More Acids and Bases (2)

More Acids and Bases 2 Link to Data Page Your feedback on these self-help problems is appreciated. Click here to send an e-mail. For on-line help on this topic, please see the following Chemcal modules: Acids and Bases (strong acids and bases; conjugate acid/base pairs; pH;self-dissociation of water) Weak Acids and Bases (weak acids; weak bases; K_a and K_b) Calculations with Weak Acids and Bases (calculations involving weak acids and bases and buffers) Acid-base Titrations (titration curves for acid/base titrations; indicators)		Shortcut to Questions Q: 1 2 3 4 5 6 7 8
1	Suggest conjugate acid/base pairs to give the following buffered pH values: (1.) 3.0 (2.) 5.0 (3.) 7.0 (4.) 9.0 (5.) 10.3 (6.) 13.0
2	What buffered pH would be obtained from solutions of equimolar amounts of the following conjugate acid/base pairs: (1.) HPO₄^2- / PO₄^3- (2.) H₃BO₃ / H₂BO₃^- (3.) HF / F^- (4.) HS^- / S^2- (5.) ClCH₂COOH / ClCH₂CO₂^-
3	What is the pH of 1.00 M ammonia solution (100 cm³) in which ammonium chloride (9.30 g) has been dissolved?
4	(1.) What is the molarity of a water-solution of propionic acid of pH 3.00? (2.) What mass of potassium propionate must be added to one litre of this solution to change the pH to 5.00?
5	What volume of 10.0 M hydrochloric acid must be added to 0.100 M sodium lactate (1.00litre) to give a solution with pH = 3.86?
6	(1.) The major buffering action in blood plasma is due to the hydrogencarbonate ion / carbonic acid system. The hydrogencarbonate ion concentration in a sample of blood plasma was found to be 0.0230 M. Calculate the concentration of carbonic acid in this sample if its pH is 7.40 at 37 ° C. (pK_A H₂CO₃ at 37 ° C = 6.10) (2.) To this blood plasma (100 cm³) is added a 0.056 M solution of sodium hydroxide (0.20 cm³). What is the new pH of the plasma?
7	Calculate the pH at the equivalence point for the following acid/base titrations with 0.20 M solutions and choose a suitable indicator for each titration. (1.) NH₃ with HCl (2.) NaOH with HCN (3.) KOH with HOCl (4.) KOH with HNO₃ (5.) NH₂OH with HCl (6.) NaOH with HNO₂
8	A 0.10 M solution of an acid / base indicator has a pH of 5.00. Would this indicator be suitable for the titration of a weak acid with a strong base or a weak base with a strong acid? Give a clear reason for your answer.
More Acids and Bases 2 (Answers)
1	A buffer is most effective when [weak acid] = [conjugate base]. Under this condition, pH of the solution = pK_A of the weak acid. To select a buffer, choose a weak acid for which pK_A ~ pH desired, and fine tune the buffer to give the precise pH required by adjusting the acid / base ratio. From the list of acid pK_A values, the following acid / base pairs are suitable: (1.) pH = 3.0 HF has a pK_A of 3.17, so a HF / F^- pair would be suitable. As the required pH is on the acidic side of pK_A(HF), an excess of HF would be required. (2.) pH = 5.0 CH₃COOH has a pK_A of 4.76, so a CH₃CHOOH / CH₃CHOO^- pair would be suitable. As the required pH is on the basic side of pK_A(CH₃COOH), an excess of CH₃COO^- would be required. (3.) pH = 7.0 H₂PO₄^- has a pK_A of 7.20, so a H₂PO₄^- / HPO₄^2- pair would be suitable. As the required pH is on the acidic side of pK_A(H₂PO₄^-), an excess of H₂PO₄^- would be required. (4.) pH = 9.0 NH₄⁺ has a pK_A of 9.24, so a NH₄⁺ / NH₃ pair would be suitable. As the required pH is on the basic side of pK_A(NH₄⁺), an excess of NH₃ would be required. (5.) pH = 10.3 HCO₃^- has a pK_A of 10.33, so a HCO₃^- / CO₃^2- pair would be suitable. As the required pH is on the acidic side of pK_A(HCO₃^-), an excess of HCO₃^- would be required. (6.) pH = 13.0 HS^- has a pK_A of 12.89, so a HS^- / S^2- pair would be suitable. As the required pH is on the basic side of pK_A(HS^-), an excess of S^2- would be required.
