Doc Brown's Chemistry  Theoretical Chemistry - Equilibria - Chemical Equilibrium 5.4

5.4 Definition, examples, pH, K_a & pK_a weak acid calculations

Revision notes for GCE Advanced Subsidiary Level AS Advanced Level A2 IB Revise AQA GCE Chemistry OCR GCE Chemistry Edexcel GCE Chemistry Salters Chemistry CIE Chemistry revising courses for pre-university students (equal to US grade 11 and grade 12 and Honours/honors level courses) KS4 Science GCSE/IGCSE Chemistry reversible reactions-equilibrium * KS4 Science GCSE/IGCSE notes acids and bases * KS4 Science GCSE/IGCSE notes acid-base theory

Equilibria Part 5 sub-index:  5.1 Lewis and Bronsted-Lowry acid-base theories * 5.2 self-ionisation of water and pH scale * 5.3 strong acids-examples-calculations * 5.4 weak acids-examples & pH-K_a-pK_a calculations * 5.5 strong bases-examples-pH calculations * 5.6 weak bases- examples & pH-K_b-pK_b calculations * 5.7 A level notes on Acids, Bases, Salts, uses of acid-base titrations - upgrade from GCSE! * EMAIL query?comment

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle-rules * Part 2. K_c and K_p equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * 4. Partition, solubility product and ion-exchange * Part 5. pH, weak-strong acid-base theory and calculations * Part 6. Salt hydrolysis, Acid-base titrations-indicators, pH curves and buffers * Part 7. Redox equilibria, half-cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria-vapour pressure, boiling point and intermolecular forces

5.4 Definition, examples and pH, K_a and pK_a calculations of weak acids

• 5.4.1 Definition and examples of WEAK ACIDS

  • Weak acids are only partially ionised in water.

  • In principle the reaction is: HA_(aq) + H₂O_(l) H₃O⁺_(aq) + A^-_(aq)

    • or more simply: HA_(aq) H⁺_(aq) + A^-_(aq)

    • The low % of ionisation gives a less acidic solution of higher pH than for strong acids, but still pH < 7.

• 5.4.2a Examples include organic carboxylic acids like ethanoic acid which are just a few % ionized.

  • CH₃COOH_(aq) + H₂O_(l) H₃O⁺_(aq) + CH₃COO^-_(aq)

  • or more simply: CH₃COOH_(aq) H⁺_(aq) + CH₃COO^-_(aq)

  • Ethanoic acid is the B-L acid and the ethanoate ion its conjugate base.

    • In general, a weak acid has a strong conjugate base.

  • Water is the base and the hydrogen/oxonium ion is its conjugate acid.

  • Since the water concentration is essentially constant, the equilibrium expression for a monobasic acid is written as:

  • Ka =	[H+(aq)] [A-(aq)]       [H+(aq)] [CH3COO-(aq)]
    	---------------------- = ------------------------------------
    	[HA(aq)]              [CH3COOH(aq)]

  • K_a is called the acid ionisation or dissociation constant with units of mol dm^-3.

    • Note:

    • Since water is the solvent, [H₂O_(l)], it is effectively constant and omitted from K_a expressions.

    • Ethanoic acid pK_a = 4.76,  K_a = 1.74 x 10^-5 mol dm^-3 and is only about 2% ionised.

• 5.4.2b The equilibrium can also be expressed as the acid-base reaction of the conjugate base with water (below) but the above expression is invariably used in problem solving.

  • e.g. for ethanoic acid: CH₃COO^-_(aq) + H₂O_(l) CH₃COOH_(aq) + OH^-_(aq)

  • Kb =	[CH3COOH(aq)] [OH-(aq)]
    	---------------------------------------
    	[CH3COO-(aq)]

  • Note that: K_a-acid x K_{b-conj. base} = K_w and pK_a + pK_b = pK_w, check it out for yourself.

• 5.4.2c Ionic acid-base equilibrium can be more complicated in the case of dibasic/diprotic acids.

  • e.g. , ethanedioic acid, more simply shown as HOOC-COOH.

  • HOOC-COOH_(aq) H⁺_(aq) + HOOC-COO^-_(aq)

    • K_a1 = [H⁺_(aq)] [HOOC-COO^-_(aq)]_/[HOOC-COOH_(aq)] mol dm^-3

  • HOOC-COO^-_(aq) H⁺_(aq) ^-OOC-COO^-_(aq)

    • K_a2 = [H⁺_(aq)] [^-OOC-COO^-_(aq)]_/[HOOC-COO^-_(aq)] mol dm^-3

  • and K_a1 > K_a2, showing, not surprisingly, the 1st proton is released more readily than the 2nd.

