Equilibrium – NEET UG ChemistryPractice Questions & PYQs

Solubility cannot be determined by comparing $K_{sp}$ values alone because different salts have distinct dissociation stoichiometries that influence the relationship between $K_{sp}$ and molar solubility ($S$). For the 1:1 electrolyte $\text{AgBr}$, the dissociation $\text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^-$ leads to $K_{sp} = S^2$, resulting in $S = \sqrt{5.0 \times 10^{-13}} \approx 7.07 \times 10^{-7} \, \text{mol/L}$. Both $\text{Zn}(\text{OH})_2$ and $\text{Hg}_2\text{Cl}_2$ dissociate into three ions, following the expression $K_{sp} = 4S^3$. Applying this, $\text{Zn}(\text{OH})_2$ yields $S = \sqrt[3]{\frac{1.0 \times 10^{-15}}{4}} \approx 6.3 \times 10^{-6} \, \text{mol/L}$, while $\text{Hg}_2\text{Cl}_2$ yields $S = \sqrt[3]{\frac{1.3 \times 10^{-18}}{4}} \approx 6.9 \times 10^{-7} \, \text{mol/L}$. These calculations reveal the solubility order as $\text{Zn}(\text{OH})_2 > \text{AgBr} > \text{Hg}_2\text{Cl}_2$, since $6.3 \times 10^{-6} \, \text{mol/L}$ exceeds $7.07 \times 10^{-7} \, \text{mol/L}$, which in turn exceeds $6.9 \times 10^{-7} \, \text{mol/L}$.

The auto-ionization of water is an endothermic process, meaning the equilibrium constant $K_w =$[H^+]$[OH^-]$depends directly on thermal energy. According to Le Chatelier's principle, an increase in heat shifts the reaction equilibrium toward the products, which raises the concentration of ions and leads to a higher value for the equilibrium constant. Given that the standard value is$K_w = 10^{-14}$at 25 degrees Celsius, heating the system to 90 degrees Celsius must result in a value where$K_w > 10^{-14}$. Among the options,$K_w = 1 \times 10^{-12}$ is the only value that reflects this expected increase.

The dissociation of a weak acid in an aqueous solution is defined by the initial concentration of the acid and its degree of ionization. Multiplying the $0.1 \text{ M}$ acid concentration by the dissociation fraction of $0.02$ gives the concentration of hydrogen ions, $[\mathrm{H}^{+}]$, as $2 \times 10^{-3} \text{ M}$. Because the product of hydrogen and hydroxide ion concentrations remains constant at $10^{-14}$ in aqueous solutions, the concentration of hydroxide ions is determined by dividing this ionic product of water by the $[\mathrm{H}^{+}]$. Dividing $10^{-14}$ by $2 \times 10^{-3} \text{ M}$ yields an hydroxide ion concentration, $[\mathrm{OH}^{-}]$, of $5 \times 10^{-12} \text{ M}$.

When chemical reactions are combined, the overall equilibrium constant for the resulting process is determined by applying specific rules to the equilibrium constants of the individual steps. If a reaction is reversed, its new equilibrium constant becomes the reciprocal of the original value, and whenever multiple reactions are added together, the equilibrium constant of the net reaction is the product of the individual equilibrium constants.

The target equation $\mathrm{XeO}_{4(g)} + 2\mathrm{HF}_{(g)} \rightarrow \mathrm{XeO}_{3}\mathrm{F}_{2(g)} + \mathrm{H}_{2}\mathrm{O}_{(g)}$ results from first reversing the equation $\mathrm{XeF}_{6} + \mathrm{H}_{2}\mathrm{O}_{(g)} \rightarrow \mathrm{XeOF}_{4(g)} + 2\mathrm{HF}_{(g)}$, which gives a new equilibrium constant of $\frac{1}{K_1}$. This reversed process is then added to the second provided reaction, $\mathrm{XeO}_{4(g)} + \mathrm{XeF}_{6(g)} \rightarrow \mathrm{XeOF}_{4(g)} + \mathrm{XeO}_{3}\mathrm{F}_{2(g)}$, which has an equilibrium constant of $K_2$. Combining these two steps causes the intermediates $\mathrm{XeF}_{6}$ and $\mathrm{XeOF}_{4}$ to cancel out, yielding the desired chemical process. Because the net reaction is the sum of these two steps, the overall equilibrium constant is the product of the individual values, $\frac{1}{K_1}$ and $K_2$, which simplifies to $\frac{K_2}{K_1}$.

The pH of a buffer solution consisting of a weak acid and its conjugate base is determined using the Henderson-Hasselbalch equation, expressed as \mathrm{pH} = \mathrm{p}K_a + \log \frac{[\mathrm{X^-}]}}{[\mathrm{HX}]}$. At 298 K, the acid dissociation constant$K_a$for$\mathrm{HX}$is derived from the given base dissociation constant$K_b$for$\mathrm{X^-}$using the relationship$K_a × K_b = 10^{-14}$, which results \in$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.

