CHAPTER 7IONIC EQUILIBRIACHEMISTRY UNIT, KMNS
7.1 Acids And Bases
ACIDS AND BASESACIDS BASES• LITMUS PAPER CHANGES COLOUR FROM:Blue to red Red to blue• REACTION WITH METALAcid + metal → salt + H2(g) No reaction• REACTION WITH METAL OXIDEAcid + metal oxide → salt + H2O No reaction• REACTION WITH CARBONATE, CO32-IONAcid + CO32- → Salt + H2O + CO2 No reaction• REACTION WITH ACIDNo reaction Base + acid → salt + H2O
41. ARRHENIUS THEORY Definition:• Acid : a substance which can produce H+ or H3O+ when dissolves in water.Example : HCl (aq) H+(aq) + Cl-(aq)• Base : a substance which can produce OH- when dissolves in water.Example : NaOH (aq) Na+(aq) + OH-(aq)Limitations:• Applicable to compounds containing hydrogen and hydroxide.• Restricted to aqueous solution.ACID-BASE THEORIES
2. BRONSTED-LOWRY THEORYDefinitions:• Acid : a substance which donates H+(proton). When an acid loses H+, conjugate base is formed.• Base : a substance which accepts H+(proton). When a base accepts H+, conjugate acid is formed.CH3COO- + HCN CH3COOH + CNConjugate pair Conjugate pairAcid Conj. base Base Conj. acidHCN CN- CH3COO- CH3COOH 5
H2PO4– + NH3 HPO42– + NH4+CO32– + H2O HCO3– + OH–Conjugate pair Conjugate pairAcid Conj. base Base Conj. acidH2PO4– HPO42– NH3 NH4+Conjugate pair Conjugate pairAcid Conj. base Base Conj. acidH2O OH– CO32– HCO3– 6
COMPARISON OF STRONG ACID & WEAK ACIDSTRONG ACID:• Examples: HCl, HNO3, H2SO4• Degree of dissociation in water, = 1 (@ % ionisation = 100%)• Dissociation equation: HCl(aq) H+(aq) + Cl-(aq)1.0M 1.0MWEAK ACID:• Examples: CH3COOH (Ka = 1.8 X 10-5), HCN (Ka = 4.9 X 10-10)• Degree of dissociation in water, 1 (@ % ionization 100%)• Dissociation equation: CH3COOH(aq) CH3COO-(aq) + H+(aq)0.01M < 0.01M7
COMPARISON OF STRONG BASE & WEAK BASESTRONG BASE:• Examples: NaOH, KOH, Mg(OH)2• Degree of dissociation in water, = 100%• Example of dissociation equation: NaOH(aq) Na+(aq) + OH-(aq)2.0M 2.0MWEAK BASE:• Examples: NH3 (Kb = 1.8 X 10-5), (C2H5)2NH (Kb = 9.6 X 10-4)• Degree of dissociation in water, 100%• Example of dissociation equation:NH3(aq) + H2O(?) NH4+(aq) + OH-(aq)5 x 10-3M 5 x 10-3M 8
1. Write equations for the acid-base reactions that occurred in the aqueous solutions for each of the following speciesi. NH4+ii. S2-2. At 25 °C, 2.20% of 0.125 M benzoic acid, C6H5COOH ionises. Write the dissociation equation for benzoic acid in water. NH4+(aq) + H2O(?) NH3(aq) + H3O+(aq)S2-(aq) + H2O(?) HS-(aq) + OH-(aq)C6H5COOH(aq) + H2O(?) C6H5COO-(aq) + H3O+(aq)9
Equation for auto-ionisation of H2O:At equilibrium :Kc x [H2O]2 = [OH-] x [H3O+] Kw = [OH-] x [H3O+] At 25oC, [OH-] = [H3O+] = 1 x 10-7 M Kw = (1 x 10-7) x (1 x 10-7)= 1 x 10-14pKw = - log Kw = 14IONIC PRODUCT OF WATER, KWH2O(?) + H2O(?) OH-(aq) + H3O+(aq)Kc =OH− H3O+H2O 2X 10010
DISSOCIATION CONSTANT Ion product of water, Kw For Weak Acid, Ka For Weak Base, KbDissociation equationH2O(?) + H2O(?) H3O+(aq) + OH-(aq)@2H2O(?) H3O+(aq) + OH-(aq)HA(aq) + H2O(?) H3O+(aq) + A-(aq)@HA(aq) H+(aq) + A-(aq)B(aq) + H2O(?) BH+(aq) + OH-(aq)Expression Kw = [H+] [OH-]- log Kw = pKw- log Ka = pKa- log Kb = pKbAt 25oC, pKa x pKb = pKw = 14 Ka x Kb = Kw = 1 x 10-14Kw = [H3O+][OH-] = 1 x 10-14- log Kw = - log [H3O+] – (-log [OH-] ) = - 1x10-14pKw = pH + pOH = 14pKa , Ka , acidity pKb Kb , basicity Ka = [H+] [OH-][HA]Kb = [BH+] [OH-][B]pKa 1KapKb 1Kb11