  • K_a1 =  5.89 x 10^-2 mol dm^-3 (pK_a1 = 1.23), and K_a2 = 5.24 x 10^-5 mol dm^-3 (pK_a2 = 4.28)

• 5.4.2d There are many examples of inorganic weak acids e.g.

  • (a) Hydrofluoric acid, HF: pK_a = 3.25, K_a = 5.6 x 10^-4 mol dm^-3

    • HF_(aq) + H₂O_(l) H₃O⁺_(aq) + F^-_(aq)

    • The strong hydrogen-fluorine bond and the intermolecular HF-H₂O hydrogen bonding are mainly responsible for the lack of dissociation into ions in dilute solution. HF (562), HCl (431), HBr (366) and HI (299) have progressively weaker bonds as the halogen atom gets bigger and the bond length increases, so bar HF they are all very strong acids and virtually completely ionised and don't hydrogen bond with water. (endothermic bond enthalpies in kJ mol-1)

  • (b) Hydrocyanic acid, HCN: pK_a = 9.31, K_a = 4.9 x 10^-10 mol dm^-3

    • HCN_(aq) + H₂O_(l) H₃O⁺_(aq) + CN^-_(aq)

    • The strong hydrogen-carbon bond is mainly responsible for the lack of ionisation.

  • (c) Phosphoric(V) acid, H₃PO₄, is a tribasic acid.

    • (a1) H₃PO_4(aq) H⁺_(aq) + H₂PO₄^-_(aq)  (K_a1 = 7.9 x 10^-3 mol dm^-3, pK_a1 = 2.1)

    • (a2) H₂PO₄^-_(aq) H⁺_(aq) + HPO₄^2-_(aq)  (K_a2 = 6.2 x 10^-8 mol dm^-3, pK_a2 = 7.2)

    • (a3) HPO₄^2-_(aq) H⁺_(aq) + PO₄^3-_(aq)  (K_a3 = 4.4 x 10^-13 mol dm^-3, pK_a3 = 12.4)

    • The subsequent ions H₂PO₄^- and HPO₄^2- are, not surprisingly, increasingly weak acids.

  • (d) Hexa-aqua complex ions can donate protons to water

    • e.g. [M(H₂O)₆]²⁺_(aq) + H₂O_(l) [M(H₂O)₅(OH)]⁺_(aq) + H₃O⁺_(aq) 

    • where M = Mn, Fe, Co, Ni, Cu, Mg etc. give very weak acid solutions with pH's just less than 7.

    • or [M(H₂O)₆]³⁺_(aq) + H₂O_(l) [M(H₂O)₅(OH)]²⁺_(aq) + H₃O⁺_(aq) 

      • e.g. when M = Ti, V, Cr, Fe, Al etc. give very weak acids solutions of pH's in the 3-5 region, but generally stronger than for M²⁺ because of the greater polarising power of the more highly charged central cation.

• 5.4.2e Carbon dioxide is a weakly acidic gas.

  • It dissolves in water to give 'carbonic acid' (fizzy 'carbonated water'!). Unpolluted rainwater has a pH of about 5.5 when in equilibrium with the 0.03-0.04% of CO₂ in air.

  • The carbon dioxide may exist as (a) dissolved CO₂ or (b) 'carbonic acid', which complicates matters a bit, but either should get you the marks in the exam! So the possible equilibria are:

    • (a) CO_2(g) CO_2(aq)  and (b)  CO_2(g) + H₂O_(l) H₂CO_3(aq),

      • prior to acid ionisation/dissociation.

    • (c) CO_2(aq) + 2H₂O_(l) HCO₃^-_(aq) + H₃O⁺_(aq)

      • or (d) H₂CO_3(aq) + H₂O_(l) HCO₃^-_(aq) + H₃O⁺_(aq)

      • and more simply:

        • (c) CO_2(aq) + H₂O_(l) HCO₃^-_(aq) + H⁺_(aq)

        • or (d) H₂CO_3(aq) HCO₃^-_(aq) + H⁺_(aq)

        • so the 1st ionization gives the hydrogencarbonate ion and hydrogen ion.

