Acid - Base

Acid - Basepage 6(e) Calculate the value of the ionization constant, Answer:Ka, of the acid HA.(a) at equivalence point, moles HA moles NaOH MbVb (0.0274 L)(0.135 M) 3.70 10–3 molHAmass HA0.682g 184 g/mol(b) molec.wt . mol HA3.70 10 3 mol(c) HA OH– A– H2Oinitial:0.00370 moladded:(0.0106L)(0.135M) 0.00143 moleremaining: (0.00370 – 0.00143) 0.00227 mol(d) pH –log[H3O ]; [H3O ] 10–pH 10–5.65 2.2 10–6M[H3O ][A ] (2.2 10 6 )(0.00143 / v) [HA](0.00227 / v)–6 1.4 10(e) Ka 1978 DPredict whether solutions of each of the followingsalts are acidic, basic, or neutral. Explain yourprediction in each case(a) Al(NO3)3(b) K2CO3(c) NaBrAnswer:(a) acidic; Al3 H2O AlOH2 H ;hydrolysis of Al3 ;Al(OH2)n3 as Brönsted acid, etc.(b) basic; CO32– H2O HCO3– OH– ; orhydrolysis of CO32– as conjugate to a weakacid, etc.(c) neutral; Na from strong base; Br– from strongacid1979 BA solution of hydrochloric acid has a density of 1.15grams per milliliter and is 30.0% by weight HCl.(a) What is the molarity of this solution of HCl?(b) What volume of this solution should be taken inorder to prepare 5.0 liters of 0.20 molar hydrochloric acid by dilution with water?(c) In order to obtain a precise concentration, the0.20 molar hydrochloric acid is standardizedagainst pure HgO (molecular weight 216.59)by titrating the OH– produced according to thefollowing quantitative reaction.Copyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - BaseHgO(s) 4I– H2O HgI42– 2OH–page 7In a typical experiment 0.7147 grams of HgO required 31.67 milliliters of the hydrochloric acidsolution for titration. Based on these data what isthe molarity of the HCl solution expressed tofour significant figures.Answer:1.15g 1000mL 30.0gHCl 1mol 9.5M(a)1mL1L100g35.5g(b) MfVf MiVi(0.20M)(5.0L) (9.5M)(V)V 0.11 L0.7147g 0.003300 mol HgO(c)216.59g molmol OH– prod. 2 (mol HgO) 0.006600 molmol HCl req. mol OH– prod. 0.006600 mol0.006600molM HCl 0.2084M0.03167L1979 DNH4 OH– NH3 H2OH2O C2H5O– C2H5OH OH–The equations for two acid–base reactions are givenabove. Each of these reactions proceeds essentiallyto completion to the right when carried out inaqueous solution.(a) Give the Brönsted–Lowry definition of an acidand a base.(b) List each acid and its conjugate base for each ofthe reactions above.(c) Which is the stronger base, ammonia or theethoxide ion. C2H5O–? Explain your answer.Answer:(a) acid proton donor; base proton acceptor(b)AcidConjugate base1st reactionNH4 NH3H2 OOH–2nd reactionH2 OOH–C2H5OHC2H5O–(c) ethoxide is a stronger base than ammonia.A stronger base is always capable of displacinga weaker base. Since both reactions are quantitative, in terms of base strength, OH– NH3 in 1streaction; C2H5O– OH– in 2nd rxn.Copyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Basepage 81980 AMethylamine CH3NH2, is a weak base that ionizes insolution as shown by the following equation.CH3NH2 H2O CH3NH3 OH–(a) At 25 C the percentage ionization in a 0.160molar solution of CH3NH2 is 4.7%. Calculate[OH–], [CH3NH3 ], [CH3NH2], [H3O ], and thepH of a 0.160 molar solution of CH3NH2 at25 C(b) Calculate the value for Kb, the ionization constant for CH3NH2, at 25 C.