Titration Curve: Titration Between Strong Acid – Strong Base(Titrant: Strong Base; Analyte: Weak Acid)End point pH rangeInitial pH Before equivalence point(Buffered solution)After equivalence point (Excess of strong base)Initial pH is determined using the Ka of the weak acid.pH after base is added, but before equivalence point, is determined using the Henderson-Hasselbalch equation of an acidic buffer.pH at the equivalence point is determined using the Kb of the conjugate base of the weak acidpH after equivalence point is determined by the excess strong base.Equivalence point; pH > 7 due to the hydrolysis of anion which produced OH-
Titration Curve: Titration Between Strong Acid – Weak Base(Titrant: Strong Acid; Analyte: Weak Base)Initial pHBefore equivalence point(Buffered solution)End point pH rangeAfter equivalence point Initial pH is determined using the Kb of the weak base.pH after acid is added, but before equivalence point, is determined using the Henderson-Hasselbalch equation of an basic buffer.Equivalence point: pH < 7 due to the hydrolysis of cation which produced H3O+Beyond the equivalence point, the excess strong acid determines the pH.
0.1 M NaOH was gradually added to 25 mL of 0.1 M HCl solution. Calculate the pH of the solution, a) before NaOH was added.b) after the addition of 24 mL NaOH.c) at equivalence point.d) after the addition of 30 mL NaOH.pH Calculation: Strong Acid – Strong Base Reaction
HCl(aq) H+ (aq) + Cl–(aq)a) Before NaOH was added.Before titrant (a strong base) is added, pH is determined by the strong acid alone. Therefore, [H+] = 0.1 MpH = - log [H+]= 1.0[ ]i/M 0.1
b) After the addition of 24 mL NaOH.After base has been added, but before equivalence point, excess strong acid determines the pH.HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)ni/mol nchange/molnf/mol[ ]f/M0.1 x 251000= 2.5 x 10−30.1 x 241000= 2.4 x 10−30 -− 2.4 x 10−31 x 10−4(25 + 24) x 10−3= 2.04 x 10−31 x 10−40− 2.4 x 10−3 + 2.4 x 10−32.4 x 10−3HCl(aq) H+ (aq) + Cl–(aq)[H+] = [HCl] =2.04 × 10-3 MpH = –log [H+]= 2.69
c) At equivalence point. (nacid = nbase)At equivalence point, neutral salt is formed, neither cation nor anion undergoes hydrolysis, therefore pH = 7.HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)ni/mol :nchange/mol :nf/mol :0-− 2.5 x 10−3 − 2.5 x 10−30 02.5 x 10−3At 25oC, Kw = [H+] [OH-]1 × 10-14 = (x)(x)x = 1 × 10-7x = [H+] pH = - log (1 × 10-7) = 70.1 x 251000= 2.5 x 10−30.1 x 251000= 2.5 x 10−3+2.5 x 10−3
d) After the addition of 30 mL NaOH.Beyond equivalence point, excess strong base determines the pH. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)ni/mol nchange/mol nf/mol[ ]f/M0.1 x 251000= 2.5 x 10−30.1 x 301000= 3.0 x 10−30 -− 2.5 x 10−3 − 2.5 x 10−3 + 2.5 x 10−305 x 10−4 2.5 x 10−35 x 10−4(25 + 30) x 10−3= 9.09 x 10−3
NaOH(aq) Na+ (aq) + OH–(aq)[ ]f/M 9.09 × 10 9.09 × 10-3 -3Therefore, [OH-] = [NaOH] = 9.09 x 10-3 MpOH = - log [OH-]= 2.04 pH = 14 – 2.04 = 11.96
What is the pH of the resultant solution upon addition of 25 mL of 0.1 M NaOH to 25 mL of 0.1 M H2SO4in a titration.pH = 1.30
