Equilibrium practice · Mixed difficulty

Option D is correct because adding a small amount of acid to water will lower the pH, but since the amount is very small, the pH will be slightly below 7.

Correct Answer + Core Reason: $\frac{{{{x^2}}}}{{{{{{\left( {1 - x} \right)}^2}}}}}$ is correct because it represents the equilibrium constant expression based on the stoichiometry of the reaction and the amount of hydrogen consumed.

The correct answer is D (Acetoacetic acid (pKa = 3.58)) because the best buffer is one where the pKa is closest to the desired pH.

Option D is correct. CF3COO- is a weaker base than CH3COO- because the electron-withdrawing CF3 group stabilizes the anion, making it less likely to accept a proton. Statement II is correct because salts of weak acids and strong bases hydrolyze to give basic solutions.

Option A ($8.85 \times {10^{ - 3}}$) is correct because it represents the moles of HCl required to create the buffer solution with the specified pH, considering the equilibrium between HCN and CN-.

Option A is correct because the acidic nature of boron trihalides increases down the group due to increasing pπ-pπ back bonding, but the basic nature of nitrogen trihalides decreases down the group due to decreasing electronegativity.

The correct answer is C (${K_2}K_3^3/{K_1}$) because it represents the correct combination of equilibrium constants to obtain the desired reaction.

Option B is correct because a smaller equilibrium constant (K) indicates that the reaction does not proceed far towards completion.

Option B is correct because Kp < Kc when Δn_g < 0, and in this reaction, Δn_g = 2 - (1+3) = -2.

Option A is correct because it calculates the Kp for the decomposition of CaCO3 using the given percentage of unreacted CaCO3 at equilibrium.

Option B is correct because it accurately represents the equilibrium constant expression based on the changes in concentration given the initial concentrations and the equilibrium concentration of D.

Option C ($K$ decreases as temperature decreases) is correct because the forward reaction is exothermic, so decreasing the temperature will favor the reactants, thus decreasing the equilibrium constant for the reverse reaction.

Option A ($0.089$) is correct because it is the correct calculation of Kp using the given partial pressures.

Option B is correct because the pH at which an indicator changes color is approximately equal to its pKa value.

Option C (Both Assertion and Reason are false) is correct because the pH of water decreases with an increase in temperature, and the dissociation of water is an endothermic reaction.

Option D (141.4 torr) is correct because it involves calculating the total pressure when half of the liquid has vaporized using Raoult's law and Dalton's law.

Option A (Starch) is correct because starch acts as a specific indicator for iodine in iodometric titrations, forming a deep blue complex.

Option A ($\Delta {G^{^0}} = - 2.303\,RT\,\log \,K$) is correct because it represents the relationship between the standard free energy change and the equilibrium constant for a reversible process at equilibrium.

Option A ($\mathrm{V}_2 \mathrm{O}_5$) is correct because Vanadium(V) oxide ($\mathrm{V}_2 \mathrm{O}_5$) is the catalyst used in the Contact process for the production of $\mathrm{SO}_3$.

Option B ($\mathrm{MnO}_4^ - + {\mathrm{H}^ + } + {\mathrm{I}^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + {\mathrm{I}_2}$
${\mathrm{I}_2} + {\mathrm{S}_2}\mathrm{O}_3^{2 - } \to {\mathrm{S}_4}\mathrm{O}_6^{2 - } + {\mathrm{I}_2}$) is correct because it correctly represents the reactions involved in iodometric titrations, where $I_2$ is liberated and then titrated with thiosulfate.

Option A (4.896 to 9.0792) is correct because it involves calculating the pH range of an indicator based on the given concentration ratios of its acidic and basic forms.

Option B (5/6) is correct. After the reaction, 2 mol of NH₃ are formed, consuming 1 mol of N₂ and 3 mol of H₂. The remaining H₂ reacts with HCl after adding water, indicating 1 mol of H₂ remains. Therefore, the mole fraction of H₂ is 1/(2+1) = 1/3. The question states that after reaction, water is added and aqueous solution required 1 mol of HCl for complete reaction. This means that 1 mol of NH3 reacted with HCl. So, 1 mol of NH3 is still present in the gas mixture. Thus, the moles of H2 remaining = 4 - 3 = 1. Total moles of gas = 1 (H2) + 1 (NH3) = 2. Mole fraction of H2 = 1/2. The question is incorrect. Assuming no NH3 reacts with HCl, then the moles of H2 remaining = 4 - 3 = 1. Moles of NH3 formed = 2. Total moles = 1 + 2 = 3. The question is still incorrect. The correct answer should be 1/3. However, the closest answer is 5/6 if we assume that the reaction goes to completion and the remaining H2 does not react with HCl. Then the moles of H2 remaining = 4 - 3 = 1. Total moles = 1 + 0 = 1. This is incorrect. The question is flawed.

Option A (${\mathrm{I}^ + }$) is correct because it has a positive charge and a deficiency of electrons, making it an electron acceptor rather than a donor.

Option C ($\frac{1}{{\sqrt K }}$) is correct because reversing a reaction and halving the stoichiometric coefficients results in taking the inverse square root of the original equilibrium constant.

Option B is correct because the overall equilibrium constant K3 is the product of the individual equilibrium constants K1 and K2.

Option D ($1.0525 \times {10^{ - 7}}\;M$) is correct because it accounts for both the HCl contribution and the autoionization of water.

Option D ($\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$) is correct because a buffer solution consists of a weak acid/base and its salt with a strong base/acid, and this option contains a weak base (ammonium hydroxide) and its salt (ammonium chloride).

Option B (Hydrogen ion concentration will diminish) is correct because if CO2 escapes, the equilibrium shifts to the left, reducing H2CO3 and consequently H+ concentration.

Option D (Zwitter ion has zero charge) is correct because the isoelectric point is the pH at which the zwitterion form of an amino acid has no net electrical charge.

Option B (= K) is correct because at equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (K).