Equilibrium – NEET UG ChemistryPractice Questions & PYQs

According to the Brønsted-Lowry theory, an amphoteric substance must be capable of functioning as both a proton donor and a proton acceptor. While species like $\mathrm{HCO}_3^-$, $\mathrm{NH}_3$, and $\mathrm{HSO}_4^-$ possess the necessary structure to participate in reactions in both capacities, $\mathrm{HCl}$ acts strictly as a proton donor. Because $\mathrm{HCl}$ can only release a proton and lacks the ability to accept one, it cannot function as a Brønsted base.

The solubility product constant represents the equilibrium between a sparingly soluble solid and its constituent ions in a saturated solution. For calcium hydroxide, the dissociation follows the equilibrium equation $\mathrm{Ca}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}$.

Knowing that the pH of the solution is 9, the pOH is found by subtracting this value from 14, which gives 5. This pOH value indicates that the concentration of hydroxide ions is $10^{-5}$ M. Based on the stoichiometry of the dissociation, the concentration of calcium ions must be half the concentration of the hydroxide ions, resulting in a value of $\frac{10^{-5}}{2}$ M.

To find the solubility product, use the expression $K_{\mathrm{sp}} =$[\mathrm{Ca}^{2+}]$$[\mathrm{OH}^{-}]^{2}$. Substituting the molar concentrations into this formula gives $\left( \frac{10^{-5}}{2} \right) × (10^{-5})^{2}$, which evaluates to $0.5 × 10^{-15}$.

$\mathrm{NaOH}$ acts as a strong base that dissociates completely in water, meaning the $\mathrm{OH}^-$ concentration equals the molarity of the solution, which is 0.01 M. Using the definition of pOH through the expression
$pOH = -\log[\mathrm{OH}^-]$
we determine that the pOH is 2. By applying the standard equilibrium relationship
$pH + pOH = 14$
we subtract the pOH from 14 to find the pH. This calculation leads to a pH of 12.

The solubility of a sparingly soluble salt like $\mathrm{CaF}_2$ is dictated by the common ion effect when it is dissolved in a solution already containing one of its constituent ions. In a $0.1 \text{ M}$ solution of $\mathrm{NaF}$, the large concentration of fluoride ions suppresses the dissociation of $\mathrm{CaF}_2$ to ensure the solubility product constant, $K_{\mathrm{sp}}$, remains satisfied.

Mathematically, we express the equilibrium as $K_{\mathrm{sp}} =$[\mathrm{Ca}^{2+}]$$[\mathrm{F}^{-}]^2$. Because $\mathrm{NaF}$ acts as a strong electrolyte, it fully dissociates to provide a fixed fluoride ion concentration of $0.1 \text{ M}$, which is significantly higher than the amount contributed by the sparingly soluble salt. By assigning $s$ as the molar solubility of $\mathrm{CaF}_2$, we substitute $[\mathrm{Ca}^{2+}] = s$ and $[\mathrm{F}^{-}] = 0.1 \text{ M}$ into the expression to get $5.3 × 10^{-11} = s(0.1)^2$. Dividing by $(0.1)^2$, or $0.01$, yields $s = 5.3 × 10^{-9} \text{ mol L}^{-1}$.

According to the Brønsted-Lowry concept, a conjugate base is generated when an acid molecule loses a proton, which means we simply remove one $\mathrm{H}^{+}$ ion from its chemical formula. Removing a proton from $\mathrm{H}_2\mathrm{O}$ leaves behind the hydroxide ion $\mathrm{OH}^{-}$, while the same process applied to $\mathrm{HF}$ results in the fluoride ion $\mathrm{F}^{-}$. Identifying these species relies entirely on this simple subtraction of a single hydrogen cation from the acid.

