Acids and Bases (HL) - Lesson 8 Weak acid/base calculations

Refresh Based on information in the table below, which acid is the strongest? Acid pKa Ka A. HA 2.0 – B. HB – 1 × 10–3 C. HC 4.0 – D. HD – 1 × 10–5

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Lesson 7: Weak acid/base calculations Objectives: Perform calculations involving weak acids Perform calculations involving weak bases

Recap WEAK ACIDS Ka = [H+][A-]/[HA] pKa = -log10(Ka) Smaller Ka  weaker acid Smaller pKa  stronger acid WEAK BASES Kb = [B+][OH-]/[BOH] pKb = -log10(Kb) Smaller Kb  weaker base Smaller pKb  stronger base Acid Ka pKa Hydronium ion, H3O+ 1.00 0.00 Oxalic acid, HO2CCO2H 5.9x10-2 1.23 Hydrofluoric, HF 7.2x10-4 3.14 Methanoic, CHOOH 1.77x10-4 3.75 Ethanoic, CH3COOH 1.76x10-5 4.75 Phenol, C6H5OH 1.6x10-10 9.80 Base Kb pKb Diethylamine, (C2H5)2NH 1.3x10-3 2.89 Ethylamine, C2H5NH2 5.6x10-4 3.25 Methylamine, CH3NH2 4.4x10-4 3.36 Ammonia, NH3 1.8x10-5 4.74

We need to be able to solve problems such as: What is the pH of an 1.50 M solution of weak acid, X? What is the [OH-] of a solution of weak base, Y? 50 cm3 of a 0.1 M solution of acid X reacts with 25 cm3 of a 0.1 M solution of base Y, what is the resulting pH? The pH of a 0.250 M solution of weak acid Z is 5.4, what is it’s Ka and pKa?

We will need to use a variety of equations: New(ish) today: Ka × Kb = Kw = 1.00x10-14 pKa + pKb = pKw = 14.0 pH + pOH = pKw = 14.0 And from previous lessons: pKa = -log10(Ka) pKb = -log10(Kb) pH = -log10[H+] pOH = -log10[OH-]

Example 1: Calculation of [OH-] What is the concentration of OH- ions in a 0.500 mol dm-3 solution of ammonia (Kb = 1.8x10-5)? What % of the NH3 molecules have dissociated? Since it is a weak base and the equilibrium is to the right we assume that at equilibrium, [NH3] is the same as stated in the question. So: Kb = [NH4+][OH-]/[NH3] Sub all known values into equation 1.8x10-5 = [NH4+][OH-]/ 0.500 Looks like there are 2 unknowns However, since [NH4+] = [OH-]: 1.8x10-5 = [OH-]2 / 0.500 Rearrange to make [OH-] the subject [OH-] = √(1.8x10-5 x 0.500) Perform calculation [OH-] = 0.0030 mol dm-3 % Dissociation = 0.0030 / 0.500 x 100 = 0.60%

Example 2: Calculating pH What is the pH of a 0.225 mol dm-3 solution of oxalic acid (HOOCCOOH, Ka = 5.9x10-2), and how does the pH change on ten-fold dilution? Again, assume [HA] is as stated in the question Ka = [H+][A-] / [HA] Sub in all known values 5.9x10-2 = [H+][A-] / 0.225 Looks like two unknowns, but isn’t really This time, since oxalic acid produces two protons, [A-] = ½ [H+] so the expression becomes: 5.9x10-2 = ½ [H+]2 / 0.225 Rearrange to make [H+] subject [H+] = √((5.9x10-2 x 0.225) / 2) = 0.0815 mol dm-3 pH = -log10[H+] = -log10(0.0815) = 1.09 Now with the ten-fold dilution [H+] = √((5.9x10-2 x 0.0225) / 2) = 0.0257 mol dm-3 pH = -log10[H+] = -log10(0.0257) = 1.59….i.e. ten-fold dilution increased pH by 0.50

Example 3: Calculating Kb from pH A 0.0350 mol dm-3 solution of methylamine (CH3NH2) has a pH of 11.59. Determine Kb of methylamine and Ka of the methylammonium ion (CH3NH3+). Since pH = 11.59 pOH = 14 – 11.59 = 2.41 [OH-] = 10-2.41 = 3.92x10-3 mol dm-3 Remember: [BOH] at equilibrium is same as stated in question [OH-] = [B+] Kb = [B+][OH-]/[BOH] Known values subbed in Kb = (3.92x10-3).(3.92x10-3)/0.0350 Kb = 4.39x10-4 To calculate Ka of the conjugate acid use: Ka x Kb = Kw Ka = Kw / Kb = 1.00x10-14 / 4.39x10-4 = 2.78x10-11

Example 4: Calculating Ka, pKa, Kb and pKb from each other The pKa of benzoic acid is 4.20. Calculate Ka, Kb and pKb Ka = 10-pKa = 10-4.20 = 6.31x10-5 Kb x Ka = Kw Kb = Kw / Ka = 1.00x10-14 / 6.31x10-5 = 1.58x10-10 pKb = -log10(Kb) = -log10(1.58x10-10) = 9.80 OR Since pKa + pKb = pKw pKb = 14 – pKa = 14 – 4.20 = 9.80

Practice Time Work through the calculations on the worksheet If you finish early, complete a flow-chart that can be used to answer weak acid/base questions

Key Points Calculations rely on two key assumptions: Concentration of HA or BOH at equlibrium is the same as given in the question Reasonable as equilibrium effects mean dissociation is often 1% or less [H+]/[OH-] and [A-]/[B+] are not separate variables but are related to each other Expressing [A-]/[B+] in terms of [H+]/OH-] is a key step