Arrow Lab

Química Geral · Acid–base and solubility equilibria

Soluções-tampão

Uma lição resolvida gratuita e 40 problemas de prática em que você mesmo desenha cada seta curva.

Travou em soluções-tampão? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

Escaneie sua lição de graça

O que você vai aprender

  • Recognize a buffer as a weak acid and its conjugate base in comparable amounts, and explain the common-ion effect.
  • Calculate the pH of a buffer with the Henderson–Hasselbalch equation from concentrations, moles, masses or Kb.
  • Calculate the pH after a strong acid or base is added to a buffer, doing the neutralization stoichiometry before the equilibrium step.
  • Relate buffer capacity and useful pH range to the amounts of the two components and to the pKa.
  • Choose a conjugate pair for a target pH and work out how to prepare the buffer.

Lição gratuita

O texto da lição resolvida. No Arrow Lab, cada etapa vem com as estruturas e as setas desenhadas.

Ler a aula gratuita (7 partes)

What a buffer is

A buffer is a solution that resists changes in pH when a little strong acid or base is added. It contains a weak acid and its conjugate base in comparable amounts: acetic acid with sodium acetate, or ammonium chloride with ammonia (drawn). The weak acid neutralizes added OH⁻ and the weak base neutralizes added H₃O⁺. A strong acid with its anion, such as HCl with NaCl, cannot buffer: chloride is too weak a base to take a proton back. Two acids, or two bases, cannot buffer either, because one of the two jobs is left undone. A buffer can be made by mixing the two members of the pair, or by partly neutralizing a weak acid with NaOH (or a weak base with HCl) so that the conjugate partner forms in the flask.

The common-ion effect

Adding a product of an equilibrium pushes it backwards (Le Châtelier). For HA + H₂O ⇌ H₃O⁺ + A⁻, adding A⁻ from a soluble salt, or H₃O⁺ from a strong acid, suppresses the ionization of HA. 0.10 M acetic acid alone (Ka = 1.8 × 10⁻⁵) has [H₃O⁺] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M: pH 2.87, 1.3 % ionized. In the same solution made 0.10 M in sodium acetate, [H₃O⁺] = Ka × [HA]/[A⁻] = 1.8 × 10⁻⁵ M: pH 4.74, only 0.018 % ionized. Because so little reacts, the equilibrium concentrations of HA and A⁻ in a buffer are essentially the amounts that were mixed.

The Henderson–Hasselbalch equation

Taking −log of Ka = [H₃O⁺][A⁻]/[HA] gives pH = pKa + log([A⁻]/[HA]). Equal amounts give pH = pKa; each tenfold excess of base raises the pH one unit above pKa, and each tenfold excess of acid lowers it one unit below. Both species share one volume, so a ratio of moles works as well as a ratio of concentrations. For a buffer built on a weak base, use the pKa of its conjugate acid, pKa = 14.00 − pKb at 25 °C: 0.20 M NH₃ with 0.10 M NH₄Cl has pH = 9.25 + log(0.20/0.10) = 9.25 + 0.30 = 9.55. The equation is reliable when both concentrations are much larger than [H₃O⁺] and [OH⁻] and the ratio lies between about 0.1 and 10. In very dilute buffers of a fairly strong weak acid, the amount that ionizes is no longer negligible and the full equilibrium must be solved.

Adding strong acid or base: stoichiometry first

Work in two stages. First let the strong acid or base react completely with the buffer component that consumes it: H₃O⁺ + A⁻ → HA + H₂O, or OH⁻ + HA → A⁻ + H₂O. Tabulate moles before and after. Then put the new amounts into the Henderson–Hasselbalch equation. Example: 1.00 L containing 0.050 mol H₂PO₄⁻ and 0.050 mol HPO₄²⁻ (drawn; pKa2 7.20) receives 0.010 mol NaOH. Afterwards H₂PO₄⁻ is 0.040 mol and HPO₄²⁻ 0.060 mol, so pH = 7.20 + log(0.060/0.040) = 7.20 + 0.18 = 7.38. The same NaOH in 1.00 L of pure water would give pH 12.00. If the strong acid or base exceeds the component that neutralizes it, the buffer is exhausted: the excess strong acid or base, divided by the total volume, sets the pH.

