7.2. Brønsted–Lowry theory of acids and bases
A subsection of Chemistry, 9701, through 7. Equilibria
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Acidified potassium dichromate(can oxidise ethanedioic acid, H2C2O4. The relevant half-equations are shown. Cr2O7 2– + 14H+ + 6e– → 2Cr3+ + 7H2O H2C2O4 → 2CO2 + 2H+ + 2e– State the overall equation for the reaction between acidified dichromate(ions and ethanedioicacid. In an experiment a 0.242 g sample of hydrated ethanedioic acid, H2C2O4.xH2O, was reacted with a 0.0200 mol dm–3 solution of acidified potassium dichromate(. 32.0 cm3 of the acidified potassium dichromate(solution was required for complete oxidation of the ethanedioic acid. Calculate the amount, in moles, of dichromate(ions used to react with the sample of ethanedioic acid. amount = mol Calculate the amount, in moles, of ethanedioic acid in the sample. amount = mol Calculate the relative molecular mass, Mr, of the hydrated ethanedioic acid. Mr = Calculate the value of x in H2C2O4.xH2O. x =
9701_s16_qp_22
THEORY
2016
Paper 2, Variant 2
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Dicarboxylic acids dissociate in stages. stage 1 stage 2 HO2C(CH2)nCO2H HO2C(CH2)nCO2 – + H+ –O2C(CH2)nCO2 – + 2H+ The pKa values for stage 1 and stage 2 for some dicarboxylic acids are listed below. n in HO2C(CH2)nCO2H pKa(1) for stage 1 pKa(2) for stage 2 2.83 5.69 4.16 5.61 4.31 5.41 For comparison, the pKa of ethanoic acid, CH3CO2H, is 4.76. State the mathematical relationship between pKa and the acid dissociation constant Ka. With reference to the table above, suggest why the pKa(1) values • are all smaller than the pKa of ethanoic acid, • become larger as n increases. Suggest why all the pKa(2) values in the table above are larger than the pKa of ethanoic acid. The monosodium salts of edible dicarboxylic acids are added to some foodstuffs as buffers. Explain what is meant by the term buffer solution. Write two equations to show how monosodium butanedioate, HO2CCH2CH2CO2Na, acts as a buffer.
9701_s16_qp_41
THEORY
2016
Paper 4, Variant 1
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Ethanoic acid is a weak acid. Explain what is meant by the term weak acid. The pKa values of four acids are listed below. acid structural formula pKa CH3CO2H 4.8 CH3CH2CO2H 4.9 CH3CHCl CO2H 2.8 CH2Cl CH2CO2H 4.0 State the mathematical relationship between pKa and the acid dissociation constant Ka. With reference to acidity, explain the difference in pKa values between • acid 1 and acid 2, • acid 2 and acid 3, • acid 3 and acid 4. Draw a fully labelled diagram of the equipment needed to measure the voltage of an electrochemical cell consisting of the standard hydrogen electrode and the standard Cu / Cu2+ electrode. For the cell drawn in , calculate the and state which electrode is positive. = identity of the positive electrode A monobasic acid, D, has Ka = 1.23 × 10–5 mol dm–3. Calculate the pH of a 0.100 mol dm–3 solution of D. pH = An electrochemical cell similar to the one you have drawn in was set up using a 0.100 mol dm–3 solution of D in the hydrogen electrode instead of the standard solution. Use the data and the Nernst equation, E = E o + 0.059 log [H+], to calculate the new E cell in this experiment. E cell = V
9701_s16_qp_42
THEORY
2016
Paper 4, Variant 2
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Dicarboxylic acids dissociate in stages. stage 1 stage 2 HO2C(CH2)nCO2H HO2C(CH2)nCO2 – + H+ –O2C(CH2)nCO2 – + 2H+ The pKa values for stage 1 and stage 2 for some dicarboxylic acids are listed below. n in HO2C(CH2)nCO2H pKa(1) for stage 1 pKa(2) for stage 2 2.83 5.69 4.16 5.61 4.31 5.41 For comparison, the pKa of ethanoic acid, CH3CO2H, is 4.76. State the mathematical relationship between pKa and the acid dissociation constant Ka. With reference to the table above, suggest why the pKa(1) values • are all smaller than the pKa of ethanoic acid, • become larger as n increases. Suggest why all the pKa(2) values in the table above are larger than the pKa of ethanoic acid. The monosodium salts of edible dicarboxylic acids are added to some foodstuffs as buffers. Explain what is meant by the term buffer solution. Write two equations to show how monosodium butanedioate, HO2CCH2CH2CO2Na, acts as a buffer.
