Ionic Equilibrium – JEE Main ChemistryPractice Questions & PYQs

According to Kohlrausch's Law of Independent Migration of Ions, the molar conductivity of an electrolyte at infinite dilution is the sum of the ionic conductivities of its constituent ions. The ionic conductance of the ammonium ion is calculated as $\lambda^\circ(\text{NH}_4^+) = 185 - 70 = 115$ S cm$^2$ mol$^{-1}$, based on the values provided for ammonium chloride and chloride ions. Consequently, the infinite dilution molar conductance of ammonium hydroxide is $\Lambda^\circ_m(\text{NH}_4\text{OH}) = 115 + 170 = 285$ S cm$^2$ mol$^{-1}$. The degree of dissociation is determined by the ratio of the molar conductance at the given concentration to that at infinite dilution, $\alpha = \frac{85.5}{285} = 0.3$. As the degree of dissociation is expressed as $x \times 10^{-1}$, the value of $x$ is 3.

The pH of an acidic solution is defined by the formula $\text{pH} = -\log[H^+]$. Given an initial pH of 1.0, the initial concentration of hydrogen ions is $[H^+] = 10^{-1} = 0.1$ molar. Diluting the solution with an equal volume of water doubles the total volume, which reduces the concentration of the solute according to the relationship $M_1V_1 = M_2V_2$. With the volume doubling, the new concentration is:

$[H^+]_{new} = \frac{0.1 \times V}{2V} = 0.05 \text{ molar}$

The resulting pH is determined by taking the negative logarithm of the new concentration:

$\text{pH} = -\log(0.05) = -\log(5 \times 10^{-2}) = 2 - \log 5$

Using the identity $\log 5 = \log(10/2) = \log 10 - \log 2$ and the provided value $\log 2 = 0.30$, the calculation proceeds as follows:

$\text{pH} = 2 - (1 - 0.30) = 2 - 0.70 = 1.3$

The auto-ionization of water, represented by the equilibrium $H_2O \rightleftharpoons H^+ + OH^-$, is an endothermic process, meaning that heat shifts the balance toward the production of ions. When the temperature of water is raised from 25 °C to 80 °C, the increased thermal energy causes more water molecules to dissociate, which elevates the concentration of hydrogen ions in the solution. Because pH is determined by the negative logarithm of the hydrogen ion concentration, an increase in these ions leads to a decrease in the pH value. Consequently, the pH of pure water drops from 7 at room temperature to approximately 6.93 at 80 °C.

The equilibrium of a weak acid is fundamentally defined by the acid dissociation constant, which relates the concentration of the dissociated ions to the remaining undissociated molecules. For a weak acid $HA$ dissociating into $H^+$ and $A^-$ with an initial concentration $C$ and a degree of dissociation $x$, the equilibrium concentration of hydrogen ions is $[H^+] = Cx$ and the concentration of the undissociated acid is $[HA] = C(1-x)$. Substituting these into the equilibrium expression K_a = \frac{[H^+]\$[A^-]$}{[HA]}$leads to$K_a = \frac{(Cx)(Cx)}{C(1-x)}$, which simplifies to$K_a = $[H^+]$ \left( \frac{x}{1-x} \right)$. Rearranging this to solve for the hydrogen ion concentration yields$[H^+] = K_a \left( \frac{1-x}{x} \right)$. Taking the negative logarithm of both sides gives$-\log $[H^+]$ = -\log K_a - \log \left( \frac{1-x}{x} \right)$, which is equivalent to$\text{pH} = \text{pK}_a + \log \left( \frac{x}{1-x} \right)$. Subtracting$\text{pK}_a$from both sides results \in the final expression$\text{pH} - \text{pK}_a = \log \left( \frac{x}{1-x} \right)$.

The Van't Hoff factor provides a direct relationship between the theoretical behavior of a solute and its actual particle count in solution when dissociation occurs. By taking the ratio of the normal molar mass of 164 to the observed molar mass of 65.6, we determine the Van't Hoff factor $i = \frac{164}{65.6}$, which results in a value of 2.5.