2	K_A = [H₃O⁺] [A^-] / [HA] Since equimolar amounts of weak acid and its conjugate base are present in each of the buffer solutions, [A^-] / [HA] = 1 and thus K_A = [H₃O⁺]. Therefore pH = pK_A of the weak acid in each conjugate acid/base pair. So, from the data sheet: (1.) pH = pK_A = 12.36 (2.) pH = pK_A = 9.24 (3.) pH = pK_A = 3.17 (4.) pH = pK_A = 12.89 (5.) pH = pK_A = 2.86
3	Volume of solution is 100 mL. [NH₃] = 1.00 mol L^-1 Moles of NH₄Cl = 9.30 g / 53.49 g mol^-1 = 0.174 mol [NH₄⁺] = 0.174 mol / 0.100 L = 1.74 mol L^-1 pK_A(NH₄⁺) = 9.24 K_A(NH₄⁺) = 10^-9.24 K_A = [H₃O⁺] [A^-] / [HA] = [H₃O⁺][NH₃] / [NH₄⁺] [H₃O⁺] = K_A [NH₄⁺] / [NH₃] = 10^-9.24 x 1.74 / 1.00 = 1.00 x 10^-9 M pH = -log [H₃O⁺] = -log 1.00 x 10^-9 mol L^-1 = 9.00
4	(1.) H₂O + CH₃CH₂COOH H₃O⁺ + CH₃CH₂CO₂^- pK_A (CH₃CH₂COOH) = 4.87 K_A (CH₃CH₂COOH) = 10^-4.87 K_A = [CH₃CH₂CO₂^-] [H₃O⁺] / [CH₃CH₂COOH] = [H₃O⁺]² / [CH₃CH₂COOH] [H₃O⁺] for a pH of 3.00 = 0.0010 mol L^-1 ∴ [CH₃CH₂COOH] = (0.0010 )² / 1.35 x 10^-5 = 0.074 M (2.) K_A = [H₃O⁺] [A^-] / [HA] Now the desired pH is 5.00, so [H₃O⁺] = 1.00 x 10^-5 M ∴ [CH₃CH₂CO₂^-] = K_A [CH₃CH₂COOH] / [H₃O⁺] = 10^-4.87 x 0.074 / 1.00 x 10^-5 = 0.10 M The volume of the solution is one litre, so 0.10 mol CH₃CH₂CO₂K is required. Mass CH₃CH₂CO₂K = moles x molar mass = 0.10 mol x 112.14 g mol^-1 = 11 g
5	pK_A lactic acid = 3.86. Note that lactic acid is a monoprotic acid. If pH is equal to pK_A, it follows that [HA] = [A^-] [A^-] = [lactate] = 0.100 mol L^-1 Lactic acid is a weak acid, so H₃O⁺ will react with sodium lactate to give lactic acid. There is 0.100 mole of lactate ion present, so to obtain equimolar amounts of lactate ion and lactic acid, we need to add enough H₃O⁺ to react with half the lactate ion present. This will provide us with a solution containing 0.0500 mole of lactate and 0.0500 mole of lactic acid. To produce 0.0500 mole of lactic acid we need 0.0500 mole of H₃O⁺, which is obtained from 0.0500 mole of hydrochloric acid (since both lactic acid and hydrochloric acid are monoprotic). ∴ Volume 10.0 M hydrochloric acid required = moles hydrochloric acid / molarity of hydrochloric acid = 0.0500 mol / 10.0 mol L^-1 = 0.00500 L = 5.00 mL.