9701_s16_qp_43
THEORY
2016
Paper 4, Variant 3
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One atom of each of the four elements H, C, N and O can bond together in different ways. Two examples are molecules of cyanic acid, HOCN, and isocyanic acid, HNCO. The atoms are bonded in the order they are written. Draw ‘dot-and-cross’ diagrams of these two acids, showing outer shell electrons only. HOCN, cyanic acid HNCO, isocyanic acid Suggest the values of the bond angles HNC and NCO in isocyanic acid. HNC NCO Suggest which acid, cyanic or isocyanic, will have the shorter C–N bond length. Explain your answer. Isocyanic acid is a weak acid. HNCO H+ + NCO– Ka = 1.2 × 10–4 mol dm–3 Calculate the pH of a 0.10 mol dm–3 solution of isocyanic acid. pH = Sodium cyanate, NaNCO, is used in the production of isocyanic acid. Sodium cyanate is prepared commercially by reacting urea, (NH2)2CO, with sodium carbonate. Other products in this reaction are carbon dioxide, ammonia and steam. Write an equation for the production of NaNCO by this method. Barium hydroxide, Ba(OH)2, is completely ionised in aqueous solutions. During the addition of 30.0 cm3 of 0.100 mol dm–3 Ba(OH)2 to 20.0 cm3 of 0.100 mol dm–3 isocyanic acid, the pH was measured. Calculate the [OH–] at the end of the addition. [OH–] = mol dm–3 Use your value in to calculate [H+] and the pH of the solution at the end of the addition. final [H+] = mol dm–3 final pH = On the following axes, sketch how the pH changes during the addition of a total of 30.0 cm3 of 0.100 mol dm–3 Ba(OH)2 to 20.0 cm3 of 0.100 mol dm–3 isocyanic acid. volume of Ba(OH)2 added / cm3 pH The cyanate ion, NCO–, can act as a monodentate ligand. State what is meant by the terms monodentate, ligand. Silver ions, Ag+, react with cyanate ions to form a linear complex. Suggest the formula of this complex, including its charge. When heated with HCl , organic isocyanates, RNCO, are hydrolysed to the amine salt, RNH3Cl, and CO2. RNCO + H2O + HCl RNH3Cl + CO2 A 1.00 g sample of an organic isocyanate, RNCO, was treated in this way, and the CO2 produced was absorbed in an excess of aqueous Ba(OH)2 according to the equation shown. The solid BaCO3 precipitated weighed 1.66 g. Ba(OH)2+ CO2BaCO3+ H2OCalculate the number of moles of BaCO3 produced. moles of BaCO3 = Hence calculate the Mr of the organic isocyanate RNCO. Mr of RNCO = The R group in RNCO and RNH3Cl contains carbon and hydrogen only. Use your Mr value calculated in to suggest the molecular formula of the organic isocyanate RNCO. molecular formula of RNCO Suggest a possible structure of the amine RNH2, which forms the amine salt, RNH3Cl.
