This question tests the skill of calculating pH and pOH of strong acids and bases. HCl is a strong acid, meaning it fully dissociates in water to produce one H⁺ ion per molecule, so the [H⁺] equals the concentration of HCl, which is $1.0×10^{-3}$ M. The pH is calculated as -log[H⁺], so pH = $-log(1.0×10^{-3}$) = 3.00. This direct relationship holds because there is no equilibrium to consider for strong acids. A tempting distractor is 11.00, which might result from confusing pH with pOH or mistakenly calculating for a base. For strong acids and bases, the ion concentration directly corresponds to the solute concentration (adjusted for stoichiometry) before taking the negative logarithm.

This question tests the skill of calculating pH and pOH of strong acids and bases. Ba(OH)₂ is a strong base that fully dissociates in water to produce two OH⁻ ions per formula unit, so [OH⁻] = 2 × $2.0×10^{-4}$ = $4.0×10^{-4}$ M. The pOH is then -log[OH⁻] = $-log(4.0×10^{-4}$) ≈ 3.40. This calculation assumes complete ionization without any equilibrium considerations. A tempting distractor is 10.60, which could come from calculating pH instead of pOH or forgetting the factor of 2. For strong acids and bases, the ion concentration directly gives the value before taking logs, remembering to multiply by the number of ions per molecule.

This question tests the skill of calculating pH and pOH of strong acids and bases. LiOH is a strong base that fully dissociates to produce one OH⁻ ion per molecule, so [OH⁻] = $3.0×10^{-1}$ M. Using log(3.0) ≈ 0.48, pOH = $-log(3.0×10^{-1}$) = 1 - 0.48 = 0.52. This calculation assumes complete ionization. A tempting distractor is 13.48, perhaps from calculating pH = 14 - pOH incorrectly as 13.48. For strong acids and bases, concentration directly gives ion concentration before taking logs.

This question tests the skill of pH and pOH of strong acids and bases. HNO₃ is a strong acid, which means it fully dissociates in water according to HNO₃ → H⁺ + NO₃⁻, producing one H⁺ ion per molecule. Therefore, the concentration of H⁺ is equal to the concentration of HNO₃, which is $1.0×10^{-4}$ M. The pH would be calculated as -log[H⁺] = $-log(1.0×10^{-4}$) = 4.00, but the question asks for [H⁺] directly. A tempting distractor might be $1.0×10^{-14}$, if someone confused it with the ion product of water, K_w. For strong acids, the concentration directly gives ion concentration before taking logs.

This question tests the skill of pH and pOH of strong acids and bases. HBr is a strong acid, which means it fully dissociates in water according to HBr → H⁺ + Br⁻, producing one H⁺ ion per molecule. Therefore, the concentration of H⁺ is equal to the concentration of HBr, which is $1.0×10^{-2}$ M. The pH is $-log(10^{-2}$) = 2.00, and pOH = 14 - pH = 12.00. A tempting distractor might be 2.00, if someone calculated pH instead of pOH. For strong acids, the concentration directly gives ion concentration before taking logs.

This problem tests pH and pOH of strong acids and bases. HCl is a strong acid that completely dissociates in water: HCl → H⁺ + Cl⁻, meaning [H⁺] = [HCl] = 0.0010 M. To find pH, we use pH = -log[H⁺] = -log(0.0010) = -log(10⁻³) = 3.00. A common mistake would be to calculate pOH instead of pH, which would give 11.00 (choice B), but the question specifically asks for pH. For strong acids, always remember that the acid concentration directly equals the H⁺ concentration before taking the negative logarithm.

This question tests pH and pOH of strong acids and bases. HCl is a strong acid that completely dissociates in water according to HCl → H⁺ + Cl⁻, meaning a 1.0×10⁻³ M HCl solution produces [H⁺] = 1.0×10⁻³ M. The pH is calculated as pH = -log[H⁺] = -log(1.0×10⁻³) = -(-3) = 3.00. A common mistake is confusing pH with pOH, which would give 11.00 (choice B), but pH directly uses [H⁺] concentration. For strong acids, the acid concentration equals the H⁺ concentration before taking the negative logarithm.

This question tests pH and pOH of strong acids and bases. KOH is a strong base that completely dissociates as KOH → K⁺ + OH⁻, producing [OH⁻] = 1.0×10⁻² M from a 1.0×10⁻² M solution. To find [H⁺], we use the water equilibrium constant: Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C, so [H⁺] = Kw/[OH⁻] = (1.0×10⁻¹⁴)/(1.0×10⁻²) = 1.0×10⁻¹² M. A common mistake is using the OH⁻ concentration as the H⁺ concentration (choice B), forgetting they are inversely related. For strong bases, calculate [H⁺] by dividing Kw by [OH⁻].

This question tests pH and pOH of strong acids and bases. HBr is a strong acid that completely dissociates as HBr → H⁺ + Br⁻, producing [H⁺] = 0.050 M from a 0.050 M solution. The pH is calculated as pH = -log[H⁺] = -log(0.050) = -log(5×10⁻²) = 2 - log(5) ≈ 2 - 0.70 = 1.30. A common mistake is calculating pOH instead of pH, which would give pOH = 12.70 (choice B), but strong acids directly give H⁺ concentration for pH calculation. For strong acids, use the acid concentration as [H⁺] and calculate pH directly.

This question tests the skill of pH and pOH of strong acids and bases. HClO$_4$ is a strong acid, which means it fully dissociates in water according to $ \text{HClO}_4 \rightarrow \text{H}^{+} + \text{ClO}_4^{-} $, producing one H$^+$ ion per molecule. Therefore, the concentration of H$^+$ is equal to the concentration of HClO$_4$, which is $1.0 \times 10^{-3} , \text{M}$. The pH is calculated as $-\log[\text{H}^{+}] = -\log(10^{-3}) = 3.00$. A tempting distractor might be 11.00, if someone calculated pOH instead of pH. For strong acids, the concentration directly gives ion concentration before taking logs.