12pH SCALE• pH is defined as : the negative logarithm of [H+] or [H3O+].• Used to measure the [H+] or [H3O+].• pH = - log [H+] @ - log [H3O+]• pOH is defined as : the negative logarithm of [OH-]• Used to measure the [OH-].• pOH = - log [OH-]RELATIONSHIP BETWEEN KW , pH AND pOHAt 25oC, Kw = [H3O+][OH-] = 1 x 10-14- log Kw = - log [H3O+] – (- log [OH-] ) = - log (1x10-14)pKw = pH + pOH = 14
Calculate the pH of the following solution:An aqueous solution contains 0.700g NaOH in 485ml H2O, where NaOH is a strong base.13nNaOH =mNaOHmolar mass=0.700 g40 gmol−1= 1.75 x 10−2MMsoln =nNaOHVNaOH=1.75 x 10−2mol485 x 10−3L= 0.036MpOH = − log OH−= − log 0.036= 1.44pH = 14 − pOH= 14 − 1.44= 12.56NaOH (aq) Na+(aq) + OH-(aq)
Calculate the pH of 0.02 M Ba(OH)2solution.Ba(OH)2(aq) Ba2+(aq) + 2OH-(aq)(0.02M) (2 x 0.02M)pOH = - log [OH-]= - log (0.04)= 2.4Calculate the pH of a 1 x 10–3 M solution of hydrochloric acid and a 1 x 10–3 M solution of calcium hydroxide.HCl(aq) Cl-(aq) + H+(aq) (1 x 10-3M) (1 x 10-3 M) pH = - log [H+]= - log (1 x 10-3)= 3 14pH = 14 - pOH= 12.6Complete dissociation of Ba(OH)2:Complete dissociation of HCl :
The pH value of a solution containing 0.245 mol HF in 500mL water is 1.88. Calculate the acid ionisation constant for HF.HF(aq) + H2O(?) F-(aq) + H3O+(aq)[HF][F ][H O ]K = +3-apH = - log [H3O+]1.88 = - log [H3O+] [H3O+] = 0.0132 M = 0.49 M500 x 10 L0.245 molM = HF -3[ ]initial/M 0.49 - 0 0[ ]change/M - x - + x + x[ ]eqm/M 0.49 – x - x x0.477 0.0132 0.0132HF(aq) + H2O(l) F-(aq) + H3O+(aq) 0.4770.0132 x 0.0132 = = 3.65 x 10-415
At 25 °C, 2.20% of 0.125 M benzoic acid, C6H5COOH ionises. i. Write the dissociation equation for benzoic acid in water.C6H5COOH(aq) + H2O(?) C6H5COO-(aq) + H3O+(aq)ii. Determine the acid dissociation constant, Kafor benzoic acid. C6H5COOH + H2O C6H5COO - + H3O+[ ]initial/M 0.125 - 0 0[ ]change/M - x - + x + x[ ]eqm/M 0.125 - x - x x[C H COOH][C H COO ][H O ] K = 6 5+3- 6 5a? =[ ]?ℎ????[ ]???????x 100% 2.20 =[ ]?ℎ????0.125 x 100 [ ]change = 2.75 x 10-3 = x = [H3O+]16= 2.75 x 10-3 = 2.75 x 10-3 = 0.122
0.122(2.75 x 10 ) (2.75 x 10 ) K = -3 -3a= 6.186 x 10-5iii. Calculate the pH of the solution. pH = - log [H3O+]= - log (2.75 x 10-3)= 2.5617
Write the dissociation equation of base for a 0.5 M aqueous solution of CH3NH2andcalculate the pH. [Kb= 3.7 x 10–4M]CH3NH2(aq) + H2O(?) CH3NH3+(aq) + OH-(aq)CH3NH2 + H2O CH3NH3+ + OH-[ ]initial/M 0.5 - 0 0[ ]change/M - x - + x + x[ ]eqm/M 0.5 – x - x xx1 = 0.0134x2 = - 0.014 (neglected)= [OH-] pH = 14 – 1.87= 12.13 [CH NH ][CH NH ][OH ] K = 3 2+ - 3 3b(0.5 - x)(x)(x) 3.7 x 10 = -4x2 + 3.7 x 10-4x – 1.85 x 10-4 = 0 18pOH = - log [OH-]= - log (0.0134)= 1.87
19REMARKS• If the value of Ka @ Kbis < x 10 -5: can use approximation.• If the value of Ka @ Kbis x 10 -4: cannot use approximation.
