This question tests understanding of weak acid and base equilibria. Both acetic acid and HCN are weak acids that establish equilibria with their conjugate bases and H+ ions. The strength of a weak acid is determined by its Ka value - larger Ka means stronger acid and more dissociation. Since acetic acid has Ka = 1.8 × 10^-5 while HCN has Ka = 6.2 × 10^-10, acetic acid is the stronger acid and will produce more H+ ions at equilibrium. Choice D incorrectly assumes both acids fully dissociate, which only happens with strong acids like HCl or HNO3. For comparing weak acids at the same concentration, the one with larger Ka always produces more H+ and thus has lower pH.

This question assesses understanding of weak acid and base equilibria. Weak acids like acetic acid partially dissociate, forming H+ and CH3COO- in equilibrium. Ka = $1.8×10^{-5}$ indicates limited ionization, but the reaction stoichiometry ensures [H+] = [CH3COO-] from each dissociation event. This equality governs the concentrations, neglecting water's autoionization. Choice A is a common misconception, suggesting $[CH3COO-]{eq}$ > $[CH3COOH]{eq}$ because ions are favored, but small Ka means the opposite. For weak species, write the equilibrium first before thinking about pH.

This question tests understanding of weak acid and base equilibria. For the dissociation of benzoic acid $$ \text{HC}_6\text{H}_5\text{COO} \rightleftharpoons \text{H}^{+} + \text{C}_6\text{H}_5\text{COO}^{-} $$, the acid dissociation constant Ka is defined as the ratio of products to reactants: $ K_a = \frac{[\text{H}^{+}][\text{C}_6\text{H}_5\text{COO}^{-}]}{[\text{HC}_6\text{H}_5\text{COO}]} $. This follows the general pattern for equilibrium constants where products appear in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients. Choice A incorrectly inverts the expression, placing reactant over products, which would represent 1/Ka instead. For any weak acid equilibrium, always write Ka with ion products in the numerator and the molecular acid in the denominator.

This question tests understanding of weak acid and base equilibria. For a weak acid HA with Ka = 1.0×10⁻⁶, the very small equilibrium constant indicates minimal ionization, meaning the equilibrium HA + H₂O ⇌ H₃O⁺ + A⁻ lies far to the left. Since only a tiny fraction of HA molecules ionize, the equilibrium concentration [HA]eq remains very close to the initial concentration of 0.10 M. The concentrations of H₃O⁺ and A⁻ will be much smaller, approximately equal to each other but far less than 0.10 M. A common error (choice E) is thinking the conjugate base concentration equals the initial acid concentration, which would only occur for complete dissociation. For very weak acids (Ka << 1), assume [HA]eq ≈ initial concentration as a first approximation.

This question tests understanding of weak acid and base equilibria. The ratio $[\text{A}^{-}]$/$[\text{HA}]$ at equilibrium is directly related to the extent of ionization - a larger ratio means more of the acid has ionized to form its conjugate base. For weak acids at the same initial concentration, the acid with the larger $K_a$ will have greater ionization and thus a larger $[\text{A}^{-}]$/$[\text{HA}]$ ratio. Since HX has $K_a = 1.0\times 10^{-2}$ (much larger than HY's $K_a = 1.0\times 10^{-6}$), HX ionizes to a much greater extent, producing more $\text{X}^{-}$ relative to the remaining HX. A common error is thinking smaller $K_a$ means greater ionization, but $K_a$ is the equilibrium constant for ionization - larger $K_a$ means the equilibrium favors products (ions) more. For comparing weak acid ionization, remember that larger $K_a$ always means greater percent ionization and larger $[\text{A}^{-}]$/$[\text{HA}]$ ratio.

This question tests understanding of weak acid and base equilibria. For the weak base ammonia, the equilibrium NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ is described by the base ionization constant Kb. The equilibrium constant expression is always products over reactants, with each species raised to its stoichiometric coefficient. Since water is the solvent and its concentration remains essentially constant, it is not included in the Kb expression, giving Kb = [NH₄⁺][OH⁻]/[NH₃]. A common error (choice A) is including water in the expression, but pure liquids and solids are never included in equilibrium constant expressions. For weak base equilibria, remember to write the balanced equation first, then form the Kb expression excluding pure liquids.

This problem involves weak acid and base equilibria. Acetic acid (CH₃COOH) is a weak acid with Ka = 1.8×10⁻⁵, which is much less than 1. This small Ka value indicates that the equilibrium CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻ lies far to the left, meaning the reverse reaction is favored. Consequently, most CH₃COOH molecules remain undissociated at equilibrium, with only a small fraction converting to CH₃COO⁻ and H₃O⁺. The incorrect choice A claims extensive dissociation, which would require Ka >> 1, characteristic of a strong acid. For weak acids, remember that Ka << 1 means the equilibrium strongly favors the undissociated form.

This problem involves weak acid and base equilibria. Both HA and HB are weak acids that partially dissociate according to their Ka values. Since Ka(HA) = 1.0×10⁻³ is larger than Ka(HB) = 1.0×10⁻⁶, acid HA dissociates to a greater extent than HB. A larger Ka means more H₃O⁺ is produced at equilibrium, so the HA solution will have a higher [H₃O⁺] concentration. The incorrect choice B assumes equal [H₃O⁺] concentrations, which would only be true if the acids had identical Ka values. When comparing weak acids at the same initial concentration, always remember that larger Ka means greater dissociation and higher [H₃O⁺].

This question tests understanding of weak acid and base equilibria. Methylamine ($\text{CH}_3\text{NH}_2$) is a weak base with $K_b = 4.4 \times 10^{-4}$, establishing equilibrium $\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-$. Since $K_b$ is small (much less than 1), the equilibrium lies far to the left, meaning most methylamine molecules remain unreacted. Therefore, at equilibrium $[\text{CH}_3\text{NH}_2] > [\text{CH}_3\text{NH}_3^+]$. Choice B incorrectly assumes complete reaction to produce $[\text{OH}^-] = 0.10,\text{M}$, which would only occur for a strong base like NaOH. For weak bases, use the $K_b$ expression to recognize that small $K_b$ means little reaction, so the molecular form predominates.

This question tests understanding of weak acid and base equilibria. For weak acids at the same initial concentration, the acid with larger Ka dissociates more and produces higher [H+]. Since HA has Ka = 1.0 × 10^-3 (larger) and HB has Ka = 1.0 × 10^-6 (smaller), HA dissociates more extensively. This means the HA solution has higher [H+] and therefore lower pH than the HB solution. Choice A incorrectly states that higher Ka leads to higher pH, confusing the inverse relationship between [H+] and pH. For weak acid comparisons, remember: larger Ka → more dissociation → higher [H+] → lower pH.