Calculating the final pH of a mixture containing a strong acid and a strong base requires identifying the remaining concentration of hydrogen ions, governed by the formula $pH = -\log[H^+]$. Achieving a $pH$ of 1 requires the resulting molarity of hydrogen ions to be 0.1 M. We determine the milliequivalents of each component by multiplying their respective volumes by their concentration. In the third mixture, the acid provides 15 milliequivalents and the base provides 5 milliequivalents. The excess acid remaining after neutralization is calculated as $15 - 5 = 10$ milliequivalents. With a total volume of 100 mL, the resulting molarity is defined by the ratio $10 / 100 = 0.1$. Substituting this into the expression gives $pH = -\log(0.1) = 1$, which confirms the required acidity level.

Le Chatelier's principle dictates that systems at equilibrium shift to counteract imposed external changes, allowing us to optimize the yield for the reaction $A_2(g) + B_2(g) \rightleftharpoons X_2(g)$. Because the forward process results in a decrease from two moles of gaseous reactants to one mole of gaseous product, increasing the system's pressure compels the equilibrium to shift toward the product side to reduce the total number of gas molecules. Furthermore, since the reaction is exothermic with an enthalpy change of $\Delta_r H = -X\ \mathrm{kJ}$, lowering the temperature encourages the system to move in the forward direction to release heat and re-establish equilibrium.

The solubility product constant, $K_{\text{sp}}$, characterizes the equilibrium established when a sparingly soluble salt dissolves until the solution becomes saturated. Converting the mass solubility of $\mathrm{BaSO}_4$ from $2.42 \times 10^{-3} \text{gL}^{-1}$ to molarity requires dividing by the molar mass of $233 \text{gmol}^{-1}$, resulting in a value of $1.04 \times 10^{-5} \text{molL}^{-1}$. Because the dissociation reaction $\mathrm{BaSO}_4(\text{s}) \rightleftharpoons \mathrm{Ba}^{2+}(\text{aq}) + \mathrm{SO}_4^{2-}(\text{aq})$ produces one mole of each ion for every mole of dissolved salt, the solubility product is determined by squaring this molar solubility, which gives $(1.04 \times 10^{-5})^2$, equalling $1.08 \times 10^{-10} \text{mol}^2\text{L}^{-2}$.

The solubility product constant, $K_{sp}$, provides a measure of the equilibrium between a solid ionic solute and its saturated solution. For silver oxalate, the dissociation follows the reaction $\mathrm{Ag}_2\mathrm{C}_2\mathrm{O}_4 \rightleftharpoons 2\mathrm{Ag}^{+} + \mathrm{C}_2\mathrm{O}_4^{2-}$. According to this balanced equation, the dissolution of the salt releases two silver ions for every single oxalate ion present in the solution.

With the silver ion concentration given as 2.2 x 10^-4 mol L^-1, the stoichiometric ratio requires the oxalate ion concentration to be exactly half that value, which is 1.1 x 10^-4 mol L^-1. The expression for the solubility product is $K_{sp} =$[\mathrm{Ag}^{+}]^2 [\mathrm{C}_2\mathrm{O}4^{2-}]$. By substituting the specific ion concentrations into the formula$K{sp} = (2.2 × 10^{-4})^2 \cdot (1.1 × 10^{-4})$, the calculated product is 5.3 x 10^-12.

The equilibrium pressure of carbon dioxide in the decomposition of strontium carbonate is determined solely by the equilibrium constant $K_p = 1.6$. Before reaching this state, the initial quantity of carbon dioxide in the 20 litre container at 400 K is found using the ideal gas law $PV = nRT$, which yields $n = \frac{0.4 \times 20}{400 \times R} = \frac{0.02}{R}$. As the container volume is reduced, the moles of gas remain constant, and the volume required to maintain the equilibrium pressure is determined by rearranging the ideal gas law to $V = \frac{nRT}{P}$. Substituting the values $n = \frac{0.02}{R}$, $T = 400$, and $P = 1.6$ results in a final volume of $V = \frac{0.02 \times R \times 400}{1.6 \times R} = 5$ litre.

