Equilibrium practice · Medium

Option D ($\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$) is correct because it is a salt of a weak acid and a strong base, which will hydrolyze in water to produce OH- ions, resulting in a basic solution.

Option B ($2.37 \times 10^{-3} M$) is correct because it is calculated using the acid dissociation constant and the given hydrogen ion concentration.

Option D ($PO_4^{ - 3}\,\mathrm{and}\,{C_2}O_4^{ - 2}$) is correct because Bronsted bases are proton acceptors, and $PO_4^{ - 3}$ and ${C_2}O_4^{ - 2}$ accept protons in the reaction.

Option D ($\Delta G^{\circ}=-2.303 \,{RT} \,\log {K}$) is correct because it represents the standard Gibbs free energy change in terms of the equilibrium constant K.

The correct answer is C ($\frac{{K_p^{1/2}{P^2}}}{{16}}$) because it correctly calculates the partial pressure of ammonia at equilibrium when pure ammonia dissociates.

Option D is correct because a buffer solution is formed by a weak acid and its conjugate base, which is produced when NaOH partially neutralizes the weak acid.

Option C (11) is correct because NH4OH is a weak base, and the pH can be calculated using the Kb value and the concentration of the base.

Option B (${K_p} = {K_c}$) is correct because the change in the number of moles of gas ($\Delta n$) is zero in this reaction, making $K_p$ equal to $K_c$.

Option B ($1.25 \times 10^{-5}$) is correct because it represents the calculated dissociation constant using the given equivalent conductance values.

The correct answer is A ($\frac{1}{{K_1^2}} + \frac{1}{{{K_2}}}$) because it represents the equilibrium constant for the reverse of the given reactions to form N2O4 from N2 and O2.

Option A ($\mathrm{CaO} < CuO < {\mathrm{H}_2}\mathrm{O} < C{\mathrm{O}_2}$) is the correct order because it reflects the increasing non-metallic character and acidity.

Option A (Phenolphthalein) is correct because phenolphthalein changes color in the pH range of 8.3 to 10.0, which includes the range of 8 - 9.8.

Option A (Dissociation of $PC{l_5}$ increases) is correct because adding an inert gas at constant pressure increases the volume, favoring the side with more moles of gas, which is the dissociated side.

The correct answer is C (${K_2}/{K_1}$) because it represents the equilibrium constant obtained by manipulating the given reactions to match the target reaction.

Option C ($\mathrm{PC}{\mathrm{l}_5}$ is less) is correct because adding CO shifts the second equilibrium to the left, decreasing Cl2, which then shifts the first equilibrium to the left, decreasing PCl5.

Option A (4.67, 9.61) is correct because it is calculated using the Henderson-Hasselbalch equation for polyprotic acids.

Option B (8) is correct because in an amphoteric buffer where [H+] = [HX], pH = pKa.

Option C is correct because at equilibrium, the rate of the forward reaction equals the rate of the backward reaction.

Option C ($S{n^{4 + }}$) is correct because $S{n^{2 + }}$ loses two electrons to get oxidized to $S{n^{4 + }}$ when ${K_2}C{r_2}{O_7}$ is added.

Option C (Strong base) is correct because ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa}$ is the sodium salt of ethanol, and it completely dissociates in ethanol to form ethoxide ions ($\mathrm{C}_2{\mathrm{H}_5}\mathrm{O^-}$), which are strong bases.

Option B (${H_2}PO_4^ - \,and\,PO_4^{3 - }$) is correct because the conjugate acid is formed by adding a proton and the conjugate base by removing a proton.

Option C ($\frac{{{K_2}}}{{{K_1}}}$) is correct because the equilibrium constant for the target reaction can be obtained by dividing the second reaction's constant by the first.

Option C is correct because the hydrolysis constant of a salt of a weak acid and strong base is given by Kw/Ka.

Option C (No effect) is correct because adding an inert gas at constant volume does not change the partial pressures of the reactants or products, so the equilibrium position remains unchanged.

Option B ($40\%$) is correct because 0.8 moles of AB formed from 2 moles of A represents a 40% conversion.

Option A (Decreases) is correct because adding $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ to $\mathrm{N}{\mathrm{H}_4}\mathrm{OH}$ introduces a common ion ($\mathrm{N}{\mathrm{H}_4}^+$), suppressing the ionization of $\mathrm{N}{\mathrm{H}_4}\mathrm{OH}$ and decreasing the pH.

Option B ($6 \times 10^5$ ions) is correct because it represents the calculated number of amide ions per cm³ based on the self-ionization constant.

Option D (In all cases) is correct because in all the given cases, the pH will not change significantly upon dilution. For a buffer, the ratio of salt to acid remains constant. For CH3COONH4, it's a salt of a weak acid and weak base. For NaH2PO4, it's an amphoteric salt.

Option D ((i), (iv)) is correct because both solutions have the same final concentration of $H^+$ ions after considering neutralization in (iv).

Option B (Decrease in ${H^ + }$ concentration by 10 times) is correct because pH is a logarithmic scale, and an increase of 1 pH unit represents a 10-fold decrease in H+ concentration.