Equilibrium practice · Mixed difficulty

Option B (3) is correct because the pH of a weak acid solution is calculated using the formula pH = -log[H+], where [H+] = √(Ka * C).

Option C ($\frac{4}{27} P^3$) is correct because the $K_p$ is calculated using the partial pressures of the gaseous products at equilibrium.

Option C ((A) is true but (R) is false) is correct because $AgBr$ precipitates first due to its lower solubility product ($K_{sp}$), meaning $K_{sp}$ of $AgCl$ is greater than $AgBr$.

Option B ($4 \times 10^{-12}$) is correct because it is calculated by dividing the solubility product by the square of the hydroxide ion concentration.

The correct answer is D (Both S and E are correct but E is not correct explanation for S) because human blood is slightly alkaline, and its pH at body temperature is around 7.4, not 6.9. The alkaline nature is due to buffer systems, not just temperature.

Option A (${10^{ - 4}}\;\mathrm{M}\;\mathrm{ KOH}$) is correct because a KOH concentration of 10^-4 M corresponds to a pOH of 4, and thus a pH of 10.

Option B ($1 \times {10^{ - 2}}\;M$) is correct because pH is defined as the negative logarithm of hydrogen ion concentration, so if pH is 2, the hydrogen ion concentration is $10^{-2}$ M.

Option D ($1.9 \times {10^{ - 5}}$) is correct because it is calculated by setting up an equilibrium expression involving the solubility product (Ksp) of AgCN, the acid dissociation constant (Ka) of HCN, and the hydrogen ion concentration at the given pH.

Option D (3.5) is correct because adding a strong acid to a buffer will decrease the pH, but the buffer resists large changes, resulting in a pH slightly lower than 4.

Option B (2.496) is correct because the equilibrium moles of HI are calculated by subtracting the dissociated moles (22% of initial) from the initial moles.

Option D ($\mathrm{F}{\mathrm{e}^{3 + }}$) is correct because $\mathrm{F}{\mathrm{e}^{3 + }}$ ions undergo hydrolysis to produce $H^+$ ions, making the solution acidic (pH < 7).

The correct answer is C ($S{n^{4 + }}$) because in the redox reaction, $S{n^{2 + }}$ is oxidized to $S{n^{4 + }}$ by acidified ${K_2}C{r_2}{O_7}$.

Option C (0.04) is correct because the equilibrium constant is calculated by dividing the product of the concentrations of the products by the concentration of the reactant.

Option A (Starch) is correct because starch acts as an indicator in iodometric titrations, forming a blue complex with iodine, which disappears at the endpoint.

Option D (${CH}_3 {COOH}$ and ${Ca}({OH})_2$) is correct because a weak acid (${CH}_3 {COOH}$) and a base can form a buffer if the base is not in excessive amount to completely neutralize the acid.

Option D is correct because the ratio of [A+] to [AB2-] is inversely proportional to the square of [B-], as derived from the equilibrium constants K1 and K2.

Option A is correct because decreasing the concentration of CO shifts the equilibrium towards the forward side to counteract the change, according to Le Chatelier's principle.

Option D (Yellow colour in alkaline medium and red colour in acid medium) is correct because methyl orange exhibits these color changes in different pH environments.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ is a weak acid) is correct because acetic acid is a weak acid that only partially dissociates in water.

Option C is correct because both statements are true. Statement I correctly describes the relationship between concentrations at equilibrium, and Statement II accurately defines dynamic equilibrium.

Option B (Statement I, II are incorrect) is correct because if $Q_c < K_c$, the reaction moves towards the products, and the equilibrium constant is defined at equilibrium, not at the \"end stage\" in the way described.

The correct answer is C because the pH of x M HCl is indeed less than that of x M CH₃COOH, but the degree of ionization of HCl and CH₃COOH are not equal, as HCl is a strong acid and CH₃COOH is a weak acid.

Option B (1) is correct because when $\Delta G^{\circ} = 0$, the equilibrium constant K is equal to 1, as derived from the equation $\Delta G^{\circ} = -RT \ln K$.

Option D (${HNO}_2$ and ${NaNO}_2$) is correct because it consists of a weak acid (${HNO}_2$) and its conjugate base (${NaNO}_2$).

Option C ($2 \times {10^{ - 4}}$) is correct because it is derived from the solubility product expression for $A{X_2}$, where $K_{sp} = 4S^3$.

Option B (The pH of the solution increases) is correct because the addition of sodium acetate suppresses the dissociation of acetic acid due to the common ion effect, decreasing [H+] and increasing pH.

The correct answer is B ($5 \times 10^{-3}\, {M}$) because the concentration of H2SO4 is half the concentration of H+ ions in a solution with pH=2.

Option B (e^-10) is correct because it is the calculated dissociation constant of NH3 based on the given thermodynamic data.

The correct answer is C ($N{a_2}C{O_3}$) because it is a salt of a strong base (NaOH) and a weak acid ($H_2CO_3$), resulting in a higher pH due to hydrolysis.

Option D is correct because for a reaction to be reversible, it must be carried out in a closed vessel to prevent the escape of gaseous products, allowing the reverse reaction to occur.