Equilibrium practice · Mixed difficulty

Option C is correct because the pH of a salt of a weak acid and a weak base is determined by the difference between the pKa and pKb values.

Option C (${F^ - }$) is correct because the strongest conjugate base comes from the weakest acid, and HF is the weakest hydrohalic acid.

Option D ($2.00\\ \times \\ {10^{ - 16}}$) is correct because the ionization constant is calculated from the degree of dissociation and the concentration of water.

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 (${k_b}[{N_2}]{[{H_2}O]^2}\,/\,[{H_2}]$) is correct because it is derived from the equilibrium constant expression using the given forward rate law.

Option B ($\mathrm{O}{\mathrm{H}^ - }$) is correct because oxide ion (${\mathrm{O}^{2 - }}$) is a strong base and readily accepts a proton from water, forming hydroxide ions.

Option A ($\frac{1}{y} \times 10^{-14}$) is correct because the equilibrium constant (${K_a}$) for the conjugate acid of ${NH}_3$ is related to the hydrolysis constant (${K}_{{h}}$) by the equation ${K_a} = \frac{{{K_w}}}{{{K_h}}}$.

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 C (0.04375) is correct because it represents the molar ratio of $Al_2O_3$ required by A and B, calculated by considering the moles of HCl and NaOH neutralized by $Al_2O_3$.

Option A (Increase in temperature) is correct because the reaction is endothermic (ΔH > 0), so increasing the temperature will shift the equilibrium towards the products.

Option B (Solubility product) is correct because it is the product of ionic concentrations in a saturated solution at a given temperature.

Option B (Dissociation of $NH_3$ increases) is correct. Adding an inert gas at constant pressure increases the volume, favoring the side with more moles of gas, thus shifting the equilibrium towards dissociation of $NH_3$.

Option C (${K_p} = {K_c}$) is correct because for the given reaction, the change in the number of moles of gas (Δn) is zero, making Kp equal to Kc.

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 D (8) is correct because it represents the equilibrium constant Kc calculated using the given equilibrium concentrations.

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.

The correct answer is C (Have no effect) because pressure favors the phase with smaller volume. Diamond has higher density hence smaller volume, so increasing pressure will favor diamond formation, but the question asks about the equilibrium, so the change in pressure will shift the equilibrium but not affect it.

Option D (All of these) is correct because aprotic solvents are solvents that cannot donate protons (H+) and all listed solvents (CCl4, C6H6, SO2) fit this definition.

Option A ($\left( {\frac{\alpha }{{1 - \alpha }}} \right) = \frac{{1.75}}{{1.3}} \times \left( {\frac{\beta }{{1 - \beta }}} \right),0$ where $\alpha$ and $\beta$ are ionised fractions of the acids) is correct because it relates the ratio of acetate and propionate ions to their respective $K_a$ values and ionization fractions.

Option A ((i), (ii), (iii)) is correct because NaCl and NaOH, NaOH and NH4OH, and CH3COONH4 and HCl do not form buffer solutions; a buffer requires a weak acid/base and its conjugate.

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 A (Quantity of solid will reduce) is correct because according to Le Chatelier's principle, adding heat to a solid-liquid system at equilibrium will favor the endothermic process, which is the melting of the solid.

Option A ([I2(g)] > [I(g)]) is correct because the equilibrium constant is very small, indicating that the dissociation of I2 is minimal.

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 B ($0.06\mathrm{ }\;mol$) is correct because it represents the amount of unreacted ${\mathrm{I}_2}$ at equilibrium, calculated using the equilibrium constant K.

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 B (${K}_2 {K}_1^2=1$) is correct because the second reaction is the reverse of the first reaction multiplied by 2, thus affecting the equilibrium constant.