20GUIDELINE IN PROBLEM SOLVING RELATED TO WEAK ACIDS @ BASESWrite the dissociation equation of the weak acid @ base Dissociate partially01Tabulate the data Enter the [ ]Initial of all species Determine the change in concentration,[ ]change Write the [ ]eqm0203 Enter the [ ] eqm into Ka @ Kb expressionSolve for x by assuming a – x a (if Ka @ Kb < 10-5) Determine [H+] @ [OH-] & the pH value0405
21SALT• Preparation: Acid reacts with baseGeneral equation: Acid + Base Salt + WaterExample: HCl + NaOH NaCl + H2O• Properties: i. An ionic compound.ii. Strong electrolyte (dissociate completely in water to form cation and anion).Example: NaCl Na+ + Cl-• Classification: Salt is classified into neutral or basic or acidic through hydrolysis reaction. Hydrolysis reaction is a chemical reaction between anion or cation (of a salt) with H2O.
NEUTRAL SALTFormation : Strong acid + strong baseExample : HCl + NaOH NaCl + H2ODissociation equation : NaCl Na+ + ClHydrolysis equation :Neither the cation nor anion undergoes hydrolysispH 7ACIDIC SALTFormation : Strong acid + weak baseExample : HCl + NH3 NH4ClDissociation equation :NH4Cl NH4+ + ClHydrolysis equation :Cation reacts with H2ONH4+ + H2O NH3 + H3O+pH < 7 (due to the presence of H3O+)BASIC SALTFormation : Weak acid + strong baseExample : CH3COOH + NaOH CH3COONa + H2ODissociation equation : CH3COONa CH3COO- + Na+Hydrolysis equation : Anion reacts with H2OCH3COO - + H2O CH3COOH + OHpH > 7 (due to the presence of OH-)22
1. Write the hydrolysis equations for sodium carbonate, Na2CO3andammonium nitrate, NH4NO3solutions. State whether the solutions areacidic, basic or neutral.Ans: Na2CO3 2Na+ + CO32-Ans: CO32- + H2O H2CO3 + OH- ~ Basic saltAns: NH4NO3 NH4+ + NO3-Ans: NH4+ + H2O NH3 + H3O+ ~ Acidic salt2. OBr -ion is the conjugate base for a weak acid, HOBr. Write the hydrolysis equation of NaOBr. Ans: NaOBr Na+ + OBrAns: OBr- + H2O HOBr + OH-
3. Sodium cyanide, NaCN is a salt formed when a strong base, NaOH is reacted with a weak acid, HCN.i. Write a balanced equation to show the reaction between NaOH and HCN. Classify the salt formed.ii. What would be the expected pH of the NaCN solution? Explain the answer using appropriate equation(s). Balanced equation: HCN + NaOH NaCN + H2OSalt dissociation equation: NaCN 2Na+ + CNHydrolysis equation: CN- + H2O HCN + OHClassification of salt: Basic salt.pH > 7 : Due to the presence of the OHCN- + H2O HCN + OHSolution:Solution:
Type of Acid/Base Type of SaltSpecies Which HydrolysepH Solution ExamplesStrong Acid + Strong BaseStrong Acid + Weak BaseWeak Acid + Strong BaseWeak Acid + Weak BaseSUMMARYNeutral NoneAcidic CationBasic AnionBothNeutral @ Acidic @Basic7< 7> 7NaCl, KI, KNO3, NaSO4NH4Cl, (NH4)2CO3, CuSO4CH3COONa, KCN, NaCN, Na2SDepends On Type Of SaltC2H3O2NH4, (NH4)2C2O4
BUFFER SOLUTION• Definition: Solution that has the ability to resist changes in pH when a small amount of strong acid or strong base is added to the solution.• Type:i. Acidic buffer ii. Basic buffer• Preparation of an acidic buffer: When weak acid is mixed with its conjugate salt.Example:Weak acid: CH3COOH CH3COO- + H+Conjugate salt: CH3COONa CH3COO- + Na+• Preparation of basic buffer: When weak base is mixed with its conjugate salt.Example: Weak base: NH3 + H2O NH4+ + OHConjugate salt: NH4Cl NH4+ + Cl-
Buffer solution has 2 components: A component which is able to neutralise the acids added to the buffer solution. A component which is able to neutralise the bases added to the buffer solution.