Since the buffer contains equal concentrations of $\mathrm{HX}$ and $\mathrm{X^-}$, the ratio of these species is 1, which causes the logarithmic term in the equation to become zero. Consequently, the $\mathrm{pH}$ of the system simplifies to the $\mathrm{p}K_a$ of the acid, defined as the negative logarithm of the $K_a$. Calculating $-\log(10^{-4})$ results in a pH of 4.

The equilibrium of a sparingly soluble substance in water is defined by its solubility product constant, which links the concentration of dissolved ions to the solution's acidity. For the dissociation of $\mathrm{BiO}(\mathrm{OH})(\mathrm{s}) \rightleftharpoons \mathrm{BiO}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq})$, the equilibrium constant $K =$[\mathrm{BiO}^{+}]$[\mathrm{OH}^{-}]$is$4 \times 10^{-10}$. Because the dissociation produces ions in a 1:1 molar ratio, we denote the concentration of both ions as$s$, leading to$s^2 = 4 \times 10^{-10}$and an$[\mathrm{OH}^{-}]$of$2 \times 10^{-5}$M. Determining the alkalinity requires calculating the pOH via the expression$\mathrm{pOH} = -\log[\mathrm{OH}^{-}]$, which results in$\mathrm{pOH} = -\log(2 \times 10^{-5}) = 5 - \log 2$. Using the provided value$\log 2 = 0.3010$, the pOH is$4.699$. Subtracting this value from$14$, as defined by the relationship$\mathrm{pH} + \mathrm{pOH} = 14$at$298$K, yields a final pH of$9.301$.

Sodium hydroxide acts as a strong base, fully dissociating in water to make the hydroxide ion concentration equal to the solution normality of 0.1 M, which is $10^{-1}$ M. Determining the hydronium ion concentration requires the ionic product of water, $K_w = 10^{-14}$, and the relationship $[H_3O^+] = \frac{K_w}{[OH^-]}$}$:

$[H_3O^+] = \frac{10^{-14}}{10^{-1}} = 10^{-13}$

The pH is calculated using the expression $\mathrm{pH} = -\log [H_3O^+]$, which simplifies to $\mathrm{pH} = -\log(10^{-13})$, resulting in 13.

A conjugate acid is simply the species formed when a base accepts a proton, which is represented as adding an $\mathrm{H}^{+}$ ion to the starting material. Applying this principle to the amide ion, $\mathrm{NH}_{2}^{-}$, requires the addition of one proton to balance the charge, resulting in the formation of $\mathrm{NH}_{3}$. This relationship is expressed by the equation $\mathrm{NH}_{2}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{NH}_{3}$, identifying the ammonia molecule as the conjugate acid of the amide ion.

The dissociation of hydrogen iodide, represented by the balanced equation $2\mathrm{HI} \rightarrow \mathrm{H}_2 + \mathrm{I}_2$, requires tracking mole changes using the dissociation constant $\alpha$. By setting the initial amount of reactant to 2 moles, we find that the quantity of HI left at equilibrium is $2(1-\alpha)$, while the products formed are $\alpha$ moles of $\mathrm{H}_2$ and $\alpha$ moles of $\mathrm{I}_2$. Combining these components results in the equation:
$2(1-\alpha) + \alpha + \alpha = 2$
This confirms that the total number of moles remains unchanged as the reaction reaches equilibrium.

Phenolphthalein acts as an acid-base indicator that remains colourless in acidic or neutral environments and transitions to a pink shade when the medium becomes basic. In the titration of a strong base like sodium hydroxide with a weak acid like oxalic acid, governed by the reaction $\mathrm{2\,NaOH + H_2C_2O_4 \rightarrow Na_2C_2O_4 + 2\,H_2O}$, the solution reaches a slightly basic state at the equivalence point due to the hydrolysis of sodium oxalate. Since the indicator shifts colour within the pH range of 8.3 to 10.0, the solution effectively turns from colourless to pink as it nears this endpoint.

The solubility product constant, $K_{sp}$, is defined by the product of the equilibrium concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient. For calcium fluoride, the dissociation reaction is $\mathrm{CaF}_{2} \rightleftharpoons \mathrm{Ca}^{2+} + 2\mathrm{F}^{-}$, meaning for a solubility of $s$, the concentration of calcium ions is $s$ and the concentration of fluoride ions is $2s$.

The resulting expression is $K_{sp} =$[\mathrm{Ca}^{2+}]$$[\mathrm{F}^{-}]^2$, which simplifies to $K_{sp} = s × (2s)^2$, or $K_{sp} = 4s^3$. Inserting the given solubility of $s = 2 × 10^{-4}$ moles per litre, we calculate $K_{sp} = 4 × (2 × 10^{-4})^3$. This calculation yields $4 × 8 × 10^{-12}$, which equals $32 × 10^{-12}$.

