Equilibrium practice · Medium

Option B ($BB{r_3} > BC{l_3} > B{F_3}$) is correct because the Lewis acidity decreases as the extent of pπ-pπ backbonding increases, which is greatest in BF3.

Option C is correct because, with $\Delta n_g = 0$, pressure has no effect, and the forward reaction rate depends only on temperature, increasing with temperature due to the endothermic nature.

Option B ($2.5 \times 10^{-12} \mathrm{M}$) is correct because the solubility of AgBr is reduced due to the common ion effect from the $Br^-$ ions in the $CaBr_2$ solution.

Option C (= 4.76) is correct because, according to the Henderson-Hasselbalch equation, when the concentrations of the acid and its conjugate base are equal, the pH equals the pKa.

Option D (10.8) is correct because for a solution of a salt of weak acid and weak base, if the dissociation constants are the same, pH + pOH = 14.

Option D (1) is correct because the percentage of ionization is calculated using the formula derived from the equilibrium constant Ka and concentration.

The correct answer is 3 because the complex has Fe in the +1 oxidation state with a 3d7 configuration, resulting in three unpaired electrons.

The correct answer is Endothermic because the equilibrium constant increases with increasing temperature, indicating that heat is absorbed, favoring product formation.

Option A (${10^{ - 9}}$) is incorrect. Option C (${10^{ - 19}}$) is the correct answer because it is calculated using the formula for the hydrolysis constant, $K_h = \frac{{K_w}}{{K_b}}$.

Option D (Statement I is incorrect, Statement II is correct) is correct because Statement I is incorrect as the ionic product is less than Ksp, so no precipitation occurs, and Statement II is correct as precipitation occurs only when the ionic product exceeds Ksp.

Option A (100) is correct because using the formula $\Delta G^\circ = -2.303RT \log K$, and substituting the given values, we find that the equilibrium constant K is 100.

Option A ($2.08 \times 10^{-13}; 12.64 \times 10^{-8}$) is correct because it provides the ionic product and degree of dissociation calculated for heavy water.

Option D ($5.0 \times {10^{ - 7}}{M^3}$) is correct because the Ksp is calculated from the pH of the saturated solution of Ba(OH)2.

Option D (0.01) is correct because the degree of dissociation is calculated using the formula $\alpha = \frac{{[H^+]}}{C}$, where [H+] is the hydrogen ion concentration and C is the initial concentration of the weak acid.

Option B ($\frac{1}{1+{K}_1+{K}_2}$) is correct because it represents the mole fraction of A at equilibrium, derived from the equilibrium constants for the given gas-phase isomerization reactions.

Option B (12) is correct because the pH is calculated from the hydroxide ion concentration of the strong electrolyte Ca(OH)2.

Option B ($4.14 \times {10^{ - 11}}$) is correct because it is calculated using the solubility product expression, ${K_{sp}} = 4{s^3}$, where s is the molar solubility of $\mathrm{Ca}{\mathrm{F}_2}$.

Option C ($\frac{1}{64}$) is correct because the equilibrium constant for the reverse reaction is the inverse of the square of the original equilibrium constant.

Option D is correct because the pH of the resulting buffer solution is calculated to be 11.6, which falls within the color transition range of Malachite green (11.4 to 13).

Option A ($\frac{5 \times 10^{-5}}{R}$) is correct because Kc is calculated from Kp using the relationship Kp = Kc(RT)^Δn.

Option A (Law of mass-action) is correct because the law of mass action states that the rate of a reaction is proportional to the active masses of the reactants.

Option A (6) is correct because the pH at the first equivalence point is the average of the two pKa values.

The correct answer is C. Increasing pressure does not disturb the equilibrium because the number of moles of gaseous reactants and products are equal.

Option B ($8.64 \times 10^{-7} mol^2 L^{-2}$) is correct because it involves calculating the solubility product (Ksp) from the given solubility of calcium fluoride.

Option C (${HPO}_4^{2-}$) is correct because it can both donate a proton (act as an acid) and accept a proton (act as a base).

Option A (8.7) is correct because the pH of the buffer solution is calculated using the Henderson-Hasselbalch equation for a basic buffer.

Option C (6.8) is correct because it accurately calculates the pH of a neutral solution at 37°C using the given Kw value.

Option B is wrong because a mixture of a weak acid (CH3COOH) and a strong base (NaOH) will result in a pH greater than 7 at equivalence point.

The correct answer is C because both the statement and the explanation are correct, and the explanation correctly justifies why the [H+] in 10^-8 N HCl is slightly higher than 10^-7 M due to the contribution from water.

Option C ($4{S^3}$) is correct because the solubility product for a salt of type $A_2B$ is given by $4S^3$, where S is the solubility.