Further Acids & Bases Calculations

## Define the acid dissociation constant - For HA ⇌ H⁺ + A⁻, Ka=[H⁺][A⁻]/[HA] in the dilute concentration convention. All concentrations are equilibrium values. - Water is the solvent; its approximately constant activity is absorbed into this convention. Neglecting water-derived H⁺ is a separate calculation assumption. - For a polyprotic acid, write one Ka for each proton-loss step. H₂CO₃ first gives HCO₃⁻, not CO₃²⁻ and two protons at once. ## Calculate weak-acid pH with stated assumptions - For a pure monoprotic acid with negligible water ionisation, [H⁺]≈[A⁻]=x and [HA]=C−x. Thus Ka=x²/(C−x). - If x is much smaller than C, use x≈√(KaC). For C=0.0800 and Ka=2.00×10⁻⁵, x≈1.265×10⁻³ mol dm⁻³. - Then pH≈2.90 to two decimal places. Retain guard digits and check that ionisation is small before accepting the approximation. ## Test the small-ionisation shortcut - The square-root estimate must be checked against the original acid concentration. A five-percent check is a useful stated classroom criterion, not a law of equilibrium. - Here x/C≈0.316, or 31.6%, so replacing C−x by C is poor. Solve x²+Kax−KaC=0 instead. - The positive root is 2.70×10⁻⁴ mol dm⁻³ and pH≈3.57. If water ionisation is also important, a fuller charge-balance treatment is needed. ## Find Ka from a measured pH - Convert pH to [H⁺] first: pH 3.00 gives x=1.00×10⁻³ mol dm⁻³. For the specified pure-acid model, [A⁻]≈x. - The remaining acid is 0.0500−0.00100=0.0490 mol dm⁻³. Thus Ka=x²/(C−x)=2.04×10⁻⁵ mol dm⁻³. - Using the initial concentration in the denominator would be an approximation. Do not present an approximate denominator as the exact equilibrium concentration. ## Use pKa to compare acids - pKa=−log₁₀Ka using the specified concentration reference convention. Ka=2.00×10⁻⁵ gives pKa=4.70 to two decimal places. - A larger Ka corresponds to a smaller pKa and greater ionisation at matched concentration and conditions. Acid Y is stronger than acid X in this comparison. - Dilution can increase the fraction ionised without changing Ka at fixed temperature. Weak-acid pKa values do not all lie between 3 and 7. ## Use half-neutralisation to estimate pKa - The plotted model starts with 25.0 cm³ of 0.100 mol dm⁻³ HA and adds 0.100 mol dm⁻³ strong base; Ka=1.80×10⁻⁵ and Kw=1.00×10⁻¹⁴ at 298 K using the rounded Kw. - Equivalence is 25.0 cm³, so half-neutralisation is 12.5 cm³. With appreciable HA and A⁻ and small equilibrium corrections, their concentrations are approximately equal and pH≈pKa≈4.74. - Read pH at half the equivalence volume, then calculate Ka≈10^(−pH). The graph includes water and equilibrium shifts, so formal half-neutralisation does not assert exact equality of all equilibrium species. ## Calculate concentration from a titre - Convert the acid volume to dm³: n(HCl)=0.0600×0.02000=1.20×10⁻³ mol. The equation gives the same amount of NaOH. - Divide by the delivered base volume in dm³: C(NaOH)=0.00120/0.01500=0.0800 mol dm⁻³. - Use the balanced equation rather than assuming every titration is 1:1. Diprotic neutralisation can require two moles of OH⁻ per mole of acid. ## Find pH after a strong reagent is in excess - Neutralise amounts first: acid supplies 2.00 mmol and base supplies 3.00 mmol. There is 1.00 mmol of excess hydroxide. - Use the total 50.0 cm³ volume, giving [OH⁻]≈0.0200 mol dm⁻³. Then pH=pKw+log₁₀[OH⁻]=12.30. - Before equivalence, excess strong acid controls pH; after equivalence, excess strong base controls it. At equivalence use the appropriate water or salt equilibrium. ## Measure a pH curve in required practical 9 - Transfer a fixed aliquot using suitable volumetric apparatus and record the initial pH with a calibrated probe. Keep the sensing region immersed without striking the glass base. - Add measured reagent increments from a rinsed burette, mix and wait for stable readings. Use smaller increments around the rapid pH change. - Plot pH against cumulative added volume. Remove the filling funnel before readings, avoid a jet air bubble, read the meniscus at eye level and follow the specified eye-protection and