49 \times 10^{-10} \ \mathrm{M}$.

The solubility of a sparingly soluble salt like silver carbonate is significantly reduced in the presence of a common ion, a phenomenon known as the common ion effect. The dissolution equilibrium for this salt is represented by the following equation:

$\mathrm{Ag_2CO_3}(s) \rightleftharpoons 2\mathrm{Ag}^+(aq) + \mathrm{CO_3^{2-}}(aq)$

The equilibrium constant expression is K_{sp} = \[\mathrm{Ag}^+]^2[\mathrm{CO_3^{2-}}]$.

Since the solution already contains 0.1 M silver nitrate, the initial silver ion concentration is 0.1 M. If S is the molar solubility of silver carbonate, the concentration of carbonate ions at equilibrium is S, and the silver ion concentration becomes $0.1 + 2S$. Because the value of the solubility product is extremely low, the contribution of $2S$ to the total silver ion concentration is negligible, allowing us to approximate the concentration as 0.1. Substituting these values into the equilibrium expression yields $8 \times 10^{-12} = (0.1)^2 \times S$. Dividing $8 \times 10^{-12}$ by 0.01 provides the molar solubility, resulting in a value of $S = 8 \times 10^{-10}$ M.

Rainwater is naturally slightly acidic because it absorbs carbon dioxide from the atmosphere, which reacts with the water droplets to form weak carbonic acid. While pure water is typically neutral, this atmospheric interaction lowers the pH of falling rain to approximately 5.6. Consequently, unpolluted rainwater consistently exhibits this characteristic acidity rather than a strictly neutral value.

The dissolution of aluminum hydroxide is governed by the common ion effect, which limits the solubility of the salt when introduced into a solution already containing hydroxide ions. Because sodium hydroxide is a strong electrolyte, it dissociates completely, establishing a fixed hydroxide concentration of $0.2 \text{ M}$ in the mixture. Using the solubility product expression K_{sp} = \[\mathrm{Al^{3+}}]$$[\mathrm{OH^-}]^3$, we can substitute the provided values into the equation$2.4 × 10^{-24} = \mathrm{Al^{3+}}^3$. Calculating$(0.2)^3$results \in$0.008$, and dividing$2.4 × 10^{-24}$by this value yields a molar solubility of$3 × 10^{-22}$ for the aluminum ions.

The reaction of 0.04 mol of $H_2SO_4$ with 0.04 mol of $NaOH$ results in 0.04 mol of excess $H^+$ ions in 0.8 L of solution, yielding $[H^+] = 0.05 \text{ M}$ and $pH = -\log(0.05) \approx 1.3$. The auto-ionization of water, $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$, is an endothermic process; increasing the temperature shifts the equilibrium to the right, thereby increasing the ionic product $K_w$. For a monobasic acid, the equilibrium constant is expressed as K_a = \frac{[H^+]\\alpha}{1-\alpha}$, where$10^{-5} = \frac{10^{-5} × 0.5}{0.5}$confirms that a degree of dissociation of 50$pH$ of 5. Le Chatelier's Principle applies directly to the common-ion effect, as adding a common ion causes the equilibrium of a weak electrolyte to shift toward the undissociated form to counteract the increase in concentration, making statement d incorrect.

Since the titration begins with a sodium hydroxide solution in the flask, the initial pH is high because hydroxide ions, denoted as $\text{OH}^-$, dominate the mixture. As the 0.1 M $\text{HCl}$ solution is gradually added from the burette, the neutralization reaction consumes these hydroxide ions, causing the pH to decrease steadily before dropping sharply at the equivalence point and eventually flattening out in the acidic region.

A salt formed from a strong base and a weak acid will consistently yield the most basic aqueous solution because the anion undergoes hydrolysis to release hydroxide ions. Among the given options, $CH_3COOK$ is derived from the strong base potassium hydroxide and the weak acid acetic acid, causing the $CH_3COO^-$ ion to undergo anionic hydrolysis and resulting in a higher pH. Other listed salts, such as $FeCl_3$, are derived from strong acids or involve weak bases, preventing them from creating a basic solution comparable to that of $CH_3COOK$.