6	The acid / base pair in this system is H₂CO₃ / HCO₃^-. (1.) [HCO₃^-] = 0.0230 mol L^-1 pH = 7.40 ∴ [H₃O⁺] = 10^-7.40 pK_A (H₂CO₃) = 6.10 (at 37 ° C) ∴ K_A (H₂CO₃) = 10^-6.10 K_A = [H₃O⁺] [HCO₃^-] / [H₂CO₃] [H₂CO₃] = [H₃O⁺] [HCO₃^-] / K_A = 10^-7.40 x 0.0230 / 10^-6.10 M = 1.2 x 10^-3 M (2.) The buffer system reacts to the strong base OH^- by consuming H₂CO₃ and producing HCO₃^- : OH^- + H₂CO₃ → HCO₃^- + H₂O Assuming that all OH^- is converted to HCO₃^-, we can calculate the new equilibrium amounts after addition of NaOH: Initial [HCO₃^-] = 2.30 x 10^-2 mol L^-1 Initial [H₂CO₃] = 1.2 x 10^-3 mol L^-1 Volume of solution = 0.100 L, so in this volume initial moles H₂CO₃ = volume x molarity = 1.2 x 10^-4 mol and initial moles HCO₃^- = 2.30 x 10^-3 mol Moles of hydroxide ion added = volume x molarity = 0.20 x 10^-3 x 0.056 mol = 1.12 x 10^-5 mol. Addition of the 1.12 x 10^-5 mole of NaOH leads to new amounts of the buffer pair forming in response to the strong base OH ^-. The result at equilibrium is an increase in the moles of HCO₃^- and an equal decrease in the moles of H₂CO₃. Eqb. moles HCO₃^- = 2.30 x 10^-3 mol + 1.12 x 10^-5 mol = 2.31 x 10^-3 mol. Eqb. moles H₂CO₃ = 1.2 x 10^-4 mol - 1.12 x 10^-5 mol = 1.09 x 10^-4 mol. Volume = 0.100 L ∴ Eqb. [HCO₃^-] = 2.3 x 10^-2 M ∴ Eqb. [H₂CO₃] = 1.2 x 10^-3 M¹ Calculating pH with these new amounts of acid and base: K_A = [H₃O⁺] [HCO₃^-] / [H₂CO₃] [H₃O⁺] = K_A [H₂CO₃] / [HCO₃^-] ∴ [H₃O⁺] = K_A x [H₂CO₃] / [HCO₃^-] = (10^-6.10 x 1.09 x 10^-3) / (2.31 x 10^-2) = 3.75 x 10^-8 M ∴ pH = -log (3.75 x 10^-8) = 7.43
7	The equivalence point is reached when stoichiometric amounts of acid and base have reacted. (1.) NH₃ + H₃O⁺ → NH₄⁺ + H₂O There is a change in volume during titration. At the equivalence point equal volumes of 0.20 M HCl and 0.20 M NH₃ have reacted because the NH₃ and hydrochloric acid solutions are of the same molarity. This leads to a solution which is 0.10 M in NH₄⁺ - not 0.20 M! This is the concentration used to calculate pH below. At the equivalence point the solution is 0.10 M in NH₄⁺ and Cl^-. Finding the pH at the equivalence point is the same as finding the pH of a 0.10 M NH₄Cl solution. Cl^- is the conjugate base of a strong acid, HCl, so it does not hydrolyse water and plays no role in altering pH. The ammonium ion however is a weak acid and thus does alter pH: NH₄⁺ + H₂O NH₃ + H₃O⁺ Finding [H₃O⁺]: pK_A(NH₄⁺) = 9.24 ∴ K_A(NH₄⁺) = 10^-9.24 Using K_A = [H₃O⁺] [A^-] / [HA] : K_A(NH₄⁺) = [H₃O⁺] [NH₃] / [NH₄⁺] From the stoichiometry of the equation for the dissociation of ammonium ion given above, [NH₃] = [H₃O⁺]. Therefore K_A(NH₄⁺) = [H₃O⁺]² / [NH₄⁺] ∴ [H₃O⁺] = (K_A(NH₃) x [NH₄⁺])^½ = (5.75 x 10^-10 x 0.10)^½ = 7.58 x 10^-6 M. pH = -log [H₃O⁺] = -log (7.58 x 10^-6) = 5.12. A suitable indicator will have a pK_A about the same as the pH at the titration equivalence point. Referring to a table of indicators and their