(c) If 0.050 mole of crystalline lanthanum nitrate isadded to 1.00 liter of a solution containing 0.20mole of CH3NH2 and 0.20 mole of its saltCH3NH3Cl at 25 C, and the solution is stirreduntil equilibrium is attained, will any La(OH)3precipitate? Show the calculations that proveyour answer. (The solubility constant forLa(OH)3, Ksp 1 10–19 at 25 C)Answer:(a) CH3NH2; 0.160M 4.7% 7.5 10–3 M ionizing(0.160M – 0.0075M) 0.0152M @ equilibrium[CH3NH3 ] [OH–] 7.5 10–3 MKw 12[H3O ] M 3 1.3 107.5 10pH –log [H3O ] 11.89(b) K b [CH3 NH 3 ][OH ] (7.5 10 3 )2 [CH3 NH2 ]0.152 3.7 10–4 4(c) K b 3.7 10 (0.20 X)(X) X(.020 X) [OH–]Q [La3 ][OH–]3 (0.050)(3.7 10–4)3 2.5 10–12Q Ksp, therefore, La(OH)3 precipitatesCopyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

1981 DAl(NO3)3Acid - Basepage 9(b) The pH remains unchanged because the ratio ofthe formate and formic acid concentration staysthe same.(a) Predict whether a 0.10 molar solution of each ofthe salts above is acidic, neutral or basic.(b) For each of the solutions that is not neutral,write a balanced chemical equation for areaction occurring with water that supports yourprediction.Answer:(a) Al(NO3)3 – acidicK2CO3 – basicK2CO3 NaHSO4NaHSO4 – acidicNH4ClNH4Cl – acidic(b) Al3 H2O Al(OH)2 H Al(H2O)63 H2O [Al(H2O)5OH]2 H3O Al3 3 H2O Al(OH)3 3 H CO32– H2O HCO3– OH–HSO4– H2O SO42– H3O NH4 H2O NH3 H3O 1982 AA buffer solution contains 0.40 mole of formic acid,HCOOH, and 0.60 mole of sodium formate,HCOONa, in 1.00 litre of solution. The ionizationconstant, Ka, of formic acid is 1.8 10–4.(a) Calculate the pH of this solution.(b) If 100. millilitres of this buffer solution isdiluted to a volume of 1.00 litre with pure water,the pH does not change. Discuss why the pHremains constant on dilution.(c) A 5.00 millilitre sample of 1.00 molar HCl isadded to 100. millilitres of the original buffersolution. Calculate the [H3O ] of the resultingsolution.(d) A 800.–milliliter sample of 2.00–molar formicacid is mixed with 200. milliliters of 4.80–molarNaOH. Calculate the [H3O ] of the resulting solution.Answer:(a) using the Henderson–Hasselbalch equation [A ] pH pK a log [HA] 0.60 log(1.8 10 4 ) log 0.40 3.92{other approaches possible}Copyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Basepage 10Conductivity, λ(c) initial concentrationshydroxide originally present in the solution thatis titrated.5.00mL1.00M HCl 0.0476M(d) Explain why the conductivity does not fall to105mLzero at the equivalence point of this titration.100mL0.40M HCOOH 0.38MAnswer:105mL100mL(a) Ba2 2 OH– 2 H SO42– BaSO4(s) 20.60M HCOO- 0.57M105mLH2 O –concentrations after H reacts with HCOO(b) The initial conductivity is high because of the0.38M 0.05M 0.43M HCOOHpresence of Ba2 and OH– ions. Theconductivity decreases because Ba2 forms0.57M – 0.05M 0.52M HCOO–insoluble BaSO4 with the addition of SO42–.0.43M[H3O ] 1.8 10 4 1.5 4 MThe conductivity also decreases because OH–0.52Mcombines with the addition of H ions by(d) 0.800L 2.00M HCOOH 1.60 molforming H2O.0.200L 4.80M NaOH 0.96 mol OH–Beyond the equivalence point conductivity inat equil., (1.60 – 0.96) 0.64 mol HCOOH andcreases as H and SO42– ions are added.0.96 mol HCOO–(c) # mol Ba(OH)2 # mol H2SO40.64M[H3O ] 1.8 10 4 1.2 4 M0.96M 0.1M 0.04L 0.004 mol(d) BaSO4(s) dissociates slightly to form Ba2 and1982 DSO42–, while the water ionizes slightly to formA solution of barium hydroxide is titrated with 0.1–H and OH–.M sulfuric acid and the electrical conductivity