When $MX_2$ dissociates according to the reaction $MX_2 \rightleftharpoons M^{2+} + 2X^-$, the total number of particles after dissociation is governed by the degree of ionization $\alpha$, expressed as $i = 1 + 2\alpha$. Setting this theoretical expression equal to the calculated value gives $1 + 2\alpha = 2.5$. Solving this equation leads to $2\alpha = 1.5$, which simplifies to $\alpha = 0.75$, corresponding to a percent degree of ionization of 75%.

The solubility product constant, $K_{\mathrm{sp}}$, quantifies the equilibrium state of a saturated solution of a sparingly soluble salt, where the product of the ion concentrations, each raised to the power of its stoichiometric coefficient, equals $K_{\mathrm{sp}}$. Zirconium phosphate dissociates according to the reaction $\mathrm{Zr}_3(\mathrm{PO}_4)_4(\mathrm{s}) \rightleftharpoons 3\mathrm{Zr}^{4+}(\mathrm{aq}) + 4\mathrm{PO}_4^{3-}(\mathrm{aq})$, which means that for a molar solubility $s$, the concentration of $\mathrm{Zr}^{4+}$ is $3s$ and the concentration of $\mathrm{PO}_4^{3-}$ is $4s$. Substituting these equilibrium concentrations into the solubility product expression gives $K_{\mathrm{sp}} = (3s)^3 \cdot (4s)^4$, which evaluates to $K_{\mathrm{sp}} = 27s^3 \cdot 256s^4$, simplifying to $K_{\mathrm{sp}} = 6912s^7$. Solving this equation for $s$ results in the expression $\left( \frac{K_{\mathrm{sp}}}{6912} \right)^{\frac{1}{7}}$.

The precipitation of a sparingly soluble salt from a solution occurs when the ionic product of its constituent ions exceeds its solubility product, $K_{\mathrm{sp}}$. By calculating the exact concentration of hydroxide ions required for each compound to begin precipitating, one can determine which salt will form first as $\mathrm{NH}_4\mathrm{OH}$ is added.

For $\mathrm{A}(\mathrm{OH})_2$, the equilibrium condition is [A^{2+}]\[\mathrm{OH}^-]^2 = 9 × 10^{-10}$, and given that the concentration of$A^{2+}$is$1\ \mathrm{M}$, the precipitation threshold is$[\mathrm{OH}^-] > \sqrt{9 × 10^{-10}} = 3 × 10^{-5}\ \mathrm{M}$. Similarly, for$\mathrm{B}(\mathrm{OH})_3$, the equilibrium condition is$[B^{3+}]$[\mathrm{OH}^-]^3 = 27 \times 10^{-18}$, and with$[B^{3+}] = 1\ \mathrm{M}$, the threshold for precipitation is$[\mathrm{OH}^-] > \sqrt[3]{27 \times 10^{-18}} = 3 \times 10^{-6}\ \mathrm{M}$. Because the solution reaches the lower hydroxide concentration of$3 \times 10^{-6}\ \mathrm{M}$before it reaches the higher concentration of$3 \times 10^{-5}\ \mathrm{M}$,$\mathrm{B}(\mathrm{OH})_3$will precipitate before$\mathrm{A}(\mathrm{OH})_2$.

Buffer solutions maintain a relatively constant pH by utilizing the equilibrium between a weak acid and its conjugate base to neutralize added strong acids or bases. This buffering capacity can be quantified using the Henderson-Hasselbalch equation, which determines the shift in $pH$ based on the logarithmic ratio of salt to base concentrations as the system adjusts to the new chemical environment.

