Equilibrium practice · Easy

Option B (${K_p}/RT$) is correct because it correctly relates Kp and Kc based on the change in the number of moles of gas.

Option C ($2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}$) is correct because $K_p < K_c$ when $\Delta n < 0$, and in this reaction, $\Delta n = 2 - (2+1) = -1$.

Option C ((A) is true but (R) is false) is correct because acetic acid is indeed a weak acid, but its conjugate base (acetate ion) is a relatively strong base, not a weak one.

Option B is correct because if CO2 escapes, the equilibrium shifts to the left, decreasing [H+] and increasing pH.

Option A is correct because HCl and NaCl cannot form a buffer solution as buffer solutions require a weak acid and its conjugate base.

Option D (${\text{li}}{{\text{t}}^2}\,{\text{mo}}{{\text{l}}^{ - 2}}$) is correct because the unit of $K_c$ is determined by the change in the number of moles of gas in the balanced chemical equation.

Option D (I, II, III) is correct because water can act as both a Bronsted acid and a Bronsted base (amphoteric) and is an amphoteric oxide.

Option D ($Ca^{2+}] = \frac{{K_{Sp}}}{{4}}^{1/3}$) is correct because it accurately represents the relationship between the calcium ion concentration and the solubility product for CaF₂.

Option C (Reduced by 1000 times) is correct because each unit increase in pH represents a tenfold decrease in H+ ion concentration.

The correct answer is A (Decrease in pressure suddenly which results in a decrease in solubility of $C{O_2}$ gas in water) because the solubility of a gas in a liquid is directly proportional to the pressure above the liquid (Henry's Law).

Option C (${H_2}C{O_3}$) is correct because the conjugate acid of a base is formed by adding a proton ($H^+$) to the base, so adding $H^+$ to $HCO_3^ -$ gives ${H_2}C{O_3}$.

Option A is correct because the weakest conjugate base corresponds to the strongest acid, and HClO4 is the strongest acid among the given options.

Option D (6) is correct because at $90^{\circ}C$, pure water is neutral with $[H^+] = 10^{-6}$ M, and the addition of equal equivalents of HCl and KOH results in a neutral solution.

Option D is correct because the decrease in the equilibrium constant with increasing temperature indicates that the reaction is exothermic, and this is a property of the system.

The reaction is retarded by the addition of hydrogen ion because adding $H^+$ shifts the equilibrium to the left, according to Le Chatelier's principle.

Option C (T=12.5 K) is correct because when Kp = Kc, RT = 1, and solving for T gives 1/R = 1/0.08 = 12.5 K.

Option A (${\mathrm{I}^ + }$) is correct because Lewis bases are electron pair donors, and I+ is electron deficient, thus it is least likely to act as a Lewis base.

Option B is correct because HCl is a strong acid, NH4Cl is an acidic salt, NaCl is a neutral salt, and NaCN is a basic salt, resulting in the pH increasing in the order HCl < NH4Cl < NaCl < NaCN.

The correct answer is $\mathrm{HSO}_4^ -$ because it can both donate a proton (acting as an acid) and accept a proton (acting as a base).

Option A (Ammonium chloride) is correct because ammonium chloride ($\mathrm{NH}_4\mathrm{Cl}$) is a salt of a weak base ($\mathrm{NH}_4\mathrm{OH}$) and a strong acid (HCl), resulting in an acidic solution upon dissolution in water.

Option A (Weak, low) is correct because acetic acid is a weak acid and a less polar solvent than water, leading to weaker and lower ionization of strong electrolytes.

Option A ($2 \times {10^{ - 3}}\;M$) is correct because the solubility is calculated using the solubility product expression for PbI2.

Option C (Increase of temperature) is correct because the degree of dissociation for this reaction is influenced by temperature as it affects the equilibrium constant.

Option A (Quantity of solid will reduce) is correct because according to Le Chatelier's principle, adding heat to a solid-liquid system shifts the equilibrium towards the endothermic direction, favoring the dissolution of the solid.

Option D (HI is less stable than H₂ and I₂) is correct because the equilibrium constant decreases with increasing temperature, indicating that the reverse reaction (formation of H₂ and I₂) is favored at higher temperatures, meaning HI is less stable.

Option B is correct because adding more PCl5(s) will shift the equilibrium to the right, increasing the concentration of Cl2(g).

Option D ($\frac{1}{2}p{K_w} + \frac{1}{2}p{K_a} + \frac{1}{2}\log C$) is correct because it represents the pH of a salt of a weak acid and a strong base.

Option B ($4.76$) is correct because, in this case, the concentrations of the acid and conjugate base are equal, making pH equal to pKa.

Option C ($\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$) is correct because sodium carbonate hydrolyzes in water to produce a strong base (NaOH) and a weak acid ($H_2CO_3$), resulting in an alkaline solution.

Option C ($6.95$) is correct because after dilution, the pH is calculated considering the contribution of hydrogen ions from both the acid and water.