A Lewis base is defined by its ability to donate a lone pair of electrons to an electron-acceptor, so identifying one requires locating a central atom that possesses such a pair. In molecules like $BF_3$, $SiF_4$, and $CF_4$, the central atoms are either electron-deficient or have completely utilized their valence electrons in bonding, leaving them unable to act as donors. Phosphorus in $PF_3$, however, retains a lone pair because it only utilizes three of its five valence electrons for bonding, which makes it the only species among the choices capable of acting as a Lewis base.

Pyridine behaves as a weak base when dissolved in water, reacting to form the pyridinium ion and hydroxide ions. Because it only partially dissociates, the degree of dissociation $\alpha$ is related to the base dissociation constant $K_b$ and the initial concentration $c$ by the standard approximation $\alpha = \sqrt{\frac{K_b}{c}}$. Substituting the provided values $K_b = 1.7 \times 10^{-9}$ and $c = 0.10$ M gives $\alpha = \sqrt{\frac{1.7 \times 10^{-9}}{0.10}}$. Solving this expression results in $\alpha = \sqrt{1.7 \times 10^{-8}}$, which simplifies to $\alpha = 1.3 \times 10^{-4}$. Converting this degree of dissociation into a percentage by multiplying by $100$ gives $1.3 \times 10^{-4} \times 100 = 0.013\%$.

The common ion effect describes how the solubility of a sparingly soluble salt like $AgCl$ decreases when it is dissolved in a solution already containing one of its constituent ions. Sodium chloride acts as a strong electrolyte, providing a constant concentration of chloride ions that effectively shifts the equilibrium of $AgCl$ dissociation to the left. The solubility product constant for this system is defined by $K_{sp} =$[Ag^{+}]$[Cl^{-}]$. By substituting the solubility$S$for the silver ion concentration and$S + 0.1$for the chloride ion concentration, the relationship becomes$1.6 \times 10^{-10} = S(S + 0.1)$. Given that the contribution of$S$is negligible relative to the 0.1 concentration of chloride ions provided by the sodium chloride, the expression simplifies to$1.6 \times 10^{-10} = S(0.1)$, which yields a final solubility of$S = 1.6 \times 10^{-9}$.

To compare the solubilities of these two salts, we must recognize that molar solubility depends not just on the value of $K_{sp}$, but also on the stoichiometry of the dissociation process. For the salt $MY$, the dissociation $MY \rightleftharpoons M^+ + Y^-$ leads to $K_{sp} = S^2$. Calculating the solubility, $S = \sqrt{6.2 \times 10^{-13}}$, gives approximately $7.9 \times 10^{-7}$ mol/L. In contrast, $NY_3$ dissociates as $NY_3 \rightleftharpoons N^{3+} + 3Y^-$, which results in $K_{sp} = [S][3S]^3 = 27S^4$. Solving $27S^4 = 6.2 \times 10^{-13}$ reveals that $S \approx 10^{-3.5}$ mol/L, a significantly higher value than the solubility of $MY$. Because the common ion $Y^-$ from the added salt $KY$ would suppress the dissolution of both salts due to the common ion effect, any claim that solubility remains unchanged or increases is incorrect, leaving the conclusion that the molar solubility of $MY$ in water is lower than that of $NY_3$.