handling procedure. ## Treat titres and uncertainty consistently - The titre is final minus initial reading: 24.60−1.20=23.40 cm³. It is a delivered volume, not the final burette reading alone. - With ±0.05 cm³ per reading, the worst-case absolute titre uncertainty is ±0.10 cm³. The corresponding percentage is about 0.43%. - Use the stated concordance rule to select repeat results before averaging; exclude the rough trial. Instrument uncertainty and the permitted spread between repeats are different quantities. ## Interpret a strong-acid/strong-base curve - Model A starts with 25.0 cm³ of 0.100 mol dm⁻³ strong monoprotic acid and adds 0.100 mol dm⁻³ strong base; Kw=1.00×10⁻¹⁴ at 298 K. - The equivalence volume is 25.0 cm³. With inert salt ions, [H⁺]=[OH⁻] there and the specified neutral pH is 7.00. - The model solves water ionisation continuously through equivalence. Neutrality is not universally pH 7 at other temperatures, and the vertical region depends on concentration. ## Explain weak-acid/strong-base equivalence - Model B uses the same 25.0 cm³ and 0.100 mol dm⁻³ reagents, with HA Ka=1.80×10⁻⁵, a strong base and Kw=1.00×10⁻¹⁴. - Before equivalence, HA and A⁻ form a buffer region. At equivalence the conjugate base hydrolyses: A⁻+H₂O⇌HA+OH⁻. - The equivalence solution is alkaline, about pH 8.72 in this model. Stoichiometric equivalence does not mean every equilibrium molecule of HA has disappeared. ## Explain strong-acid/weak-base equivalence - Model C uses a strong monoprotic acid and a weak base with Kb=1.80×10⁻⁵; initial and added reagent concentrations are 0.100 mol dm⁻³, with 25.0 cm³ initially and Kw=1.00×10⁻¹⁴. - At equivalence the weak base’s conjugate acid reacts with water: BH⁺+H₂O⇌B+H₃O⁺. The equivalence pH is about 5.28 in this model. - Beyond equivalence the excess weak base does not follow [OH⁻]=its formal concentration. Its ionisation equilibrium must still be considered. ## Evaluate a weak-acid/weak-base curve - Model D has 25.0 cm³ of 0.100 mol dm⁻³ weak acid and adds 0.100 mol dm⁻³ weak base, with Ka=Kb=1.80×10⁻⁵ and Kw=1.00×10⁻¹⁴. - Both conjugate partners participate in equilibrium. Matching Ka and Kb gives neutral equivalence here; unequal constants can give acidic or alkaline equivalence. - This curve has a gradual transition, so a simple colour indicator gives a poorly defined endpoint for this example. A weak/weak titration does not universally have a sharp vertical step. ## Read the addition direction before interpreting a curve - Model E starts with 25.0 cm³ of 0.100 mol dm⁻³ strong base and adds 0.100 mol dm⁻³ strong acid; Kw=1.00×10⁻¹⁴. - The x axis still records added volume, but the pH falls as acid is added. The initial reagent and addition direction determine which excess species is present. - Do not mechanically reflect arbitrary weak/strong curves. Recalculate diluted totals and equilibrium for the stated reagents and concentrations. ## Explain an indicator colour transition - An acid–base indicator has differently coloured conjugate forms. In the stated model HIn⇌H⁺+In⁻, increasing acidity favours HIn. - At equal equilibrium concentrations of the two forms, pH=pKIn in the dilute model. A visible endpoint need not occur at exactly equal form concentrations. - The observed colour changes over a range because the form ratio changes continuously. The endpoint is the chosen experimental colour change; equivalence is the stoichiometric point. ## Choose an indicator from the actual curve - Compare the indicator’s transition range with the steep pH region around equivalence for the actual titration. The supplied example supports indicator P. - A transition inside this region changes colour over a small delivered volume, reducing endpoint error. Universal rules such as every steep region being pH 4–10 are unreliable. - Phenolphthalein often suits a weak-acid/strong-base titration at suitable concentrations. Acidic-range indicators can suit a strong-acid/weak-base example; use the supplied curve and range. ## Recognise the components of a buffer - A buffer resists large pH changes when small quantities of