Fundamentals Of Acids & Bases

## Identify proton donation and acceptance - A Brønsted–Lowry acid donates a proton and a Brønsted–Lowry base accepts one in the reaction being considered. - In NH₃+H₂O⇌NH₄⁺+OH⁻, ammonia accepts H⁺ and water donates it. The proton transfer changes each partner’s hydrogen count and charge. - Aqueous H⁺ is shorthand for a hydrated proton; H₃O⁺ makes water’s role explicit. Acid–base proton transfer does not by itself require a change in oxidation states. ## Recognise genuine conjugate acid–base pairs - A conjugate acid and base differ by exactly one H⁺. Removing H⁺ lowers the species charge by one unit. - HCO₃⁻/CO₃²⁻ and NH₄⁺/NH₃ are conjugate pairs. H₃O⁺/OH⁻ differ by two hydrogens and are not one conjugate pair. - A pair can contain molecules or ions. Use atom and charge bookkeeping rather than choosing species only because both occur in an acid–base equation. ## Explain water’s amphiprotic behaviour - Water can accept a proton from HCl to form H₃O⁺. In that reaction water acts as a Brønsted base. - Water can donate a proton to ammonia to form OH⁻. In that reaction water acts as a Brønsted acid. - Amphiprotic means able to act as a proton donor or acceptor. The role depends on the reaction partner, not on water having one permanent acid/base label. ## Separate proton transfer from a later chemical change - In the supplied first step, bicarbonate accepts H⁺ to form H₂CO₃. The conjugate pair is H₂CO₃/HCO₃⁻. - The subsequent conversion H₂CO₃⇌CO₂+H₂O is a separate chemical process. It does not turn the two-product collection CO₂+H₂O into one conjugate species. - Apply the same care when a protonated species later loses water: identify the immediate one-proton partner before interpreting the overall multi-step equation. ## Distinguish acid strength from concentration - Strength describes the extent of ionisation under the stated conditions. Concentration describes how much solute is present per unit volume. - The supplied X is strong and relatively dilute, while Y is weak and more concentrated. A concentrated weak acid is not automatically a strong acid. - pH depends on the resulting hydrogen-ion activity or its dilute concentration approximation. It is not determined by the word strong alone, and it is not total neutralisation capacity. ## Calculate pH using a base10 logarithm - In the dilute AQA concentration approximation, pH=−log₁₀([H⁺]/(1 mol dm⁻³)). Use the numerical concentration in mol dm⁻³. - The supplied concentration1.60×10⁻⁴ gives pH3.795880..., which rounds to 3.80 to two decimal places. - pH has no unit. The more precise thermodynamic definition uses dimensionless hydrogen-ion activity; the concentration approximation has limits outside its applicable range. ## Recover hydrogen-ion concentration from pH - Rearrange the logarithmic relation to[H⁺]/(1 mol dm⁻³)=10^(−pH). The exponent is negative. - For pH3.10, the concentration is 7.943282...×10⁻⁴ mol dm⁻³, or7.94×10⁻⁴ to three significant figures. - Check direction: a lower pH corresponds to a higher hydrogen-ion concentration under the same approximation. Do not use a natural exponential instead of a base10 antilogarithm. ## Translate a pH difference into a concentration ratio - One unit lower pH means ten times the hydrogen-ion concentration in the same concentration model. - The supplied pH2.00 versus5.00 differs by three units, giving[H⁺]A/[H⁺]B=10³=1000. - This ratio compares hydrogen-ion concentrations. It does not by itself compare solution volumes, total acid moles, strength or the amount of alkali needed for complete neutralisation. ## Use the strong monoprotic-acid approximation - For a completely ionised monoprotic acid with negligible water contribution, each mole supplies one mole of aqueous H⁺. - The supplied0.0200 mol dm⁻³ HCl therefore gives[H⁺]≈0.0200 mol dm⁻³ and pH1.698970...