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Revision notes on chemical equilibrium - buffer solution pH calculations - for Advanced A/AS Level Theoretical-Physical Chemistry

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Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 6.4

Equilibria Part 6.4 Buffer solution pH calculations

How do you calculate the pH of a buffer solution?

How do you calculate the quantities required to make up a buffer solution of a desired pH?

Chemical Equilibrium Notes Parts 5 & 6 Index

6.4 Buffer pH calculations – theoretical calculation of a buffer solution

  • 6.4.1 Calculations involving a buffer made from a weak acid and its salt with a strong base.

    • Consider the mixture is made from a monobasic weak acid HA and an alkali metal salt M+A–

      • e.g. A = CH3COO, M = K or Na

    • it is reasonable to assume for simple approximate calculations that ..

      1. [A–(aq)] = [salt(aq)] since salt fully ionised and M+ is a spectator ion, and

      2. [HA(aq)]equilib., = [HA(aq)]initial since little of the weak acid is ionised.

    • Therefore the weak acid Ka expression is ...

      • (i) Ka =

        [H+(aq)] [A–(aq)]
        ––––––––––––––––––––––––– mol dm–3
              [HA(aq)]
      • becomes

      • (ii) Ka =

        [H+(aq)] [salt(aq)]
        –––––––––––––––––––––––––– mol dm–3
           [acid(aq)]
      • therefore: [H+(aq)] = Ka [acid(aq)] / [salt(aq)] mol dm–3, and taking –log10 of both sides gives

      • (iii) pHbuffer = –log10(Ka x [acid(aq)] / [salt(aq)])

      • or (iv) pHbuffer = pKa + –log10([acid(aq)] / [salt(aq)])

      • or (v) pHbuffer = pKa + log10([salt(aq)] / [acid(aq)])

        • which is how the equation is usually quoted, sometimes called the Henderson Equation.

        • Note: For a given conjugate pair (HA and A–), the pH of the buffer is determined by the acid/salt ratio, though the more concentrated the buffer, the greater its capacity to neutralise larger amounts of added/formed in a reaction medium.

      • Another point is how to choose which weak acid is best for a desired buffer?

        • The useful range of a buffer is decided by the weak acid's Ka and the ratio of the salt and weak acid concentrations.

        • The buffer will be most useful when the ratio [salt]/[acid] is equal to one i.e. when both active ingredients are at their maximum concentrations at no expense to the other – by the principles of related chemical equilibrium, if you increase one concentration you would decrease the other.

        • Therefore the maximum buffer capacity is when [salt] = [acid]

        • Now Ka = [H+(aq)] when [salt] = [acid] in equation (i) or (ii) above

        • therefore taking –log10 of both sides gives ...

        • pKa = pH when [salt] = [acid], because log10(1) = 0 in equations (iv) or (v)

        • and this simple mathematical argument gives the necessary guidance ...

        • So for example, supposing you wanted a buffer to cover a pH range of 4.5 to 6.0,

        • and your choice of weak acid pKa's was 2.8, 4.2, 5.5 and 6.5,

        • you would choose the weak acid with a pKa of 5.5 because its pKa is well in the desired pH range.

        • You would then formulate it with its sodium (or potassium) salt – note that sodium ion and potassium ion salts are usually used because they have virtually no acidic or basic character to complicate matters e.g. the hydrated ions Na+(aq) and K+(aq) do not donate protons in the way, for example, the hexa–aqua ions of aluminium can ...

          • [Al(H2O)6]2+(aq) + H2O(l) [Al(H2O)5(OH)]+(aq) + H3O+(aq)

          • because the polarising power of the central metal ions, Na+ or K+ is too small to effect this process.

  • 6.4.2 Buffer calculations

  • Calculating the pH of a buffer – calculating amounts of salt and acid needed

    • Buffer calculation example 6.4.2a

      • A buffer solution was prepared which had a concentration of 0.20 mol dm–3 in ethanoic acid and 0.10 mol dm–3 in sodium ethanoate. If the Ka for ethanoic acid is 1.74 x 10–5 mol dm–3, calculate the theoretical hydrogen ion concentration and pH of the buffer solution.