Precipitation begins when the ionic product of a solution reaches the solubility product constant, $K_{sp}$. Since 50 mL of a 1 M solution of $\mathrm{Na_2SO_4}$ is added to reach a final volume of 500 mL, the total amount of sulfate ions introduced is 50 mmol. Dividing this amount by the final volume results in a sulfate concentration of $[\mathrm{SO_4^{2-}}] = 0.1 \text{ M}$.

At the threshold of precipitation, the product of the concentrations of barium and sulfate ions must satisfy the solubility product expression, K_{sp} = \[\mathrm{Ba^{2+}}]$$[\mathrm{SO_4^{2-}}]$ = 1 \times 10^{-10}$. Substituting the known sulfate concentration into this equation yields the barium ion concentration in the final mixture:
$[\mathrm{Ba^{2+}}] = \frac{1 \times 10^{-10}}{0.1} = 10^{-9} \text{ M}$

Because the solution was originally 450 mL before the sulfate solution was added to it, the initial concentration of barium ions can be determined using the dilution principle $M_1 V_1 = M_2 V_2$. With the final concentration $M_2 = 10^{-9} \text{ M}$, a final volume $V_2 = 500 \text{ mL}$, and an initial volume $V_1 = 450 \text{ mL}$, the original concentration $M_1$ is calculated as:
$M_1 = \frac{10^{-9} \times 500}{450} = 1.1 \times 10^{-9} \text{ M}$

The pH of a mixture of strong acid and strong base is determined by the final concentration of excess hydrogen ions, calculated using the formula $[H^+] = \frac{M_1V_1 - M_2V_2}{V_1 + V_2}$. To achieve a pH of 1, the resulting concentration $[H^+]$ must be equal to $10^{-1} = 0.1\ \mathrm{M}$. By using a concentration of $0.2\ \mathrm{M}$ for the solutions in option A, the calculation yields $[H^+] = \frac{0.2\ \mathrm{M}(75\ \mathrm{mL}) - 0.2\ \mathrm{M}(25\ \mathrm{mL})}{75\ \mathrm{mL} + 25\ \mathrm{mL}} = \frac{15\ \mathrm{mmol} - 5\ \mathrm{mmol}}{100\ \mathrm{mL}} = 0.1\ \mathrm{M}$, which results in a pH of $-\log(0.1) = 1$.

A Lewis acid is formally defined as any chemical species that functions as an electron-pair acceptor. Both $BCl_3$ and $AlCl_3$ contain central atoms that are electron-deficient, meaning their valence shells contain fewer than eight electrons and can readily accommodate incoming electron pairs to achieve a stable octet configuration.

Methyl orange functions as an indicator that transitions from yellow in basic conditions to pinkish red in acidic conditions, typically operating within a pH range of 3.1 to 4.4. When a weak base is titrated against a strong acid, the equivalence point occurs within this acidic range, meaning the solution begins as a weak base and ends as an acidic mixture. Consequently, the titration proceeds from the yellow color of the initial basic solution to the pinkish red color of the acidic endpoint, which characterizes the correct combination for this titration.

A Lewis acid is defined as a chemical species that possesses an incomplete octet or vacant orbitals, allowing it to act as an electron-pair acceptor. In the molecule $B(CH_3)_3$, the central Boron atom is bonded to three methyl groups and is surrounded by only six valence electrons, leaving it electron-deficient with an empty p-orbital that is ready to accept a lone pair. In contrast, $PH_3$ and $NF_3$ possess lone pairs on their central atoms and function as Lewis bases, while $NaH$ acts as a source of the hydride ion $H^-$, which behaves as a strong base.