≈1.70. - State the assumptions. This shortcut does not automatically apply to every polyprotic acid or to solutions so dilute that water’s ionisation is comparable to the added acid. ## Use the final total volume in a dilution - With no solute loss, n=cV is conserved during dilution. Use the final total solution volume, not just the volume of water added. - The supplied volume rises tenfold from 25.0 to 250.0 cm³, so HCl concentration falls from.0400 to.00400 mol dm⁻³. - With the stated strong-acid approximation, pH changes from 1.40 to 2.40, an increase of one unit. This simple logarithmic dilution rule does not describe all weak-acid or extreme-dilution cases. ## Convert amounts and volumes before taking logarithms - Find amount from n=m/M. The supplied.365 g HCl corresponds to.0100 mol using M=36.5 g mol⁻¹. - Convert500.0 cm³ to.5000 dm³ before c=n/V. The resulting formal concentration is.0200 mol dm⁻³. - Only after establishing concentration and ionisation assumptions should you calculate pH. Mixing cm³ with mol dm⁻³ introduces a factor-of 1000 error. ## Separate diprotic neutralisation from ionisation assumptions - A diprotic acid can supply two protons during complete neutralisation. The supplied equation therefore consumes two moles NaOH per mole H₂SO₄. - This stoichiometry does not prove both ionisation steps are complete in the original acid solution. Sulfuric acid’s second ionisation is condition dependent. - For the supplied complete-first/partial-second model, do not set[H⁺]=2C or ignore the second step without instruction. A numerical pH needs an appropriate equilibrium constant or an explicit approximation. ## Use the ionic product at the stated temperature - Water maintains the equilibrium2H₂O⇌H₃O⁺+OH⁻. In the concentration convention, Kw=[H⁺][OH⁻]. - Using the supplied rounded Kw=1.00×10⁻¹⁴ mol² dm⁻⁶ and[OH⁻]=2.00×10⁻³ gives[H⁺]=5.00×10⁻¹² mol dm⁻³. - Kw depends on temperature. Acidic and alkaline aqueous solutions still contain both ions, with one often present at much lower concentration. ## Define neutrality by equal ion concentrations - Neutrality requires[H⁺]=[OH⁻] in this aqueous concentration model. For pure water each concentration equals√Kw. - At the supplied higher temperature, Kw=4.00×10⁻¹⁴ gives both ions2.00×10⁻⁷ mol dm⁻³ and neutral pH6.70. - Heating pure water increases Kw over the ordinary range because ionisation is endothermic. Both ions increase together; lower pH does not make this pure water acidic relative to neutrality at that temperature. ## Calculate pH from a strong alkali - Fully dissolved NaOH supplies one hydroxide per formula unit. With water contribution negligible, [OH⁻]≈.150 mol dm⁻³. - Using the supplied Kw gives[H⁺]=6.666...×10⁻¹⁴ mol dm⁻³. Its pH is 13.176091..., rounding to 13.18. - Keep guard digits until the end. Truncating to 13.17 is not rounding to two decimal places, and the temperature-specific Kw must remain consistent. ## Count hydroxide ions per dissolved formula unit - Fully dissociated Ca(OH)₂ gives Ca²⁺+2OH⁻. The supplied dissolved concentration.00500 mol dm⁻³ therefore gives[OH⁻]=.0100 mol dm⁻³. - With Kw=1.00×10⁻¹⁴ this implies[H⁺]=1.00×10⁻¹² mol dm⁻³ and pH12.00. - The given amount is already dissolved. Do not assume an arbitrary added mass of a sparingly soluble hydroxide all dissolves; a solubility problem would need separate information. ## Relate pH and pOH using p Kw - Define pOH=−log₁₀([OH⁻]/(1 mol dm⁻³)). Taking negative logarithms of Kw gives pH+pOH=p Kw. - For the supplied rounded Kw=10⁻¹⁴, p Kw=14.00. A pOH of 3.20 therefore gives pH10.80. - The sum is not always14 at every temperature. Use the given Kw to find p Kw if the temperature or model value changes. ## Classify a solution at its own temperature - Compare hydrogen and hydroxide concentrations: more H⁺ is acidic, equal concentrations are neutral, and more OH⁻ is alkaline in this model. - The supplied[H⁺]=1.00×10⁻⁷ and[OH⁻]=4.00×10⁻⁷ satisfy Kw=4.00×10⁻¹⁴. The solution is alkaline even though its pH is 7.00. - Use the neutral point for the stated temperature, not a universal assumption that every pH7 solution is neutral. ## Recognise when water’s contribution matters - For very dilute strong acid, setting[H⁺]=C can become invalid. The supplied C=10⁻⁸ mol dm⁻³ is comparable to water’s own ions. - Extension: charge balance[H⁺]=C+[OH⁻] and Kw=[H⁺][OH⁻] give[H⁺]=(C+√(C²+4Kw))/2 for this ideal HCl model. - The supplied case gives pH≈6.98 rather than 8.00. Acid dilution approaches neutrality; it does not turn pure strong acid plus pure water alkaline in this closed ideal model. ## Round the final result rather than truncating - Use all justified intermediate digits when dividing by Kw or taking logarithms. Round the final requested quantity once. - The supplied unrounded13.176091259... becomes 13.18 to two decimal places because the third decimal digit is 6. - pH is dimensionless, whereas[H⁺] and[OH⁻] carry concentration units. Follow the requested precision and keep the logarithm convention explicit.