Buffer capacity and range

Buffer capacity is the amount of strong acid or base a buffer can take before its pH changes appreciably. It grows with the amounts of the two components and is largest, for a given total, when they are equal, that is at pH = pKa, where the buffer resists acid and base equally. Diluting a buffer leaves the ratio, and so the pH, nearly unchanged but lowers its capacity. Example: add 0.0020 mol HCl to 1.0 L of an acetate buffer (pKa 4.76). With 0.10 M of each component the pH falls to 4.76 + log(0.098/0.102) = 4.74; with 0.010 M of each it falls to 4.76 + log(0.008/0.012) = 4.58. The useful range is about pKa ± 1, where the ratio [A⁻]/[HA] lies between 0.1 and 10; beyond it one component is so scarce that a small addition changes the pH sharply.

Choosing and preparing a buffer

Pick the conjugate pair whose pKa is closest to the target pH, and certainly within one unit. For a polyprotic acid, choose the pair whose pKa lies nearest the target: H₂PO₄⁻/HPO₄²⁻ (pKa2 7.20) for pH 7. Then find the ratio from 10^(pH − pKa) and supply it in one of three ways: weigh out both members of the pair; add NaOH to the weak acid; or add HCl to the weak base. Example: to bring 0.100 mol acetic acid to pH 4.50, the ratio [A⁻]/[HA] must be 10^(4.50 − 4.76) = 0.55. Adding x mol NaOH gives x/(0.100 − x) = 0.55, so x = 0.0355 mol. When the total concentration is fixed, split it in the ratio: the fraction present as base is ratio/(1 + ratio).

What to carry forward

A buffer is a weak conjugate pair in comparable amounts; the common-ion effect keeps the amounts you mix equal to the amounts at equilibrium. Use pH = pKa + log([A⁻]/[HA]) with moles or concentrations. When strong acid or base is added, do the neutralization stoichiometry first and the Henderson–Hasselbalch step second, and check that the buffer is not exhausted. Capacity follows the amounts; the useful range is pKa ± 1, so choose the pair whose pKa is closest to the pH you need.

Problemas de prática (40)

Praticar é grátis com uma conta.

Desenhe o mecanismo 2

Adding strong acid or base

  • A phosphate buffer absorbs added acidDraw hydrogen phosphate, the base component of a phosphate buffer, taking a proton from hydronium, and calculate the small pH change that results.
  • A Tris buffer absorbs added baseIdentify the protonated amine as the component of a Tris buffer that neutralizes hydroxide, draw the proton transfer, and calculate the new pH.

Preveja o produto 3

Buffer composition and the common ion

  • Add acetate to acetic acidPredict how a common ion changes the percent ionization and pH of a weak acid solution.

Adding strong acid or base

  • What happens to added hydroxide?Predict the reaction by which the weak-acid component of a buffer consumes added hydroxide.

Buffer capacity and range

  • The useful range of a phosphate bufferPredict the effective pH range of a buffer pair as pKa ± 1.

Classificar 2

Buffer composition and the common ion

  • Which species takes up added acid?Identify the base component of an ammonia–ammonium buffer as the species that consumes added hydronium.
  • Spot the bufferClassify a solution as a buffer only when it contains a weak acid and its conjugate base in comparable amounts.

Comparar 3

Buffer capacity and range

  • Same pH, different capacityCompare the buffer capacity of two buffers with the same pH but different concentrations.
  • Rank buffers by acid capacityRank buffers by how much strong acid they can neutralize by computing the moles of the base component from volume and concentration.

Buffer composition and the common ion

  • Rank by pH: the common-ion effectRank strong acid, weak acid and acetic acid–acetate mixtures by pH using the common-ion effect.

Calcular 19

Adding strong acid or base

  • Ammonia buffer plus HClCalculate the pH of an ammonia–ammonium buffer after strong acid converts part of the ammonia into ammonium.
  • When the buffer runs outCalculate the pH when added strong acid exceeds the base component of a buffer, using the excess acid and the total volume.
  • Formate buffer plus NaOH solutionCalculate a buffer pH after adding a measured volume of strong base, working in millimoles.
  • Stoichiometry firstTrack the moles of each buffer component after a strong base reacts completely with the weak acid.
  • Acid in, then base inApply the stoichiometry step to successive additions of strong acid and strong base before using the Henderson–Hasselbalch equation once.

