Equilibrium practice · Medium

Option D (32.0) is correct because Kp is calculated as the ratio of the partial pressure of products to reactants, each raised to the power of their stoichiometric coefficients.

Option C (${H_2}O(g){H_2}(g) + \frac{1}{2}{O_2}(g)$) is correct because increasing the temperature will shift the equilibrium to the right side only if the forward reaction is endothermic.

Option B is incorrect. The conjugate acid of $HSO_4^-$ is formed by adding a proton ($H^+$), resulting in $H_2SO_4$. The conjugate base is formed by removing a proton, resulting in $SO_4^{2-}$.

Option A (HA) is correct because HA has a smaller dissociation constant, indicating it's a weaker acid. Weaker acids have higher pH values at a given molarity.

Option A ($0.21$) is correct because after calculating the equilibrium concentrations, the equilibrium constant Kc is found to be approximately 0.21.

Option D (The value of ionic product is greater than the value of solubility product) is correct because a precipitate forms when the solution is supersaturated, meaning the ionic product exceeds the solubility product.

Option D is correct because it accurately calculates the equilibrium constant based on the given concentrations and stoichiometry of the reaction.

Option A (Monoacidic base) is correct because urea can accept one proton due to the presence of two nitrogen atoms, each with a lone pair of electrons.

The correct answer is C because at a constant temperature, the equilibrium pressure of CO2 will remain constant regardless of the amount of CaCO3 or the container size.

Option B (0.128 M) is correct because it represents the equilibrium concentration of I2 calculated using the equilibrium constant expression.

Option A (Less) is correct because the pH scale is logarithmic, so a solution with a pH of 8 is less basic than a solution with a pH of 12.

Option A ($3 - 4$) is correct because methyl orange is a pH indicator that changes color in the acidic range of approximately 3.1 to 4.4.

Option A (Statement I, II are correct) is correct because the pH of a buffer solution is determined by the ratio of the concentrations of the salt and acid (or base), which remains unchanged upon dilution.

Option A ($4\times 10^{-14}$) is correct because the ionic product of water ($K_w$) increases with temperature, and at 50°C, it's approximately $4 \times 10^{-14}$.

Option A ($(A) > (D) > (C) > (B)$) is correct because it correctly orders the solutions based on their pOH values, considering the acidity/basicity of each solution.

Option D (In a 0.001 M solution of $\mathrm{NaOH}$ the concentration of ${\mathrm{H}^ + }\,\mathrm{is}\,{10^{ - 3}}$ mol/L) is correct because in a 0.001 M NaOH solution, which is a strong base, the $[H^+]$ concentration should be $10^{-11}$ M, not $10^{-3}$ M.

Option C is correct. Alkyl amines are more basic than ammonia due to the +I effect of the alkyl group. HS- is more basic than I- because S is more electronegative than I.

Option C ($\mathrm{NH_4OH + NH_4Cl}$) is correct because both solutions contain the common ion NH4+, which causes the common ion effect.

Option A (Increase in $pH$) is correct because the addition of sodium acetate, the conjugate base of acetic acid, will shift the equilibrium, increasing the $pH$.

Option C (14 g, 3 g) is correct because it is calculated by determining the amount of reactants consumed to form the given amount of product at equilibrium.

Option A (${\mathrm{A}{\mathrm{g}^ + }} \left[ {\mathrm{C}{\mathrm{l}^ - }} \right] < {K_{sp}}$) is correct because an unsaturated solution exists when the ion product is less than the solubility product.

Option B ($\frac{{{K_w}}}{{{K_a} \times {K_b}}})$ is correct because $\mathrm{N}{\mathrm{H}_4}\mathrm{CN}$ is a salt of a weak acid and a weak base, and the hydrolysis constant is given by this formula.

Option A ($\mathrm{HCl}{\mathrm{O}_4}$) is correct because perchloric acid is one of the strongest known acids, readily donating its proton in solution.

Option C (K₃ = K₁K₂) is correct because reaction III can be obtained by adding reactions I and II. When reactions are added, their equilibrium constants are multiplied.

Option D ($2/3$ mole of ethyl acetate is formed) is correct because it is calculated using the equilibrium constant and initial moles of reactants.

Option D ($1M \,\,\,{H_2}S{O_4}$) is the best conductor because it is a strong acid that dissociates completely into ions in solution, leading to higher conductivity.