• Use of indicator: to detect the equivalence point of an acid-base titration.• A good choice of indicator is one at which it changes colour coincides with the equivalence point.• How and why an indicator changes colour can be explained by the equilibrium principle.• The weak acid HIn and its conjugate base In– have different colours.ACID – BASE INDICATORSHIn(aq) H+(aq) + In–(aq)Colour A Colour B
Indicator Colour in AcidColourin BasepH range of colour changeThymol blueBromophenol blueMethyl orangeMethyl redBromocresol purpleBromothymol blueCresol redPhenolphthaleinAlizarin yellowRedYellowOrangeRedYellowYellowYellowColorlessYellowYellowBlueYellowYellowPurpleBlueRedPinkRed1.2 – 2.8 3.0 – 4.6 3.2 – 4.4 4.8 – 6.0 5.2 – 6.8 6.0 – 7.6 7.0 – 8.8 8.2 – 10.0 10.1 – 12.0 SOME COMMON ACID – BASE INDICATORS
Type of Titration End point pH Range Suitable IndicatorStrong Acid –Strong Base 3 – 10 All indicatorsWeak Acid –Strong Base 7 – 11 Phenolphthalein, Cresol RedStrong Acid –Weak Base 3 – 7 Methyl Orange, Methyl RedWeak Acid –Weak Base None NoneTYPES OF ACID-BASE TITRATIONS AND ITS SUITABLE INDICATORS
In a weak acid-strong base titration, 25.00 mL KOH solution, is titrated with 21.45 mL of 0.10 M HCN to reachthe equivalence point.Explain suitable indicator for the titration.•Strong base is titrated with weak acid.•At equivalence point, pH > 7. •End point pH range: 7 – 11.•Suitable indicators: cresol red, bromothymol blue, phenolphthalein.
7.3 Solubility Equilibria
Many ionic salts are sparingly or very slightly soluble in water. When this salt is dissolved in water, a saturated solution will eventually form. A saturated solution is a solution in which no more solute will dissolve at a given temperature. SOLUBILITY EQUILIBRIA
In a saturated solution of a slightly soluble salt:a heterogeneous equilibrium is established between theundissolved solid salt and its dissolved ions (also called thesolubility equilibrium).the rate of dissolution equals the rate of precipitation.dissolutionprecipitationAg2S(s) 2Ag+(aq) + S2–(aq)Ag+Ag+S2–S2–S2–Ag+Ag+dissolved ionssolid Ag2S Saturated solution of silver sulphideS2–
Is the product of molar concentrations of ions in a saturated solution of a salt, each raised to the power of the stoichiometric coefficient of the respective ions.MxAy(s) xMy+(aq) + yAx–(aq)Ksp = [My+]x[Ax–]y(Note: The equilibrium in a saturated solution is a heterogeneous equilibrium. The concentration of pure solid is excluded as it is a constant)SOLUBILITY PRODUCT CONSTANT, Ksp The higher the Ksp, the greater the solubility. In general, for a salt with chemical formula MxAy,the equilibrium equation and Ksp are given as,
Solubility Product Constants of Sparingly Soluble SaltsAgBr 5.0 x 10-13 Fe(OH)3AgCl 1.8 x 10-10 FeS AgI 8.3 x 10-17 HgS AgIO33.1 x 10-8 Mg(OH)2Ag3PO41.3 x 10-20 MgC2O4Al(OH)32.0 x 10-32 Mn(OH)2Ba(OH)25.0 x 10-3 MnSBaSO41.1 x 10-10 NiS Bi2S31.0 x 10-97 PbCl2CaCO34.8 x 10-9 PbSO4CaC2O44.0 x 10-9 PbS CaSO41.2 x 10-6 SrSO4CdS 8.0 x 10-27 Zn(OH)2CoS 2.0 x 10-25 ZnS CuS 6.3 x 10-364.0 x 10-386.3 x 10-181.6 x 10-521.8 x 10-118.6 x 10-51.9 x 10-132.5 x 10-131.0 x 10-241.6 x 10-51.6 x 10-88.0 x 10-283.2 x 10-73.3 x 10-171.6 x 10-23Compound Ksp Compound Ksp