Henderson–Hasselbalch pH

  • Ammonia buffer pH from KbCalculate the pH of a weak-base buffer from Kb, converting through pOH or through the pKa of the conjugate acid.
  • Buffer from a weighed saltCalculate a buffer pH from a mass of the conjugate-acid salt and a volume of base solution, using mole amounts in the ratio.
  • pH of an equimolar benzoate bufferRecognize that a buffer with equal concentrations of weak acid and conjugate base has pH equal to the pKa, and find the pKa from Ka.
  • Bicarbonate in blood plasmaUse the Henderson–Hasselbalch equation with a physiological pKa to find the base concentration in the blood bicarbonate buffer.
  • pH of a formate bufferApply the Henderson–Hasselbalch equation to a buffer with excess conjugate base.
  • Pyridine–pyridinium bufferCalculate the pH of a buffer built on a weak base by converting Kb to the pKa of its conjugate acid.

Choosing and preparing a buffer

  • Titrate ammonia to pH 9.00Calculate the volume of strong acid to add to a weak base to prepare a buffer of a target pH.
  • Buffer by partial neutralizationCalculate the pH of a buffer made by neutralizing part of a weak acid with strong base.
  • Phosphate buffer with a fixed totalSplit a fixed total buffer concentration between acid and base forms to reach a target pH, and convert to a mass.
  • How much sodium acetate?Calculate the mass of conjugate-base salt needed to bring a weak acid solution to a target pH.

Buffer capacity and range

  • A dilute buffer under strainCalculate the larger pH change of a dilute buffer after strong acid is added and relate it to low capacity.
  • How much acid for one pH unit?Calculate the amount of strong acid that lowers a buffer's pH by one unit, a quantitative measure of buffer capacity.

Buffer composition and the common ion

  • Percent ionization with a common ionCalculate the percent ionization of a weak acid in the presence of its conjugate base.
  • HF in hydrochloric acidCalculate how a strong acid supplying the common ion H₃O⁺ suppresses the ionization of a weak acid.

Encontre o erro 2

Buffer capacity and range

  • An acetate buffer at pH 7?Diagnose why a buffer adjusted to a pH far from its pKa has almost no capacity.

Buffer composition and the common ion

  • A phosphate 'buffer' that isn'tDiagnose why two phosphate ions that differ by two protons do not form a buffer at the pKa between them.

Interprete as evidências 2

Buffer capacity and range

  • How much buffer was there?Interpret a measured pH change after adding strong base to work out how much of each buffer component was present.

Henderson–Hasselbalch pH

  • Read the composition from the pHInfer which buffer component is in excess, and by how much, by comparing the buffer pH with the pKa.

Trabalhe de trás para frente 1

  • pKa from a measured buffer pHWork backwards from a measured buffer pH and composition to the pKa of the weak acid.

Planeje uma síntese 3

Choosing and preparing a buffer

  • Make an acetate buffer from the saltPlan a buffer by partly protonating a weak base salt with a strong acid.
  • Pick a buffer for pH 3.80Choose the conjugate pair whose pKa is closest to an acidic target pH.
  • A phosphate buffer from phosphoric acidPlan the amount of strong base needed to turn a polyprotic acid into a buffer at its second pKa.

Explique por quê 3

Buffer capacity and range

  • Why pH = pKa is the sweet spotExplain why buffer capacity is greatest when the acid and conjugate base are present in equal amounts.

Henderson–Hasselbalch pH

  • When Henderson–Hasselbalch breaks downExplain why the Henderson–Hasselbalch equation fails for a very dilute buffer of a relatively strong weak acid.

Buffer composition and the common ion

  • Why mixed amounts are good enoughExplain why the Henderson–Hasselbalch equation can use the concentrations as mixed instead of equilibrium values.

Travou em soluções-tampão? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

Escaneie sua lição de graça