      • (a) pKa_1(CO2(aq)) = 6.4  (very weak acid)

        • Ka_1(CO2(aq)) = [HCO₃^-_(aq)] [H⁺_(aq)] / [CO_2(aq)] = 4.0 x 10^-7 mol dm^-3

      • (b) pK_a1(H2CO3) = 3.7  (weak acid)

        • K_a1(H2CO3) = [HCO₃^-_(aq)] [H⁺_(aq)] / [H₂CO_3(aq)] = 2.0 x 10^-4 mol dm^-3

    • HCO₃^-_(aq) + H₂O_(l) CO₃^2-_(aq) + H₃O⁺_(aq)

      • more simply: HCO₃^-_(aq) CO₃^2-_(aq) + H⁺_(aq)

      • The 2nd ionization gives the carbonate ion and hydrogen ion.

      • pk_a2 = pK_a(HCO3-) = 10.3  (extremely weak acid)

      • K_a2 = [CO₃^2-_(aq)] [H⁺_(aq)] / [HCO₃^-_(aq)] = 5.0 x 10^-11 mol dm^-3

• 5.4.3 Comparison of weak and strong acids in terms of equimolar aqueous solutions.

  • Due to the difference in the concentration of H⁺ ions produced. e.g. say for the sake of argument, 0.1-1.0 molar solutions of hydrochloric acid (100% ionised) and ethanoic acid (approx. 2% ionised). This means the hydrochloric acid is effectively about 50x more acidic than the ethanoic acid and results in the following sorts of observations:

• 5.4.3a pH of solution and K_a/pK_a

  • For equimolar solutions the pH of HCl_(aq) is much lower than for CH₃COOH_(aq) (about pH 0.0-1.0 and 2.5-3.0 respectively, and remember 1 pH unit change represents a 10x [H⁺] ion change in concentration.

  • The acid dissociation/ionisation constant show very different numerical value ranges.

  • The K_a for strong acids is large, typically >10² to 10¹⁰ mol dm^-3 and a negative pK_a, typically -2 to -10.

  • The K_a for weak acids is small, typically 10^-2 to 10^-10 mol dm^-3 and a positive pKa, typically 2 to 10.

• 5.4.3b Chemical reactivity

  • e.g. metal + acid ==> salt + hydrogen: Magnesium rapidly dissolves in hydrochloric acid whereas it fizzes somewhat feebly in aqueous ethanoic acid.

• 5.4.3c Electrical conductivity

  • The electrical conductivity of hydrochloric acid is much higher in ethanoic acid because there are far more ions to carry the current. The greater electrical resistance of ethanoic acid can be readily demonstrated with a simple electrolysis experiment and observe the much higher rate of hydrogen gas evolution at the positive cathode electrode of the hydrochloric acid.

• 5.4.3d Differences in enthalpy of neutralisation ΔH_{neutralisation}

  • Their widely differing values and simplified explanations.

  • The ΔH_neut for a strong acid and strong base (SA+SB) it is usually about -57.1 to -57.3 kJ mol^-1, because they are fully ionised to give the H⁺ and OH^- ions respectively, so the ΔH value essentially corresponds to the ΔH for the reaction ...

    • H⁺_(aq) + OH^-_(aq) ==> H₂O_(l) (ΔH = -57.1 kJ mol^-1)

    • e.g. for the SA/SB pairs: HCl/NaOH, HCl/KOH, HNO₃/NaOH, HNO₃/0.5Ba(OH)₂,

  • The ΔH_neut for a strong acid-weak base (SA+WB) OR a weak acid-strong base neutralisation is less exothermic than the SA+SB above, and in some cases considerable less! e.g. reacting pair and (ΔH),

    • SA/WB: HCl/NH₃ (-52.2)

      • Since NH₃ is only about 2% ionised, the energy change is 98% due to ...

      • NH_3(aq) + H⁺_(aq) NH₄⁺_(aq), which isn't quite as exothermic as H⁺ + OH^-.

      • and the neutralisation cannot be completed because of behaviour of the ammonium ion in acting as a conjugate acid, i.e. the reverse reaction.

    • WA/SB: CH₃COOH/NaOH (-55.2), HCN/KOH (-11.7)

      • Since CH₃COOH is only about 2% ionised, the energy change is 98% due to ...

      • CH₃COOH_(aq) + OH^-_(aq) CH₃COO^-_(aq) + H₂O_(l),

      • which isn't quite as exothermic as H⁺ + OH^- and incomplete due to the conjugate base behaviour of the ethanoate ion.