Write the solubility equilibrium equation and the Kspexpression for each of the following sparingly soluble salts :a) BaCrO4b) Mn(OH)2c) Ag2SO4d) Pb(IO3)2a) BaCrO4 (s)Ksp=Ba2+ (aq) CrO42– + (aq)[Ba2+][CrO42–]
c) Ag2SO4(s)Ksp=d) Pb (IO3)2(s)Ksp=b) Mn(OH)2(s)Ksp=Mn2+(aq) 2OH– + (aq)[Mn2+] [OH–]22Ag+(aq) SO42– + (aq)[Ag+]2 [SO42–]Pb2+(aq) 2IO3– + (aq)[Pb2+][IO3–]2
Molar solubility (in mol L–1) is the number of moles of solutedissolved in 1 L of a saturated solution. Solubility is the maximum amount of solute that dissolves in a given quantity of solvent to form a saturated solution.Solubility of compound (g/L)Molar Solubility of compound (mol/L)Molar concentration of ionsKsp of saltSOLUBILITY AND Ksp
Calculations involving Ksp can be divided into two categories:Calculate Kspfrom solubilitydataCalculatesolubility fromKspSolubility of compoundMolar solubility of compoundConcentrations of cations and anionsKsp of compoundKsp of compoundConcentrations of cations and anionsMolar solubility of compoundSolubility of compound
The molar solubility of calcium phosphate, Ca3(PO4)2in water at 25 oC is 2.47 x 10–6 M. Calculate the solubility product constant for this salt. Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43–(aq)3xKsp = [Ca2+]3[PO43–]2= 108 (2.47 x 10–6)5= 9.93 x 10–27 2x= (3x)3(2x)2 = 108 x5Let x = molar solubility of Ca3(PO4)2 = 2.47 x 10–6 M
At 25 oC, the solubility of CaF2in water is 1.69 x 10–2 g L–1. What is the value of Ksp for CaF2 at this temperature ?CaF2 (s) Ca2+(aq) + 2F–(aq)Ksp = [Ca2+] [F–]2Molar solubility = 1.69 x 10–2 gL–1 / 78 gmol–1= 2.16 x 10–4 mol L–1 = xx 2x= 4 (2.16 x 10–4)3= 4.03 x 10–11= (x) (2x)2
Write the solubility equilibrium equation and calculate the molar solubility of Hg(OH)2 which has a Ksp value of 3.2x10–26.Hg(OH)2 (s) Hg2+ (aq) + 2OH–(aq)x 2x4x3 = 3.2 x 10–26Let x = molar solubility of Hg(OH)2Ksp = [Hg2+] [OH–]23.2 x 10–26 = (x)(2x)2 x = 2 x 10–9 M
Ba3(PO4)2(s) 3Ba2+ (aq) + 2PO43–(aq)3x 2x108 x5 = 3.4 x 10–23Let x = molar solubility of Ba3(PO4)2Ksp = [Ba2+]3 [PO43–]23.4 x 10–23 = (3x)3(2x)2 x = 1.26 x 10–5 MThe Ksp value of Ba3(PO4)2 is 3.4x10–23. What is the molar solubility of Ba3(PO4)2in water?
When two solutions containing ions of a slightly soluble salt are mixed, prediction can be made whether a precipitate will form, by comparing the ion product, Q with Ksp.Condition DescriptionQ < Ksp Solution is unsaturated. No precipitate forms.Q = Ksp Solution is saturated. An equilibrium exists between undissolved solute and its ions.Q > Ksp Solution is supersaturated. Precipitate forms.
Concentration of cation from one soluteConcentration of anion from another soluteCalculate ion product, Q Compare Q with KspQ > Ksp precipitate forms.Q = Ksp equilibrium between undissolved salt and dissolved ions.Q < Ksp no precipitate forms.
Saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.Unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.Supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.