pK_A values, a suitable indicator for this reaction would be methyl red, which has a pK_A in the vicinity of 5.12 (2.) OH^- + HCN → CN^- + H₂O Mixing 0.20 M solutions of NaOH and HCN leads to a solution which is 0.10 M in both Na⁺ and CN^-. The sodium ions present do not affect the pH, and are not considered further. The pH of 0.10 M CN^- solution, a weak base, is found starting with the pK_A of its weak conjugate acid, HCN. ∴ K_A(HCN) = 10^-9.22 The following equilibrium will be considered: CN^- + H₂O HCN + OH^- K_B(CN^-) = K_W / K_A(HCN) = 1.0 x 10^-14 / (6.03 x 10^-10) = 1.66 x 10^-5 K_B(CN^-) = [HCN] [OH^-] / [CN^-] From the stoichiometry of the reaction above, [HCN] = [OH^-] : K_B(CN^-) = [OH^-]² / [CN^-] ∴ [OH^-] = (K_B(CN^-) x [CN^-])^½ = (1.66 x 10^-5 x 0.10 )^½ = 1.29 x 10^-3 M. [H₃O⁺] = K_W / [OH^-] = 1.00 x 10^-14 / (1.29 x 10^-3) = 7.75 x 10^-12 M pH = -log [H₃O⁺] = -log (7.75 x 10^-12) = 11.11. A suitable indicator would be alizarin yellow, which has a pK_A in the vicinity of 11.11 (3.) Reaction products are K⁺ and OCl^-. Only the OCl^- enters into hydrolysis reaction with water to form HOCl and OH^- in equal quantities as shown in the following equation: H₂O + OCl^- HOCl + OH^- i.e. OCl^- ion is a weak base. Using the same method as in (2) above, pH = 10.27, phenolphthalein indicator (4.) Reaction products are K⁺ and NO₃^-. Neither of these undergo acid / base reactions with the solvent, water, so pH = 7.00, bromothymol blue indicator (5.) Reaction products are NH₃OH⁺ and Cl^-. The chloride ion, being the conjugate of a strong acid, does not enter into a hydrolysis reaction with water but the NH₃OH⁺ ion is a weak acid which reacts slightly with water according to the equation NH₃OH⁺ + H₂O NH₂OH + H₃O⁺ Using the same method as in (1) above, pH = 3.48 and methyl yellow or methyl orange indicator would be suitable. (6.) Reaction products are Na⁺ and NO₂^-. Only the NO₂^- enters into hydrolysis reaction with water to form HNO₂ and OH^- in equal quantities as shown in the following equation: H₂O + NO₂^- HNO₂ + OH^- i.e. NO₂^- ion is a weak base. Using the same method as in (2) above, pH = 8.07, cresol red indicator.
8	The indicator, represented as HIn, is a weak acid. As the titration proceeds, the acid form of the indicator, HIn, is converted to its conjugate base, In^-, and water. The colour change of any indicator occurs when [HIn] in the solution = [In^-], ie when pH of the solution = pK_A of the indicator. From the information given, pK_A can be determined. HIn+ H₂O In^- + H₃O⁺ K_A(HIn) = [H₃O⁺] [In^-] / [HIn] pH of 0.10 M indicator solution = 5.00, ∴ [H₃O⁺] = 1.0 x 10^-5 M [In^-] = [H₃O⁺], and we assume [HIn]_init = [HIn]_eqb: K_A(HIn) = [H₃O⁺]² / [HIn]_init = (1.0 x 10^-5)² / 0.10 = 1.0 x 10^-9 ∴ pK_A(HIn) = -log K_A = 9.00. ∴ The indicator undergoes colour change at pH = 9.00. This would be suitable for the titration of a weak acid with a strong base.