of thesolution is measured as the titration proceeds. The1983 Bdata obtained are plotted on the graph below.The molecular weight of a monoprotic acid HX wasto be determined. A sample of 15.126 grams of HXwas dissolved in distilled water and the volumebrought to exactly 250.00 millilitres in a volumetricflask. Several 50.00 millilitre portions of thissolution were titrated against NaOH solution, requiring an average of 38.21 millilitres of NaOH.The NaOH solution was standardized against oxalicacid dihydrate, H2C2O4.2H2O (molecular weight:126.066 gram mol–1). The volume of NaOH solutionrequired to neutralize 1.2596 grams of oxalic acid dihydrate was 41.24 millilitres.10 20 30 40 50 60 70 80(a) Calculate the molarity of the NaOH solution.Millilitres of 0.1–M H2SO4(b) Calculate the number of moles of HX in a 50.00millilitre portion used for titration.(a) For the reaction that occurs during the titrationdescribed above, write a balanced net ionic (c) Calculate the molecular weight of HX.equation.(d) Discuss the effect of the calculated molecularweight of HX if the sample of oxalic acid dihy(b) Explain why the conductivity decreases, passesdrate contained a nonacidic impurity.through a minimum, and then increases as thevolume of H2SO4 added to the barium Answer:hydroxide is increased.(c) Calculate the number of moles of bariumCopyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Basepage 11wrong with the erroneous procedures.(No calculations are necessary, but the following126.066acidity constants may be helpful: acetic acid,9.9916 10–3 molKa 1.8 10–5; NH4 , Ka 5.6 10–10)H2C2O4 2 NaOH Na2C2O4 2 H2OAnswer:2molNaOH9.9916 10 3 mol 1.9983 10 2 mol(a) A buffer solution resists changes in pH upon the1molH 2C 2O4addition of an acid or base.1.9983 10 2 molPreparation of a buffer: (1) mix a weak acid aM NaOH 0.4846Msalt of a weak acid; or (2) mix a weak base salt0.04124Lof a weak base; or (3) mix a weak acid with(b) mol HX mol NaOHabout half as many moles of strong base; or (4)0.03821 L 0.4846 M 0.01852 mol HXmix a weak base with about half as many moles0.01852mol(c) 250.00mL 0.09260mol HXof strong acid; or (5) mix a weak acid and a50.00mLweak base.15.126gMW 163.3 g mol0.09260mol(d) The calculated molecular weight is smaller thantrue value, because:measured g H2C2O4 is larger than true value,(a) mol H2C2O4.2H2O 1.2596 ggmol calculated mol H2C2O4 is larger than true value,calculated mol NaOH is larger than true value,calculated M NaOH is larger than true valuecalculated mol HX is larger than true value,therefore,g HX (true value )MW mol HX (calculated , and too large )1983 C(a) Specify the properties of a buffer solution. Describe the components and the composition ofeffective buffer solutions.(b) An employer is interviewing four applicants fora job as a laboratory technician and asks eachhow to prepare a buffer solution with a pH closeto 9.Archie A. says he would mix acetic acid andsodium acetate solutions.Beula B. says she would mix NH4Cl and HClsolutions.Carla C. says she would mix NH4Cl and NH3solutions.Dexter D. says he would mix NH3 and NaOHsolutions.Which of these applicants has given an appropriate procedure? Explain your answer, referring toyour discussion in part (a). Explain what isCopyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Base(b) Carla