9701_s17_qp_41
THEORY
2017
Paper 4, Variant 1
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The table lists some organic acids and their pKa values. acid formula pKa ethanoic acid CH3CO2H 4.76 chloroethanoic acid Cl CH2CO2H 2.86 aminoethanoic acid H2NCH2CO2H 9.87 State the relationship between pKa and the strength of an acid. State the mathematical relationship between pKa and the acidity constant Ka. Give reasons for why the pKa value for chloroethanoic acid is smaller than that for ethanoic acid. Use the zwitterionic structure for aminoethanoic acid in aqueous solution to write an equation for its dissociation giving H+ions. Calculate the pH of a 0.100 mol dm–3 solution of aminoethanoic acid. pH = A 10.0 cm3 sample of 0.100 mol dm–3 aminoethanoic acid was titrated with 0.100 mol dm–3 NaOH. After 20.0 cm3 of NaOH, an excess, had been added, the pH was found to be 12.5. Using the following axes, sketch a graph showing how the pH changes during this titration. volume of NaOH added / cm3 pH
9701_s17_qp_42
THEORY
2017
Paper 4, Variant 2
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One atom of each of the four elements H, C, N and O can bond together in different ways. Two examples are molecules of cyanic acid, HOCN, and isocyanic acid, HNCO. The atoms are bonded in the order they are written. Draw ‘dot-and-cross’ diagrams of these two acids, showing outer shell electrons only. HOCN, cyanic acid HNCO, isocyanic acid Suggest the values of the bond angles HNC and NCO in isocyanic acid. HNC NCO Suggest which acid, cyanic or isocyanic, will have the shorter C–N bond length. Explain your answer. Isocyanic acid is a weak acid. HNCO H+ + NCO– Ka = 1.2 × 10–4 mol dm–3 Calculate the pH of a 0.10 mol dm–3 solution of isocyanic acid. pH = Sodium cyanate, NaNCO, is used in the production of isocyanic acid. Sodium cyanate is prepared commercially by reacting urea, (NH2)2CO, with sodium carbonate. Other products in this reaction are carbon dioxide, ammonia and steam. Write an equation for the production of NaNCO by this method. Barium hydroxide, Ba(OH)2, is completely ionised in aqueous solutions. During the addition of 30.0 cm3 of 0.100 mol dm–3 Ba(OH)2 to 20.0 cm3 of 0.100 mol dm–3 isocyanic acid, the pH was measured. Calculate the [OH–] at the end of the addition. [OH–] = mol dm–3 Use your value in to calculate [H+] and the pH of the solution at the end of the addition. final [H+] = mol dm–3 final pH = On the following axes, sketch how the pH changes during the addition of a total of 30.0 cm3 of 0.100 mol dm–3 Ba(OH)2 to 20.0 cm3 of 0.100 mol dm–3 isocyanic acid. volume of Ba(OH)2 added / cm3 pH The cyanate ion, NCO–, can act as a monodentate ligand. State what is meant by the terms monodentate, ligand. Silver ions, Ag+, react with cyanate ions to form a linear complex. Suggest the formula of this complex, including its charge. When heated with HCl , organic isocyanates, RNCO, are hydrolysed to the amine salt, RNH3Cl, and CO2. RNCO + H2O + HCl RNH3Cl + CO2 A 1.00 g sample of an organic isocyanate, RNCO, was treated in this way, and the CO2 produced was absorbed in an excess of aqueous Ba(OH)2 according to the equation shown. The solid BaCO3 precipitated weighed 1.66 g. Ba(OH)2+ CO2BaCO3+ H2OCalculate the number of moles of BaCO3 produced. moles of BaCO3 = Hence calculate the Mr of the organic isocyanate RNCO. Mr of RNCO = The R group in RNCO and RNH3Cl contains carbon and hydrogen only. Use your Mr value calculated in to suggest the molecular formula of the organic isocyanate RNCO. molecular formula of RNCO Suggest a possible structure of the amine RNH2, which forms the amine salt, RNH3Cl.
9701_s17_qp_43
THEORY
2017
Paper 4, Variant 3
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The three substances shown all have some acidic properties. O phenol propan-1-ol C H H H H O C H H C propanoic acid C H H H C C H H H O O H H H Write an equation for the reaction between propan-1-ol and sodium metal. Give the order of the relative acidities of propanoic acid, propan‑1‑ol and phenol, stating the most acidic first. Explain your answer to . Methanoic acid, HCO2H, has a similar acid strength to propanoicacid. Describe a chemical test to distinguish between these two acids. Name the acid which gives a positive result in this test and describe the observations that would be made. The ester phenyl propanoate, C2H5CO2C6H5, can be made from phenol and propanoicacid in a two-step synthesis. The first step produces an acyl chloride. For this two-step synthesis, ● draw the structure of the product of the first step, ● state the reagents and conditions needed for each step of the synthesis. An unknown compound, Z, is propan‑1‑ol, propanal or propanoic acid. The proton NMR spectrum of Z dissolved in CDCl 3 is shown. δ / ppm –2 From the proton NMR spectrum, identify Z. State one feature that would be seen, and why, in the proton NMR spectra of each of the two compounds that are not Z.
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THEORY
2018
Paper 4, Variant 1
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