Precipitation occurs when the concentration of silver ions added to the mixture exceeds the limit defined by the solubility product, $K_{\text{sp}}$. For the $1:1$ salts $\mathrm{AgCl}$, $\mathrm{AgBr}$, and $\mathrm{AgI}$, the threshold silver concentration is calculated as $[\mathrm{Ag}^+] = K_{\text{sp}} / [\text{anion}]$, whereas for the $2:1$ salt $\mathrm{Ag}_2\mathrm{CrO}_4$, it is calculated as $[\mathrm{Ag}^+] = \sqrt{K_{\text{sp}} / }$[\mathrm{CrO}4^{2-}]${}$. Substituting the provided$K{\text{sp}}$values-$1.8 \times 10^{-10}$for$\mathrm{AgCl}$,$5.0 \times 10^{-13}$for$\mathrm{AgBr}$,$8.3 \times 10^{-17}$for$\mathrm{AgI}$, and$1.1 \times 10^{-12}$for$\mathrm{Ag}_2\mathrm{CrO}_4$-reveals that$\mathrm{Ag}_2\mathrm{CrO}_4$ requires the highest silver ion concentration to reach its precipitation point. Consequently, this salt will remain in solution until the others have already precipitated, making it the last to form a precipitate.

The neutralization of a strong acid by a strong base relies on the stoichiometry of the reaction, where hydroxide ions and hydrogen ions react in a 1:1 ratio to form water. When equal volumes of 0.1 M sodium hydroxide and 0.01 M hydrochloric acid are mixed, the acid is entirely consumed by an equivalent amount of the base, leaving an excess of hydroxide ions in the resulting solution. Because the total volume of the mixture is double the initial volume of either reactant, the effective concentration of the remaining solute is halved.

Subtracting the 0.01 M acid from the 0.1 M base leaves 0.09 M of excess hydroxide ions relative to the original individual volumes. Diluting this into the total volume results in a final concentration of $[\mathrm{OH}^{-}] = 0.09 / 2 = 0.045$ M. Applying the definition of pOH as the negative logarithm of the hydroxide ion concentration gives $\text{pOH} = -\log(0.045) = 1.35$. Finally, determining the pH involves using the equilibrium relationship $\text{pH} = 14 - \text{pOH}$, which results in $\text{pH} = 14 - 1.35 = 12.65$.

The relationship between the Gibbs free energy change and the reaction quotient for any process is defined by the expression $\Delta G = \Delta G^{\circ} + RT \ln Q$. In a reversible process at equilibrium, the total Gibbs free energy change $\Delta G$ is zero, and the reaction quotient $Q$ becomes equivalent to the equilibrium constant $K$. Substituting these conditions into the governing equation yields $0 = \Delta G^{\circ} + RT \ln K$, which allows us to isolate the standard Gibbs free energy change as $\Delta G^{\circ} = -RT \ln K$. Converting the natural logarithm to base-10 by multiplying with the constant $2.30$ gives the final expression:
$\Delta G^{\circ} = -2.30 RT \log K$

Electrical conductivity in an aqueous solution is determined by the concentration of free ions available to transport electric charge. Because hydrochloric acid ($\mathrm{HCl}$) dissociates completely into $\mathrm{H}^{+}$ and $\mathrm{Cl}^{-}$ ions, it serves as a highly effective electrolyte for current flow. The remaining compounds are essentially non-electrolytes that fail to generate the necessary concentration of mobile ions, resulting in significantly lower conductivity.

An acidic buffer functions by maintaining a stable pH through a mixture of a weak acid and its salt, which resists significant changes when small amounts of acid or base are introduced. While pairs involving $CH_3COOH$, $H_2CO_3$, and $H_3PO_4$ act as buffers because they utilize weak acids capable of establishing the necessary chemical equilibrium, the combination in option D behaves quite differently. Since $HClO_4$ is a strong acid that dissociates completely in solution, it lacks the equilibrium properties required for buffering, rendering the pair of $HClO_4$ and $NaClO_4$ unable to function as an acidic buffer.

The equilibrium constant, $K_c$, provides a quantitative measure of how far a reaction proceeds, defined as the ratio of product concentrations to reactant concentrations at equilibrium. Because the value of $1.6 \times 10^{12}$ is exceptionally large, the numerator representing the products must be vastly greater than the denominator representing the reactants. This indicates that the chemical reaction shifts almost entirely to the right side, resulting in a mixture dominated by the products at equilibrium.