Since the starting buffer contains equal amounts of ammonia and ammonium chloride, the initial $pOH$ is equivalent to the given $pK_b$ of 4.745, which results in an initial $pH = 14 - 4.745 = 9.255$. When 0.05 mol of hydrochloric acid is added, it reacts with 0.05 mol of the ammonia to form ammonium ions, thereby reducing the ammonia content to 0.05 mol and increasing the ammonium chloride concentration to 0.15 mol. The updated $pOH$ is calculated as $pOH = 4.745 + \log \left( \frac{0.15}{0.05} \right)$, which simplifies to $4.745 + \log 3 = 5.222$. Subtracting this $pOH$ from 14 gives a new $pH = 8.778$, creating a difference of $0.477$ from the original $pH$. Expressing this shift as $47.7 \times 10^{-2}$ and rounding to the nearest integer gives 48.

Conductometric titration of a mixture containing a strong acid like HCl and a weak acid like $\mathrm{CH}_3\mathrm{COOH}$ relies on the differing ionic mobilities of the species in solution. Because the $\mathrm{H}^{+}$ ions from the strong acid have much higher ionic mobility than the $\mathrm{Na}^{+}$ ions introduced by the NaOH titrant, the reaction $\mathrm{HCl} + \mathrm{NaOH} \longrightarrow \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}$ causes a steady decrease in conductivity until the HCl is entirely consumed. Once the strong acid is neutralized, the subsequent reaction with the weak acid, $\mathrm{CH}_3\mathrm{COOH} + \mathrm{NaOH} \longrightarrow \mathrm{CH}_3\mathrm{COONa} + \mathrm{H}_2\mathrm{O}$, proceeds differently, causing the conductivity to increase as salt and hydroxide ions accumulate.

The equivalence point where the strong acid reaction concludes provides the data necessary to calculate the original concentration of the acid. Given the total volume of 40 mL and the concentration of the NaOH titrant at 0.1 M, equating the moles of HCl to the moles of NaOH consumed at the point of complete neutralization of the strong acid confirms that the concentration of HCl in the original mixture is 0.005 M.

Precipitation of the salt $AY_2$ occurs when the ionic product Q_{sp} = \[A^{2+}]$$[Y^-]^2$exceeds the solubility product constant$K_{sp} = 5.2 × 10^{-7}$. Because equal volumes of the two solutions are mixed, each component is diluted by half, resulting \in final concentrations of$1.0 × 10^{-2}$M for both$A^{2+}$and$Y^-$when starting from$2.0 × 10^{-2}$M solutions. Calculating the ionic product gives$Q_{sp} = (1.0 × 10^{-2})(1.0 × 10^{-2})^2 = 1.0 × 10^{-6}$. Since this product is greater than the given$K_{sp}$, the ions combine to form a solid precipitate.

The solubility product constant, $K_{sp}$, determines the equilibrium state of a sparingly soluble salt based on the product of its constituent ion concentrations, with each concentration raised to its stoichiometric coefficient. For the salt $\mathrm{AB}_2$, the dissolution is defined by the equilibrium $\mathrm{AB}_{2(s)} \rightleftharpoons \mathrm{A}^{2+}_{(aq)} + 2\mathrm{B}^{-}_{(aq)}$, which gives the relationship K_{sp} = \[\mathrm{A}^{2+}]$$[\mathrm{B}^{-}]^2$.

With the molar concentration of $\mathrm{A}^{2+}$ given as $1.2 \times 10^{-4} \mathrm{M}$ and $\mathrm{B}^{-}$ as $0.24 \times 10^{-3} \mathrm{M}$, we can simplify the calculation by writing the second concentration as $2.4 \times 10^{-4} \mathrm{M}$. Substituting these into the $K_{sp}$ expression results in $K_{sp} = (1.2 \times 10^{-4}) \times (2.4 \times 10^{-4})^2$. Evaluating the exponent produces $5.76 \times 10^{-8}$, and multiplying this value by $1.2 \times 10^{-4}$ yields the final result of $6.91 \times 10^{-12} \mathrm{M}^3$.