      • Ka = [H+(aq)] [salt(aq)]/[acid(aq)]

      • 1.74 x 10–5 = [H+(aq)] x 0.10 / 0.20

      • [H+(aq)] = 1.74 x 10–5 x 0.20/0.10 = 3.48 x 10–5 mol dm–3

      • pH = –log(3.48 x 10–5) = 4.46

    • Buffer calculation example 6.4.2b

      • In what ratio should a 0.30 mol dm–3 of ethanoic acid be mixed with a 0.30 mol dm–3 solution of sodium ethanoate to give a buffer solution of pH 5.6?

        • Ka for ethanoic acid is 1.74 x 10–5 mol dm–3

      • [H+(aq)] = 10–pH = 10–5.6 = 2.51 x 10–6 mol dm–3

      • Ka = [H+(aq)] [salt(aq)]/[acid(aq)]

      • [salt]/[acid] = Ka/[H+(aq)] = 1.74 x 10–5/2.51 x 10–6  = 6.93

      • Therefore volume ratio is 6.93 : 1 for salt : acid, e.g. 6.93 cm3 of 0.30M sodium ethanoate is mixed with 1.0 cm3 of 0.30 M ethanoic acid to give a buffer solution of pH 5.6.

      • Note that the pH is determined by the ratio of concentrations, but the buffering capacity of the solution can be increased by increasing the concentrations of both components in the same molar concentration ratio.

    • Buffer calculation example 6.4.2c

      • What is the pH of a buffer solution made from dissolving 2.0g of benzoic acid and 5.0g of sodium benzoate in 250 cm3 of water?

        • Ka benzoic acid = 6.3 x 10–5 mol dm–3, Ar's: H = 1, C = 12, O = 16, Na = 23

        • Molecular masses: Mr(C6H5COOH) = 122, Mr(C6H5COO–Na+) = 144, 250 cm3 = 0.25dm3

        • moles acid C6H5COOH = 2.0/122 = 0.0164 mol, molarity = 0.0656 mol dm–3

        • moles salt C6H5COO–Na+ = 5.0/144 = 0.0347 mol, molarity = 0.139 mol dm–3

        • [H+(aq)] = Ka [acid(aq)]/[salt(aq)]

        • [H+(aq)] = 6.3 x 10–5 x 0.0656 / 0.139 = 2.97 x 10–5 mol dm–3

        • pH = –log[H+(aq)] = –log(2.97 x 10–5) = 4.53

    • Buffer calculation example 6.4.2d

      • Calculate the pH of a buffer made by mixing 100 cm3 of a 0.40 M sodium propanoate and 50 cm3 of 0.2 M propanoic acid solution.

        • Ka propanoic acid = 1.3 x 10–5 mol dm–3, total volume of buffer = 150 cm3

        • molarities in the mixture:

          • [salt] = 0.40 x 100/150 = 0.267 mol dm–3

          • [acid] = 0.20 x 50/150 = 0.0667 mol dm–3

        • [H+(aq)] = Ka [acid(aq)]/[salt(aq)]

        • [H+(aq)] = 1.3 x 10–5 x 0.0667 / 0.267 = 3.24 x 10–6 mol dm–3

        • pH = –log[H+(aq)] = –log(3.24 x 10–6) = 5.48

    • Buffer calculation example 6.4.2e

      • Using the Henderson equation

      • pHbuffer = pKa + log10([salt(aq)] / [acid(aq)])

      • Calculate the pH of buffer solution made by mixing together 100 cm3 of 0.100M ethanoic acid and 50 cm3 of 0.400M sodium ethanoate,

        • given that Ka for ethanoic acid is 1.74 x 10–5 mol dm–3

      • Now because the volumes are not equal, the real concentrations in the mixture must be worked out.

        • The total volume is 150 cm3, therefore the dilutions are given by

        • [acid] = 0.1 x 100/150 = 0.06667

        • [salt] = 0.4 x 50/150 = 0.1333

      • Substituting in the Henderson Equation gives

      • pHbuffer = –log10(1.74 x 10–5) + log10(0.1333/0.0667)

      • pHbuffer = –log10(1.74 x 10–5) + log10(2)

      • pHbuffer = 4.76 + 0.3010

      • pHbuffer = 5.06

    • –

  • –


Chemical Equilibrium Notes Parts 5 & 6 Index

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Chemical Equilibrium Notes Parts 5 & 6 Index

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