      • In the 2nd pair, HCN has a strong C-H bond that must be broken, this requires considerable energy and so the ΔH for the main reaction below is considerably less exothermic and incomplete because of the strong conjugate base behaviour of the cyanide ion i.e. the reverse reaction predominates.

      • HCN_(aq) + OH^-_(aq) CN^-_(aq) + H₂O_(l)

  • The ΔH_neut for a weak acid and weak base (WA+WB) neutralisation the ΔH values are even less exothermic.

    • WA/WB: CH₃COOH/NH₃ (-50.2), HCN/NH₃ (-5.4)

      • In these cases the concentrations of H+ or OH- are both very low.

    • CH₃COOH_(aq) + NH_3(aq) CH₃COO^-_(aq) + NH₄⁺_(aq)

      • but with ethanoic acid and ammonia the neutralisation is nearly completed is if the two 'weaknesses' cancel each other out, but for

    • HCN_(aq) + NH_3(aq) CN^-_(aq) + NH₄⁺_(aq)

      • the hydrogen cyanide is so stable that very little neutralisation can take place. The cyanide ion is a very strong conjugate base and the ammonium ion is a moderately strong conjugate acid, so the reverse reaction predominates resulting in the smallest ΔH value.

  • Basically, the weaker and weaker the acid or base, the less and less the neutralisation goes to completion, hence the reaction becomes less and less exothermic.

    • For a further discussion on enthalpy values see Thermochemistry Notes.

• 5.4.4: In principle the full equilibrium expression for any weak acid HA is

  • Kc =	[H3O+(aq)] [A-(aq)]
    	--------------------------
    	[HA(aq)] [H2O(l)]

  • but since the water concentration is nearly constant, and using the simplified H⁺ symbol,  the general equilibrium expression used to solve simple weak acid ionization/dissociation problems is:

    • Ka =	[H+(aq)] [A-(aq)]
      	------------------------
      	[HA(aq)]

      • k_a is called the acid dissociation/ionisation constant with units of mol dm^-3.

      • It is sometimes quoted as a pK_a value, pK_a = -log(K_a/mol dm^-3), so K_a = 10^-pKa.

      • The bigger K_a or the smaller pK_a value, the stronger the acid.

• 5.4.5: Weak acid calculations - calculating the pH of a weak acid

  • Calculation example 5.4.5a

    • The acid dissociation constant, K_a, for ethanoic acid is 1.74 x 10^-5 mol dm^-3.

    • Calculate the hydrogen ion concentration and the pH of a 0.25 mol dm^-3 of this acid. (A = CH₃COO)

    • Ka =	[H+(aq)] [A-(aq)]
      	------------------------
      	[HA(aq)]

    • Assuming:

      1. no other common ion sources present,

      2. [H⁺_(aq)] = [A^-_(aq)], since they are formed in pairs AND the concentration of H⁺ from the self-ionisation of water will be < 10^-7 mol dm^-3 (see above),

      3. and [HA_(aq)]_init.  = [HA_(aq)]_equilib., since only a few % of HA is ionised-dissociated, this is reasonable for simple calculations.

    • therefore substituting and rearranging gives:

    • Ka =	[H+(aq)]2
      	--------------- = 1.74 x 10-5 = [H+(aq)]2/0.25
      	[HA(aq)]

    • [H⁺_(aq)] = √(1.74 x 10^-5 x 0.25) = 2.09 x 10^-3 mol dm^-3

    • pH = -log(2.09 x 10^-3) = 2.68

    • Note: pOH = pK_w - pH = 14 - 2.68 = 11.32

  • Calculation example 5.4.5b

    • A pH meter was calibrated with a buffer solution. If a 0.10 molar solution of a weak acid gave a pH of 4.2, calculate the hydrogen ion concentration and the value of the acid dissociation constant K_a and pK_a.

    • [H⁺_(aq)] = 10^-pH = 10^-4.2 = 6.31 x 10^-5 mol dm^-3

    • using the ideas explored in example 2.1 above,

    • K_a = [H⁺_(aq)]²_/[HA_(aq)] = (6.31 x 10^-5)²_/0.10 = 3.98 x 10^-8 mol dm^-3

    • pK_a = -log(3.98 x 10^-8) = 7.4

• -

Website content copyright © Dr W P Brown 2000-2010 All rights reserved on revision notes, puzzles, quizzes, worksheets, x-words etc. * Copying of website material is not permitted * I do not personally endorse the adverts - but they do pay for the site!