has the correct procedure, she has mixed aweak base, NH3, with the salt of a weak base,NH4Cl.Archie has a buffer solution but a pH of around5.Beula doesn’t have a buffer solution, her solution consists of a strong acid and a salt of a weakbase.Dexter does not have a buffer solution, since hissolution consists of a weak base plus a strongbase.1984 ASodium benzoate, C6H5COONa, is the salt of a theweak acid, benzoic acid, C6H5COOH. A 0.10 molarsolution of sodium benzoate has a pH of 8.60 atroom temperature.(a) Calculate the [OH–] in the sodium benzoatesolution described above.(b) Calculate the value for the equilibrium constantfor the reaction:C6H5COO– H2O C6H5COOH OH–(c) Calculate the value of Ka, the acid dissociationconstant for benzoic acid.(d) A saturated solution of benzoic acid is preparedby adding excess solid benzoic acid to purewater at room temperature. Since this saturatedsolution has a pH of 2.88, calculate the molarsolubility of benzoic acid at room temperature.Answer:(a) pH 8.6, pOH 5.4[OH–] 10–pOH 3.98 10–6M(b) [C6H5COOH] [OH–]page 12ions) 2.88 10–2M1984 CDiscuss the roles of indicators in the titration of acidsand bases. Explain the basis of their operation andthe factors to be considered in selecting anappropriate indicator for a particular titration.Answer:An indicator signals the end point of a titrationby changing color.An indicator is a weak acid or weak base wherethe acid form and basic form of the indicators are ofdifferent colors.An indicator changes color when the pH of thesolution equals the pKa of the indicator. In selectingan indicator, the pH at which the indicator changescolor should be equal to (or bracket) the pH of thesolution at the equivalence point.For example, when a strong acid is titrated with astrong base, the pH at the equivalence point is 7, sowe would choose an indicator that changes color at apH 7. {Many other examples possible.}1986 AIn water, hydrazoic acid, HN3, is a weak acid thathas an equilibrium constant, Ka, equal to 2.8 10–5 at25 C. A 0.300 litre sample of a 0.050 molar solutionof the acid is prepared.(a) Write the expression for the equilibriumconstant, Ka, for hydrazoic acid.(b) Calculate the pH of this solution at 25 C.(c) To 0.150 litre of this solution, 0.80 gram ofsodium azide, NaN3, is added. The salt 62[C H COH][OH ](3.98 10 )dissolved completely. Calculate the pH of theKb 6 5 6resulting solution at 25 C if the volume of the[C6H 5CO2 ](0.1 3.98 10 )solution remains unchanged. 1.58 10–10(d) To the remaining 0.150 litre of the original soluKw1.0 10 14 5tion, 0.075 litre of 0.100 molar NaOH solution is 6.33 10(c) Ka 10Kb 1.58 10 )added. Calculate the [OH–] for the resultingsolution at 25 C.(d) pH 2.88 1.32 10–3M [H ] [C6H5COO–]Answer:[H ][C6 H5CO 2 ] (1.32 10 3 )2[C6H5CO2 H] (a) HN3 H N3–Ka6.33 10 5[H ][N3 ] 2.75 10–2M [C6H5COOH]Ka [HN3 ]total dissolved (2.75 10–2M 1.32 10–3M asCopyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Base(b) [H ] [N3–] XX2; X 1.2 10 3 M0.050 pH –log[H ] 2.932.8 10 5 page 130.80g 1mol 0.082M(c) [N 3 ] 0.150L 65g[H ](0.082)2.8 10 5 ; [H ] 1.7 10 5 M0.050pH 4.77(d) (0.075L)(0.100M) 0.0075 mol NaOH(0.150L)(0.050M) 0.0075 mol HN3OH– HN3 H2O N3– ; neut. completeKN3– H2O HN3 OH– ; K b wKa1.0 10 14 [HN3 ][OH ]X2 0.0075 2.8 10 5[N 3 ] 0.225 X [OH–] 3.5 10–6M1986 DH2SO3HSO3–HClO4HClO3H3BO3Oxyacids, such as those above, contain an atombonded to one or more oxygen atoms; one or more ofthese oxygen atoms may also be bonded to hydrogen.