Will a precipitate of Mg(OH)2form when 50 mL of 0.001 M NaOH is added to 150 mL of 0.01 M MgCl2? (Ksp of Mg(OH)2 = 2 x 10–11)2NaOH(aq) + MgCl2(aq) Mg(OH)2(?) + 2NaCl(aq) In 200 mL mixture: (using M1V1 = M2V2)Ion product, Q = [Mg2+] [OH–]2= 4.7 x 10–10= (7.5x10–3) (2.5x10–4)2Mg(OH)2(s) Mg2+(aq) 2OH– + (aq)Q > KspMg(OH)2 precipitate will form.Solution is supersaturated.[Mg2+] =150 x 0.01200= 7.5 x 10−3M[OH−] =50 x 0.001200= 2.5 x 10−4M
0.02 M of CaCl2 solution is added to 0.02 M of Na2SO4 solution in equal volumes. Will CaSO4 precipitate form in the mixture? (Ksp of CaSO4 = 2x10–4)CaCl2(aq) + Na2SO4 (aq) CaSO4(?) + 2NaCl(aq) Ion product, Q = [Ca2+] [SO42–]= 1 x 10–4 CaSO4 precipitate will not form. = (0.01) (0.01)the solution is unsaturated.CaSO4(s) Ca2+(aq) + SO42–(aq)In 2V L solution:[Ca2+] =0.02 x V2V= 0.01M[SO42−] =0.02 x V2V= 0.01MQ < Ksp
Does a precipitate form when 1.0 x 10–2 mol of Ba(NO3)2 and 2.0 x 10–2 mol of NaF are dissolved in 1 Liter of solution? (Ksp of BaF2 = 1.7 x 10–6)Ba(NO3)2(aq) + 2NaF(aq) BaF2(?) + NaNO3(aq) In 1L solution:Ion product, Q = [Ba2+] [F–]2= 4 x 10–6 = (0.01) (0.02)2BaF2(s) Ba2+(aq) + 2F–(aq)solution is supersaturated[Ba2+] =0.01 mol1L= 0.01M[F−] =0.02 mol1L= 0.02MQ > KspBaF2 precipitate will form.
A 0.1 M solution of AgNO3is added dropwise to a solutioncontaining 0.1 M of NaCl and 0.01 M of K2CrO4. Which salt willprecipitate first, AgCl or Ag2CrO4 ?(Ksp AgCl = 2.0 x 10–10 ; KspAg2CrO4 = 2.4 x 10–12 )AgCl(s) Ag+(aq) + Cl–(aq)Q = [Ag+] [Cl–] = Ksp[Ag+] (0.1) = 2.0 x 10–10[Ag+] = 2.0 x 10–9 M*Precipitation begins when Q = KspAg2CrO4 (s) 2Ag+(aq) + CrO42–(aq)Q = [Ag+]2[CrO42–] = Ksp[Ag+]2(0.01) = 2.4 x 10–12[Ag+] = 1.55 x 10–5 MAgCl will precipitate first because the concentration of Ag+ion requiredfor AgCl to precipitate is lower than that required by Ag2CrO4.
Common ionAn ion in a solution that has been supplied from more than one solute.COMMON ION EFFECT AND SOLUBILITYCommon ion effectThe lowering of the solubility of a sparingly soluble salt by the addition of a common ion.
Add CrO42– [CrO42–]*The presence of a common ion decreases the solubility of the salt*Addition of Na2CrO4to a saturated solution of PbCrO4PbCrO4(s) Pb2+(aq) + CrO42-(aq)The solubility equilibrium that exists in a saturated solution of PbCrO4 is:When Na2CrO4(aq) is added to the saturated solution of PbCrO4, it provides the common ion, CrO42–, thus [CrO42–] in the solution increases.According to Le Chatelier’s principle, the solubility equilibrium will shift to the LEFT to offset the increase in [CrO42–] This means that the solubility of AgCldecreases, or more AgCl precipitate forms.
Solubility equilibrium in the saturated solution of slightly soluble saltSecond solute added Common ion BaSO4 (s) Ba2+(aq) + SO42–(aq) Na2SO4 (aq) SO42–Fe(OH)2 (s) Fe2+ (aq) + 2OH–(aq) NaOH (aq) OH–AgBr (s) Ag+ (aq) + Br–(aq) AgNO3 (aq) Ag+The net effect upon addition of common ions: the solubility equilibrium shifts to the LEFT.more precipitate forms. solubility decreases.