The solubility product constant $K_{sp}$ for the dissociation equilibrium $\mathrm{PbCl_2} \rightleftharpoons \mathrm{Pb}^{2+} + 2\mathrm{Cl}^-$ is defined as K_{sp} = \[\mathrm{Pb}^{2+}]$$[\mathrm{Cl}^-]^2$. By denoting the molar solubility as$s$mol/L, the concentrations at equilibrium are$[\mathrm{Pb}^{2+}] = s$and$[\mathrm{Cl}^-] = 2s$, which gives the expression$K_{sp} = s(2s)^2 = 4s^3$. Given$K_{sp} = 3.2 × 10^{-8}$, the molar solubility is calculated as$s = \sqrt[3]{\frac{3.2 × 10^{-8}}{4}} = \sqrt[3]{8 × 10^{-9}} = 2 × 10^{-3} \text{ mol/L}$. The molar mass of$\mathrm{PbCl_2}$is$207 + 2(35.5) = 278 \text{ g/mol}$, and the number of moles present \in$0.1\text{ g}$of the salt is$\frac{0.1}{278} \text{ mol}$. Since solubility$s$defines the maximum concentration \in mol/L, the required volume$V$is$V = \frac{\text{moles}}{s} = \frac{0.1/278}{2 × 10^{-3}} = \frac{0.1}{0.556} \approx 0.1798 \text{ L}$, which is approximately 0.18 L.

The equilibrium concentration of sulfide ions in this solution is determined by the combined dissociation constant of hydrogen sulfide, defined as $K_{net} = K_{a1} \times K_{a2} = 1.2 \times 10^{-20}$. Because $HCl$ is a strong acid, it fully dissociates to provide a high concentration of hydrogen ions, $[H^+] = 0.20$ M, which shifts the equilibrium to the left and suppresses the ionization of the weak acid $H_2S$. We can use the overall equilibrium expression to find the concentration of $S^{2-}$ ions while keeping the concentration of $H_2S$ constant at 0.10 M.

The expression for the total dissociation is:
$K_{net} = \frac{[H^+]^2 \$[S^{2-}]\$}{[H_2S]}$
Substituting the known values into this expression yields:
$1.2 \times 10^{-20} = \frac{(0.20)^2 \$[S^{2-}]\$}{0.10}$
Solving for $[S^{2-}]$ gives:
$[S^{2-}] = \frac{1.2 \times 10^{-20} \times 0.10}{(0.20)^2} = \frac{1.2 \times 10^{-21}}{0.04} = 3 \times 10^{-20}$

The pH of a salt solution formed from a weak acid and a weak base depends on the relative dissociation constants of the acid and the base, which allows the acidity or basicity of the solution to be determined quantitatively. The relationship between these values and the pH is expressed by the equation
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Using the specific constants provided, where $pK_a$ is 3.2 and $pK_b$ is 3.4, we substitute these into the equation to obtain $pH = 7 + \frac{1}{2}(3.2 - 3.4)$. This simplifies to $pH = 7 + \frac{1}{2}(-0.2)$, which further resolves to $pH = 7 - 0.1$. Completing this arithmetic results in $pH = 6.9$.

The Henderson-Hasselbalch equation for an acidic buffer is $pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$. With the ionization constant $K_a = 10^{-5}$, the $pK_a$ is calculated as $pK_a = -\log(10^{-5}) = 5$. Substituting $pH = 6$ and $pK_a = 5$ into the equation yields $6 = 5 + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$, which simplifies to $1 = \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$, indicating the ratio $\frac{[\text{Salt}]}{[\text{Acid}]} = 10^1 = 10:1$.

The reaction of a weak base $NH_3$ with a strong acid $HCl$ forms a basic buffer when the weak base is present in excess, following the chemical equation $NH_3 + HCl \rightarrow NH_4Cl$. The initial amount of $NH_3$ is $0.050 \times 0.2 = 0.010$ moles, while the amount of $HCl$ is $0.025 \times 0.2 = 0.005$ moles, which leads to $0.005$ moles of $NH_3$ remaining and $0.005$ moles of $NH_4Cl$ salt being produced.

Applying the Henderson-Hasselbalch equation for a basic buffer, $pOH = pK_b + \log \left( \frac{[\text{salt}]}{[\text{base}]} \right)$, the ratio of concentrations equals the ratio of moles because the components occupy the same total volume. Substituting the values yields $pOH = 4.75 + \log \left( \frac{0.005}{0.005} \right) = 4.75$. The final pH is calculated using $pH = 14 - pOH$, which results in $14 - 4.75 = 9.25$.