(a) Discuss the factors that are often used to predictcorrectly the strengths of the oxyacids listedabove.(b) Arrange the examples above in the order of increasing acid strength.Answer:(a) 1) As effective nuclear charge on central atomincreases, acid strength increases. ORAs number of lone oxygen atoms (oxygen atomsnot bonded to hydrogen) increases, acid strengthincreases. ORAs electronegativity of central atom increases,acid strength increases.2) Loss of H by a neutral acid moleculereduces acid strength. ORKa of H2SO3 Ka of HSO3–(b) H3BO3 HSO3– H2SO3 HClO3 HClO4H3BO3 or HSO3– weakest (must be together)1987 ANH3 H2O NH4 OH– Ammonia is a weak basethat dissociates in water as shown above. At 25 C,the base dissociation constant, Kb, for NH3 is1.8 10–5.Copyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Base(a) Determine the hydroxide ion concentration andthe percentage dissociation of a 0.150 molar solution of ammonia at 25 C.(b) Determine the pH of a solution prepared byadding 0.0500 mole of solid ammonium chlorideto 100. millilitres of a 0.150 molar solution ofammonia.(c) If 0.0800 mole of solid magnesium chloride,MgCl2, is dissolved in the solution prepared inpart (b) and the resulting solution is well–stirred,will a precipitate of Mg(OH)2 form? Showcalculations to support your answer. (Assumethe volume of the solution is unchanged. Thesolubility product constant for Mg(OH)2 is1.5 10–11.Answer:(a) [NH4 ] [OH–] X; [NH3] (0.150 – X)page 14in water and titrated with the NaOH solution. Toreach the equivalence point, 26.90 millilitres ofbase was required. Calculate the molarity of theNaOH solution. (Molecular weight: KHC8H4O4 204.2)X2[NH 4 ][OH ]–5Kb ; 1.8 10 0.150 – X[NH3]X [OH–] 1.6 10–3 M1.6 103% diss. 100% 1.1%0.150(b) [NH4 ] 0.0500 mol/0.100L 0.500M[NH3] 0.150M(0.500)(X)1.8 10 5 ; X [OH ] 5.4 10 6 M0.150pOH 5.27; pH (14 – 5.27) 8.73(c) Mg(OH)2 Mg2 2 OH–[Mg2 ] (0.0800mol/0.100L) 0.800M[OH–] 5.4 10–6MQ [Mg2 ][OH–]2 (0.800)(5.4 10–6)2 2.3 10–11Q Ksp so Mg(OH)2 precipitates1987 BThe percentage by weight of nitric acid, HNO3, in asample of concentrated nitric acid is to bedetermined.(a) Initially a NaOH solution was standardized bytitration with a sample of potassium hydrogenphthalate, KHC8H4O4, a monoprotic acid oftenused as a primary standard. A sample of pureKHC8H4O4 weighing 1.518 grams was dissolvedCopyright 1970 to 1999 by Educational Testing Service, Princeton, NJ 08541. All rights reserved. For face-to-face teaching purposes, classroomteachers are permitted to reproduce the questions. Portions opyright 1993-9 by Unlimited Potential, Framingham, MA 01701-2619.

Acid - Base(b) A 10.00 millilitre sample of the concentrated nitric acid was diluted with water to a total volumeof 500.00 millilitres. Then 25.00 millilitres of thediluted acid solution was titrated with the standardized NaOH solut

acid. Ka acetic acid 1.8 10-5 (b) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter of a 1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the buffer solution. (c) Suppose that 0.10 liter of 0.50 molar hydrochlo-ric acid is added to 0.040 liter of the buffer pre-