Equilibrium practice · Mixed difficulty

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 A (4) is correct because the pOH is calculated from the pH, which is derived from the hydronium ion concentration using the formula pH + pOH = 14.

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 A (Statement I, II are correct) is correct because acids and bases conduct electricity due to ionization in water.

The correct answer is B (Dissociation of $N{H_3}$ increases) because adding an inert gas at constant pressure increases the volume, shifting the equilibrium towards the side with more moles of gas.

The correct answer is D (All of these) because ${\mathrm{P}_2}{\mathrm{O}_5}$ can form ${\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}$, ${\mathrm{H}_4}{\mathrm{P}_2}{\mathrm{O}_7}$, and $\mathrm{HP}{\mathrm{O}_3}$ upon hydration, depending on the conditions.

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 C (1) is correct because after finding the roots and substituting them into the expression, the result simplifies to 1.

Option B (CH3OH) is the strongest acid because methanol is more acidic than the other hydrocarbons due to the electronegativity of oxygen.

Option B ($d{m^3}\\ \,mo{l^{ - 1}}\\ ,\,$) is correct because the units of the equilibrium constant are derived from the change in the number of moles in the reaction.

Option A ($5 \times 10^{-17}$) is correct because the ratio of [A^3-] to [H_3A] is calculated using the successive ionization constants.

Option A (1) is correct because the partial pressure of X at equilibrium is calculated to be 1 atm using the given Kp and dissociation percentage.

Option B ($\mathrm{HSO}_4^ -$) is the correct conjugate base because it is formed when ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ donates a proton.

Option C (5.98) is correct because after dilution, the concentration of HCl becomes very low, and the contribution of water's autoionization to the [H+] cannot be neglected. The pH is calculated considering both HCl and water's contribution.

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 ($\mathrm{CH}_3 \mathrm{COONa}$) is correct because it's a salt of a weak acid and a strong base, leading to concentration-dependent hydrolysis and pH.

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 C ($1.6 {~atm}$) is correct because it is calculated using the partial pressures of the gases at equilibrium and the expression for $K_p$.