According to Le Chatelier's principle, an equilibrium system adjusts to minimize changes in concentration by shifting toward the side that consumes any added species. The conversion between the dichromate and chromate ions is governed by the availability of hydroxide ions, expressed by the reaction $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 2\mathrm{OH}^- \rightleftharpoons 2\mathrm{CrO}_4^{2-} + \mathrm{H}_2\mathrm{O}$. Consequently, in a basic medium, the increased presence of $\mathrm{OH}^-$ ions reacts with the dichromate species and effectively forces the equilibrium to shift to the right to produce more chromate ions.

Buffer solutions consisting of a weak acid and its conjugate base follow the Henderson-Hasselbalch equation, which relates the solution $\text{pH}$ to the ratio of the salt and acid concentrations. Because the benzoic acid and sodium benzoate are provided at the same concentration of $1\text{ M}$, the ratio of their molar amounts directly corresponds to the ratio of their volumes, $V_{\text{s}}$ and $V_{\text{a}}$. Using the expression $\text{pH} = pK_a + \log \left( \frac{V_{\text{s}}}{V_{\text{a}}} \right)$, substituting the known values gives:

$4.5 = 4.2 + \log \left( \frac{V_{\text{s}}}{V_{\text{a}}} \right)$

This simplifies to $\log \left( \frac{V_{\text{s}}}{V_{\text{a}}} \right) = 0.3$, implying that the ratio $\frac{V_{\text{s}}}{V_{\text{a}}} = 2$. With the total volume constraint $V_{\text{s}} + V_{\text{a}} = 300$, replacing $V_{\text{s}}$ with $2V_{\text{a}}$ results in:

$3V_{\text{a}} = 300$

Consequently, the volume of benzoic acid is $100\text{ mL}$.

Ammonium carbonate is a salt derived from a weak base and a weak acid, and its aqueous solution is inherently basic because the dissociation constant of the base is greater than the dissociation constant of the acid. The overall acidity or basicity of such salt solutions is fundamentally determined by the values of $K_a$ and $K_b$ associated with the parent acid and base. This relationship is mathematically represented as $pH = \frac{1}{2}(pK_w + pK_a - pK_b)$, which confirms that the ionization behavior depends entirely on the relative strengths of these species.

A buffer solution is defined by the precise pairing of a weak acid or weak base with its corresponding conjugate, rendering the first statement incorrect because it erroneously suggests that these components can be combined in arbitrary quantities. In contrast, the second statement is accurate because human blood operates as a vital natural buffer system, utilizing the chemical balance between carbonic acid $\mathrm{H}_2\mathrm{CO}_3$ and the bicarbonate ion $\mathrm{HCO}_3^{-}$ to effectively stabilize its pH levels.

Solubility product constants describe the equilibrium between a solid ionic compound and its dissolved ions in a saturated solution. For calcium phosphate, the dissociation equation is $\mathrm{Ca}_3(\mathrm{PO}_4)_2 \rightleftharpoons 3\mathrm{Ca}^{2+} + 2\mathrm{PO}_4^{3-}$, where the concentration of calcium ions is three times the molar solubility $S$, and the concentration of phosphate ions is two times $S$. This results in the expression $K_{\mathrm{sp}} = (3S)^3(2S)^2$, which simplifies to $108S^5$.

Since the solubility is provided as $W$ g per 100 mL, the molarity $S$ is found by scaling the mass to 1000 mL and dividing by the molar mass $M$, which gives $S = \frac{10W}{M}$ mol/L. Substituting this value into the solubility product expression yields $108 \times (\frac{10W}{M})^5$, or $108 \times 10^5 \times (\frac{W}{M})^5$. Because $108 \times 10^5$ is equivalent to $1.08 \times 10^7$, the approximate value of the solubility product is $10^7 \left(\frac{W}{M}\right)^5$.