Mixing a 1 N solution of $CH_3COOH$ (a weak acid) with a 1 N solution of $NaOH$ (a strong base) results in a basic solution rather than a neutral one with a $pH$ of 7. The salt formed during this neutralization reaction undergoes hydrolysis in water, which produces free $OH^-$ ions. This accumulation of hydroxide ions shifts the solution's balance towards alkalinity, ensuring that the final $pH$ is strictly higher than 7.

The hydride ion ($\mathrm{H}^-$) functions as an exceptionally strong Brønsted base, possessing a far higher affinity for protons than the hydroxide ion ($\mathrm{OH}^-$). When sodium hydride ($\mathrm{NaH}$) is introduced into water, the hydride ion immediately abstracts a proton from a water molecule, which acts as the proton donor in this acid-base interaction. This process drives the irreversible formation of hydrogen gas and the production of hydroxide ions, represented by the following equation:

$\mathrm{H}_{(aq)}^{-} + \mathrm{H}_{2}\mathrm{O}_{(l)} \rightarrow \mathrm{OH}^{-} + \mathrm{H}_{2} \uparrow$

The relationship between an acid and its conjugate base is inverse; specifically, the weaker the parent acid, the stronger its corresponding conjugate base. Among the species provided, ${H}_{2} {SO}_{4}$, ${HCl}$, and ${HNO}_{3}$ act as strong acids, which means their conjugate bases are relatively weak. Conversely, ${CH}_{3} {COOH}$ is the weakest acid in this set, which necessitates that its conjugate base, ${CH}_{3} {COO}^{-}$, must be the strongest base among the options.

The efficacy of a buffer system is governed by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \left( \frac{[\text{salt}]}{[\text{acid}]} \right)$. To effectively resist changes in $pH$ with minimal concentrations of buffering agents, the chosen weak acid must have a $pK_{a}$ value that maintains the necessary equilibrium at the target $pH$ of $3.85$. 2,5-Dihydroxy benzoic acid is the most suitable agent among the provided options to establish this stable buffering environment efficiently.

The relationship between the equilibrium constants $K_P$ and $K_C$ is defined by $K_P = K_C (RT)^{\Delta n_g}$ where $\Delta n_g$ represents the net change \in the number of moles of gaseous species. Because this relation depends on the exponent $\Delta n_g$, the constants are equal only when the total moles of gaseous products equal the total moles of gaseous reactants, resulting \in $\Delta n_g = 0$. While the water gas shift, hydrogen iodide formation, and nitric oxide synthesis reactions all involve an equal number of gaseous moles on both sides, the ammonia synthesis reaction consumes four moles of gaseous reactants to produce two moles of gas, resulting \in $\Delta n_g = -2$. Since this value is non-zero, the identity $K_P = K_C$ does not hold for the ammonia synthesis reaction.

The solubility of a substance is directly dictated by its $K_{sp}$ value, where a larger constant corresponds to higher solubility. By comparing the provided values, ${CuS}$ has the largest $K_{sp}$ of $10^{-31}$, followed by ${Ag}_{2}{S}$ at $10^{-44}$, and ${HgS}$ with the smallest value of $10^{-54}$. This hierarchy of solubility products establishes the order of solubilities as ${CuS} > {Ag}_{2}{S} > {HgS}$.

A Lewis acid acts as an electron-pair acceptor, a property facilitated by either an incomplete octet or the presence of empty orbitals on the central atom. Both $BF_3$ and $FeCl_3$ function as Lewis acids because their central atoms possess incomplete octets, while $SiF_4$ qualifies because its central atom utilizes empty $d$-orbitals to accept incoming electrons. In contrast, the carbon atoms in $C_2H_4$ have already achieved a complete octet through their covalent bonds, leaving no vacancy or empty orbitals to accept an additional electron pair, which makes the molecule ineligible as a Lewis acid.

According to Le Chatelier's Principle, a system at equilibrium will shift to counteract any applied changes in conditions. Since the reaction $\mathrm{N}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g})$ is endothermic with $\Delta H = +180.7 \text{kJ mol}^{-1}$, increasing the temperature favors the forward reaction to absorb the added heat, thereby increasing the yield of the product. Additionally, increasing the concentration of either $\mathrm{N}_2$ or $\mathrm{O}_2$ drives the system in the forward direction to consume the excess reactants, which also enhances the production of NO.

Polyprotic acids like $\mathrm{H}_3\mathrm{PO}_4$ dissociate in a stepwise manner, where each subsequent ionization becomes more difficult because it requires removing a positively charged proton from an increasingly negatively charged anion. This trend naturally causes the ionization constants to decrease significantly at each stage, confirming the relationship $K_{a_1} > K_{a_2} > K_{a_3}$ and establishing that $\mathrm{H}_3\mathrm{PO}_4$ remains the strongest acid in the series compared to its intermediate conjugate bases. Because the overall chemical equation is the sum of these three discrete reaction steps, the net equilibrium constant $K$ is the product of the individual step constants, which manifests as the logarithmic sum $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$ according to standard logarithmic properties. The final statement suggesting that the first constant is an arithmetic mean of the others lacks any foundation in equilibrium principles and is therefore inaccurate.