The solubility equilibrium that exists in a saturated solution of silver chloride is:AgCl (s) Ag+(aq) + Cl–(aq)When NaCl (aq) is added to the saturated solution of AgCl, it provides the common ion, Cl–, thus [Cl–] in the solution increases.According to Le Chatelier’s principle, the solubility equilibriumwill shift to the LEFT to offset the increase in [Cl–]. This means that the solubility of AgCl decreases, or more AgClprecipitate forms.The [Ag+] decreases in the same degree as the increase in [Cl–], hence the Ksp value is unaffected.
What is the solubility of PbSO4, Ksp = 1.66x10–8in a 1.00 M solution of H2SO4?PbSO4 (s) Pb2+ (aq) + SO42–(aq)Let s = molar solubility of PbSO4in 1.00 M H2SO4(s) (s + 1.00)H2SO4 (aq) 2H+(aq) + SO42–(aq)(1.00 M)Ksp = [Pb2+] [SO42–]1.66 x 10–8 = (s)(s + 1.00)In 1M H2SO4: Assumption: Since Ksp is small, s << 1.00, s + 1.00 1.00 s = 1.66 x 10–8 M1.66 x 10–8 = (s)(1.00)
If the solubility of Ag2CrO4in water is 8.4x10–5 mol L–1, what is the solubility in 0.10 M AgNO3? Ag2CrO4 (s) 2Ag+(aq) + CrO42–(aq)(2y) (y)Ksp = [Ag+]2[CrO42–]= (2y)2(y) = 4y3= 2.37 x 10–12Solubility of Ag2CrO4 = y = 8.4 x 10–5 M= 4 x (8.4 x 10–5)3
Let x = solubility of Ag2CrO4in 0.1 M AgNO3Ag2CrO4 (s) 2Ag+(aq) + CrO42–(aq)(2x+0.1) (x)Ksp = [Ag+]2[CrO42–]2.37 x 10–12 = (2x + 0.1)2(x) x = 2.37 x 10–10 MAssumption : Since Ksp is small, x << 0.1, 2x + 0.1 0.12.37 x 10–12 = (0.1)2(x)AgNO3 (s) Ag+(aq) + NO3–(aq)(0.1 M)In 0.1M AgNO3
Calculate the molar solubility of Ag2SO4 at 25oC in a) water b) 0.1 M Na2SO4c) c) 0.1 M AgNO3Explain the difference in the value of solubilities. (Ksp of Ag2SO4 = 1.5x10–5 M3) Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)(2x) (x)a) Let x = solubility of Ag2SO4in waterKsp = [Ag+]2 [SO42–]1.5 x 10–5 = (2x)2(x)4x3 = 1.5 x 10–5 x = 1.55 x 10–2 M
b) Let y = molar solubility of Ag2SO4in 0.1 M Na2SO4Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)Na2SO4 (aq) 2Na+(aq) + SO42–(aq)(2y) (y+0.1) M(0.1 M)Ksp = [Ag+]2[SO42–]1.5 x 10–5 = (2y)2(y + 0.1)Assumption: Since Ksp is small, y << 0.1, y + 0.1 0.1In 0.1M Na2SO44y2 = 1.5 x 10–4 y = 6.12 x 10–3 M1.5 x 10–5 = (2y)2(0.1)
c) Let z = molar solubility of Ag2SO4in 0.1 M AgNO3Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)AgNO3 (s) Ag+(aq) + NO3–(aq)(z) (2z+0.1)(0.1 M)Ksp = [Ag+]2[SO42–]1.5 x 10 –5 = ( 2z + 0.1 )2(z)Assumption: Since Ksp is small, z << 0.1, 2z + 0.1 0.1 z = 1.5 x 10–3 M1.5 x 10–5 = (0.1)2 (z)In 0.1M AgNO3
Molar solubility of Ag2SO4 (M)in water 1.55 x 10 –2in Na2SO4 6.12 x 10 –3in AgNO3 1.50 x 10 –3The solubility of Ag2SO4in Na2SO4 and AgNO3 are lower than its solubility in water.The presence of common ions, SO42–in Na2SO4 and Ag+in AgNO3causes a shift in the solubility equilibrium to the LEFT.Ag2SO4 (s) 2Ag+(aq) + SO42–(aq)Hence, more Ag2SO4 precipitate form or its solubility decreases.
CHEMISTRY UNIT, KMNS THE END