acid or alkali are added. An acidic buffer contains appreciable weak acid and its conjugate base. - Ethanoic acid and soluble ethanoate supply CH₃COOH/CH₃COO⁻. A strong acid plus its inert salt is not the same kind of buffer. - There must be enough of each component to consume the intended addition. The statement pH remains exactly constant overclaims what a buffer can do. ## Write both acidic-buffer reactions - Added acid is consumed mainly by the conjugate base: A⁻+H⁺→HA. Added alkali is consumed by the weak acid: HA+OH⁻→A⁻+H₂O. - Both net equations conserve atoms and charge. Do not add an extra H⁺ to the products of the hydroxide reaction. - The component ratio changes slightly, so the pH changes slightly. Appreciable reserves of both partners explain the limited response. ## Explain a basic buffer without inventing hydroxide in the salt - An NH₃/NH₄⁺ mixture is a basic buffer under appropriate conditions. NH₄Cl dissolves into NH₄⁺ and Cl⁻; it does not dissociate into NH₄⁺ and OH⁻. - Added acid reacts with ammonia: NH₃+H⁺→NH₄⁺. Added hydroxide reacts with ammonium: NH₄⁺+OH⁻→NH₃+H₂O. - The chloride is a spectator in these net proton-transfer reactions. Distinguish salt dissolution from weak-base equilibrium with water. ## Calculate buffer pH with the correct ratio - Rearrange Ka=[H⁺][A⁻]/[HA] to [H⁺]=Ka[HA]/[A⁻]. Taking negative base-10 logarithms gives pH=pKa+log₁₀([A⁻]/[HA]). - For the supplied concentrations, [H⁺]≈1.00×10⁻⁵ mol dm⁻³ and pH≈5.00. More conjugate base than acid makes pH greater than pKa. - Use equilibrium concentrations, or analytical amounts when equilibrium corrections are negligible. Amount ratios replace concentration ratios only when both components share the same final volume. ## Neutralise first when preparing a buffer - The acid initially supplies 6.00 mmol and hydroxide supplies 2.00 mmol. After the stoichiometric reaction, approximately 4.00 mmol HA and 2.00 mmol A⁻ remain. - Both components occupy the same 60.0 cm³ final volume, so their ratio is 2.00/4.00. Then pH≈4.69897+log₁₀(0.5)=4.40. - Do not put the original acid and hydroxide amounts directly into the buffer ratio. The hydroxide has reacted and the weak-acid amount has changed. ## Update both amounts after adding acid - Added H⁺ converts A⁻ into HA. The supplied addition changes the approximate amounts to 9.00 mmol A⁻ and 11.0 mmol HA. - Then pH≈pKa+log₁₀(9/11)=4.61. The pH falls slightly because the conjugate-base/acid ratio decreases. - Account for final volume if concentrations are needed. It cancels in this ratio because both remaining components share that volume. ## Update both amounts after adding alkali - Hydroxide consumes HA and forms A⁻. The approximate component amounts become 8.00 mmol HA and 12.0 mmol A⁻. - The buffer calculation gives pH≈pKa+log₁₀(12/8)=4.88. The pH rises because the conjugate-base/acid ratio increases. - A small pH change is compatible with buffering. The same absolute addition has a larger effect when the original component amounts are smaller. ## Stop using the buffer equation after a component is depleted - The first 1.00 mmol OH⁻ consumes the available HA. Another 2.00 mmol OH⁻ remains in excess, so the buffer approximation no longer applies. - In the stated strong-excess approximation, [OH⁻]≈0.00200/0.1000=0.0200 mol dm⁻³, giving pH≈12.30. - Do not take the logarithm of a negative or zero remaining analytical acid amount. At a weak-acid titration’s exact equivalence, consider conjugate-base hydrolysis instead. ## Separate buffer pH, capacity and application - Diluting both components by the same factor preserves their ratio and approximately preserves pH while the buffer approximation remains valid. - Equal volumes of the diluted solution contain fewer moles of each component, so they resist a smaller acid or alkali addition. Extreme dilution can invalidate the simple ratio treatment. - Buffers help maintain suitable pH in food processing and enzyme assays. Biological bicarbonate buffering belongs to an open system regulated by gas exchange and ion handling; a buffer alone does not hold physiological pH fixed.