* Advanced Chemistry Revision Notes on EQUILIBRIUM 6.4 Buffer solutions and pH calculations Doc B

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 Doc Brown's Chemistry  Theoretical Chemistry - Equilibria - Chemical Equilibrium 6.4

Equilibria Part 6.4 Buffer solution pH 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 6 sub-index: 6.1 Salt hydrolysis * 6.2 Acid-base indicator theory, pH curves and titrations * 6.3 Buffers - definition, formulation and action * 6.4 Buffer calculations * 6.5 Case studies of buffer function * EMAIL query?comment

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, 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

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.

  • TOP and LINKS6.4.2 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

    • -

  • -

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(spanish) Doc Brown de Química Química Teórica - Equilibrio - 6,4 equilibrio químico Equilibrios Parte Solución tampón pH 6,4 cálculos Equilibrios parte 6 sub-índices: 6,1 hidrólisis Sal * 6,2 indicador de base de la teoría-ácido, las curvas de pH y valoraciones * 6,3 Buffers - definición, formulación y acción * 6,4 cálculos tampón 6,5 * Estudios de caso de la función de amortiguación * consulta por correo electrónico? comentario * (portuguese) de Química Teórica - Equilíbrio - Equilíbrio químico 6,4 Equilíbrio Parte solução tampão pH 6,4 cálculos 6,4 Buffer cálculos de pH teórico - cálculo de uma solução-tampão * (thai) Doc Brown 's เคมีเคมีทฤษฎี -- สมดุล -- สมดุลเคมี Part 6.4 สารละลาย Buffer pH การคำนวณ 6.4 การคำนวณ pH Buffer -- ทฤษฎีการคำนวณสารละลายบัฟเฟอร์ * (korean) 브라운 박사는 화학 이론 화학 - Equilibria - 화학 평형 6.4 파트 6.4 버퍼 솔루션 산도를 계산 6.4 버퍼 산도 계산 - 이론 솔루션 버퍼 계산 6.4.1 계산 기지와 관련된 강한 약한 산성과 소금으로 버퍼를 만든. C는 필요한 산도를 6.4.2 버퍼 계산 버퍼 산성의 소금 - 계산 금액과 *
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