Precipitation of a sparingly soluble salt like magnesium hydroxide occurs the moment the ionic product in the solution reaches its solubility product constant, $K_{\mathrm{sp}}$. For the dissociation equilibrium of the salt, the condition for the start of precipitation is defined by the equation:

$K_{\mathrm{sp}} = \$[\mathrm{Mg}^{2+}]\$\$[\mathrm{OH}^-]^2$

Substituting the values $1 \times 10^{-11}$ for the solubility product and 0.10 M for the magnesium ion concentration results in the equation 1 \times 10^{-11} = 0.10 \times \[\mathrm{OH}^-]^2$. Solving this expression for the hydroxide concentration yields$1 × 10^{-5}$ M. Since the pOH is the negative logarithm of the hydroxide ion concentration, the pOH is 5. Using the standard relationship where the sum of pH and pOH equals 14, the pH at which precipitation begins is 9.

Osmotic pressure is a colligative property that depends on the total concentration of all solute particles in the solution, requiring us to account for the partial dissociation of the weak acid HX. We determine the equilibrium concentration of the species based on the dissociation reaction $\mathrm{HX} \rightleftharpoons \mathrm{H}^+ + \mathrm{X}^-$. Given an initial concentration of $0.03 \, \mathrm{M}$ and a small degree of dissociation $x$, the equilibrium concentrations are $0.03 - x$ for $\mathrm{HX}$ and $x$ for both $\mathrm{H}^+$ and $\mathrm{X}^-$. Using the dissociation constant $K_a = 1.2 \times 10^{-5}$, we set up the equilibrium equation:
$K_a = \frac{x^2}{0.03 - x}$
Because the acid is weak, we approximate $0.03 - x \approx 0.03$, which simplifies the expression to $x^2 = 0.03 \times 1.2 \times 10^{-5} = 3.6 \times 10^{-7}$, yielding $x = 6 \times 10^{-4} \, \mathrm{M}$.

The total molar concentration of all particles in the solution is the sum of the equilibrium concentrations, represented as $(0.03 - x) + x + x = 0.03 + x$. Substituting $x = 0.0006$, we obtain a total concentration of $0.0306 \, \mathrm{M}$. The osmotic pressure is then calculated using:
$\Pi = CRT$
Substituting $C = 0.0306 \, \mathrm{M}$, $R = 0.083 \, \mathrm{L} \, \mathrm{bar} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$, and $T = 300 \, \mathrm{K}$ results in $\Pi = 0.0306 \times 0.083 \times 300 = 0.76194 \, \mathrm{bar}$. Expressing this value in the required units of $\times 10^{-2}$ bar gives $76.194 \times 10^{-2}$, which rounds to the nearest integer of 76.

Ammonium acetate behaves as a salt derived from the reaction between a weak acid, acetic acid, and a weak base, ammonium hydroxide. Because the strength of this specific weak acid and weak base are identical, as evidenced by their equal dissociation constants of $1.8 \times 10^{-5}$, the resulting solution maintains a neutral balance. The calculation for the pH of such a salt is governed by the expression $\mathrm{pH} = \frac{pK_w + pK_a - pK_b}{2}$. Since the numerical values for $K_a$ and $K_b$ are equal, their negative logarithms $pK_a$ and $pK_b$ are also equal, causing them to cancel each other out in the formula. Consequently, the equation simplifies to $\mathrm{pH} = \frac{pK_w}{2}$, which at room temperature yields a pH value of 7.

Indicators identify the equivalence point of a titration by undergoing a color change within the sharp vertical pH range of the neutralization curve. Because the reaction between a weak acid and a weak base results in a very gradual change in pH rather than a steep jump, standard indicators like methyl orange or phenolphthalein cannot effectively detect the endpoint. While methyl orange is appropriate for strong acid vs. weak base and strong acid vs. strong base titrations, and phenolphthalein is suitable for weak acid vs. strong base and strong acid vs. strong base titrations, the claim that methyl orange can be used for weak acid vs. weak base systems is false.