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Revision notes on chemical equilibrium - Explaining what a weak acid is and pH calculations

Doc Brown's Chemistry Advanced A Level Notes - Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5

Part 5.4 Definition of a weak acid, examples, pH, Ka and pKa weak acid calculations

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What is a weak acid? How do we write equilibrium expressions to show the acid dissociation–ionisation of a weak acids? What is a weak acid's pKa? How do we calculate the pH of a solution of a weak acid? How do we calculate the Ka of a weak acid?

Sub-index for Part 5

5.4 Definition, examples and pH, Ka and pKa calculations of weak acids

First some mathematical reminders

(i) pH = log10[H+(aq)],   (ii) [H+(aq)] = 10–pH,   (iii) pKa = –log10(Ka),   (iv) Ka = 10–pKa

[H+(aq)] = hydrogen ion concentration mol dm–3 and Ka = acid dissociation constant mol dm–3

• 5.4.1 Definition and examples of WEAK ACIDS

• Weak acids are only partially ionised in water.

• In principle the equilibrium 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.

• The 'formal' equilibrium equation for the reaction is

• CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO(aq)

• or more simply for calculations: CH3COOH(aq) H+(aq) + CH3COO(aq)

• Ethanoic acid is the Bronsted Lowry acid and the ethanoate ion its B-L 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)]
• Ka 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 Ka expressions.

• Ethanoic acid pKa = 4.76,  Ka = 1.74 x 10–5 mol dm–3 and is only about 2% ionised.

• Note that [H2O(l)] is assumed to be considered a constant because water is the solvent and on ionisation the concentration of water is only slightly reduced, so is not quoted in the equilibrium expression, and mathematically it is incorporated into Ka. (You can think of Kc = K[H2O], but not required for the exam!)

• Arithmetical argument to justify this assumption:

• 1000 g of water ~1000 cm3 ~ 1 dm3 and Mr(H2O) = 18.

• Molarity of water in pure water = 1000/18 = 55.56 mol dm-3

• Suppose a 1 molar solution of weak acid HA only ionises by 5% (and often less than this).

• The water concentration will decrease by 5% of 1 mol dm-3, which equals decrease of 0.05 mol dm-3.

• Therefore the molarity of the water solvent decreases from 55.56 to 55.51 ?, <0.1 % error, no problem!

• 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: Ka–acid x Kb–conj. base = Kw and pKa + pKb = pKw, check it out for yourself.

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

• Two equilibria are involved in the ionisation/dissociation processes

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

• HOOC–COOH(aq) H+(aq) + HOOC–COO(aq)

• Ka1 = [H+(aq)] [HOOC–COO(aq)]/[HOOC–COOH(aq)] mol dm–3

• HOOC–COO(aq) H+(aq) OOC–COO(aq)

• Ka2 = [H+(aq)] [OOC–COO(aq)]/[HOOC–COO(aq)] mol dm–3

• and Ka1 > Ka2, showing, not surprisingly, the 1st proton is released more readily than the 2nd.

• Ka1 =  5.89 x 10–2 mol dm–3 (pKa1 = 1.23), and Ka2 = 5.24 x 10–5 mol dm–3 (pKa2 = 4.28)

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

• (a) Hydrofluoric acid, HF: pKa = 3.25, Ka = 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: pKa = 9.31, Ka = 4.9 x 10–10 mol dm–3

• Hydrocyanic acid is a very weak acid and the equilibrium is way over on the left–hand side.

• 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, it is sometimes described as a strong acid, but on the basis of the pka1 value, it isn't really.

• The ionisation of phosphoric(V) acid is quite complicated because three acid–base equilibria are involved.

• (a1) H3PO4(aq) H+(aq) + H2PO4(aq)  (Ka1 = 7.9 x 10–3 mol dm–3, pKa1 = 2.1)

• (a2) H2PO4(aq) H+(aq) + HPO42–(aq)  (Ka2 = 6.2 x 10–8 mol dm–3, pKa2 = 7.2)

• (a3) HPO42–(aq) H+(aq) + PO43–(aq)  (Ka3 = 4.4 x 10–13 mol dm–3, pKa3 = 12.4)

• The subsequent ions H2PO4 and HPO42– are, not surprisingly, increasingly weaker acids, but stronger and stronger conjugate bases, so the equilibrium is increasingly biased to the left–hand side..

• (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.

• The proton donation process can continue until the hydroxide precipitate forms, and then can re–dissolve to form hydroxo–complexes (see Hydrated salts, acidity of hexa–aqua ions in transition metals section)

• 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) CO2(g) CO2(aq)  and (b)  CO2(g) + H2O(l) H2CO3(aq)

• prior to acid ionisation/dissociation.

• (c) CO2(aq) + 2H2O(l) HCO3(aq) + H3O+(aq)

• or (d) H2CO3(aq) + H2O(l) HCO3(aq) + H3O+(aq)

• and more simply:

• (c) CO2(aq) + H2O(l) HCO3(aq) + H+(aq)

• or (d) H2CO3(aq) HCO3(aq) + H+(aq)

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

• (a) pKa1(CO2(aq)) = 6.4  (very weak acid)

• Ka1(CO2(aq)) = [HCO3(aq)] [H+(aq)] / [CO2(aq)] = 4.0 x 10–7 mol dm–3

• (b) pKa1(H2CO3) = 3.7  (weak acid)

• Ka1(H2CO3) = [HCO3(aq)] [H+(aq)] / [H2CO3(aq)] = 2.0 x 10–4 mol dm–3

• HCO3(aq) + H2O(l) CO32–(aq) + H3O+(aq)

• more simply: HCO3(aq) CO32–(aq) + H+(aq)

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

• pka2 = pKa(HCO3–) = 10.3  (extremely weak acid)

• Ka2 = [CO32–(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 Ka/pKa

• 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 Ka for strong acids is large, typically >102 to 1010 mol dm–3 and a negative pKa, typically –2 to –10.

• The Ka 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 by observing the much higher rate of hydrogen gas evolution at the negative cathode electrode of the hydrochloric acid when applying the same voltage (p.d.) across the two electrodes (e.g. carbon or platinum).

• 5.4.3d Differences in enthalpy of neutralisation ΔHneutralisation

• Their widely differing values and simplified explanations.

• The ΔHneut 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 ΔHneut 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 ...

• NH3(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 ΔHneut 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) + NH3(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) + NH3(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 .

• 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)]
• ka is called the acid dissociation/ionisation constant with units of mol dm–3.

• It is sometimes quoted as a pKa value, pKa = –log(Ka/mol dm–3), so Ka = 10–pKa.

• The bigger Ka or the smaller pKa 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, Ka, 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 = pKw – 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 Ka and pKa.

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

• using the ideas explored in example 2.1 above,

• Ka = [H+(aq)]2 / [HA(aq)] = (6.31 x 10–5)2 / 0.10 = 3.98 x 10–8 mol dm–3

• pKa = –log(3.98 x 10–8) = 7.4

WHAT NEXT?

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product Ksp, common ion effect, ion–exchange systems * 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 watch out for sub-indexes to multiple sections or pages

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