i. A component which is able to neutralise the acid added to the buffer solution is the conjugate base of the weak acid.ii. A component which is able to neutralise the base added to the buffer solution is the weak acid.CH3COOH CH3COO- + H+CH3COONa CH3COO- + Na+(Weak acid)(Conjugate base of the weak acid)ACIDIC BUFFER SOLUTIONA buffer containing ethanoic acid (CH3COOH) and sodium acetate (CH3COONa),When acid is added to the acidic buffer solution:H+added+ CH3COO- CH3COOHWhen base is added to the acidic buffer solution:OHadded+ CH3COOH CH3COO- + H2O pH of the buffer solution remains constant
i. A component which is able to neutralise the acid added to the buffer solution is the weak base.ii. A component which is able to neutralise the base added to the buffer solution is the conjugate acid of the weak base.(Conjugate acid of the weak base)NH3 + H2O NH4+ + OHNH4Cl NH4+ + Cl-(Weak base)When acid is added to the basic buffer solution:H+added+ NH3 NH4+When base is added to the basic buffer solution:OHadded+ NH4+ NH3 + H2O pH of the buffer solution remains constantBASIC BUFFER SOLUTIONA buffer containing ammonia (NH3) and ammonium chloride (NH4Cl),
DETERMINATION OF pH of BUFFER SOLUTIONBasic Buffer Solution[weak acid][conjugate base] pH = pK + log a Acidic Buffer SolutionKnown as Henderson-Hasselbalchequation for acidic buffer solutionwhere pKa = - log Ka[weak base][conjugate acid] pOH = pK + log b Known as Henderson-Hasselbalchequation for basic buffer solutionwhere pKb = - log Kb
Definition : Solution that has the ability to resist changes in pH when a small amount of strong acid or strong base is added to the solution.Type : ACIDIC BUFFER BASIC BUFFERPreparation : • Weak acid + its conjugate salt • Weak base + its conjugate saltExample : • CH3COOH CH3COO- + H+• Conjugate base : CH3COO-• CH3COONa CH3COO- + Na+• NH3 + H2O NH4+ + OH-• Conjugate acid : NH4+• NH4Cl NH4+ + ClReaction occurs when strong acid is added to buffer solution• When H+is added into acidic buffer • When H+is added into basic bufferReaction occurs when strong base is added to buffer solution• When OHis added into acidic buffer • When OHis added into basic bufferConclusion • H+ added/OH- added is removed by the conj. base/ weak acid. Thus, pH has no significant change.• H+ added/OH- added is removed by the weak base/conj. acid. Thus, pH has no significant change.weak acidH+added + conjugate base of the weak acidH+added + CH3COO- CH3COOHconj. base of the weak acidOHadded + weakacid + H2OOHadded + CH3COOH CH3COO- + H2O conjugate acid of the weak baseH+added +weak baseH+added + NH3 NH4+OHadded + conj. acid of the weak baseweak + H2ObaseOHadded + NH4+ NH3 + H2O
A student was asked to prepare a buffer solution of pH 4.6 using 50.00 mL of 0.5 M benzoic acid, C6H5COOH and sodium benzoate, C6H5COONa. (Ka C6H5COOH = 6.5 x 10-5) a) Explain the buffering effect of adding a small amount of NaOH and HCl, respectively into the buffer solution.Dissociation equation of C6H5COOH: C6H5COOH + H2O C6H5COO- + H3O+Dissociation equation of CH3COONa: C6H5COONa C6H5COO- + Na+(Weak acid)(Conjugate base of weak acid)32
NaOH Na+ + OHOHadded + C6H5COOH C6H5COO- + H2OOH- added is removed by the weak acid (C6H5COOH). The [C6H5COOH] decreases and [C6H5COO-] increases.Thus, pH has no significant change.HCl H+ + ClH+added + C6H5COO- C6H5COOHH+ added is removed by the conjugate base (C6H5COO-). The [C6H5COO-] decreases and [C6H5COOH] increases.Thus, pH has no significant change.Adding NaOH to buffer solutionAdding HCl to buffer solution
b) Calculate the mass of sodium benzoate required to prepare the buffer solution.(Ka C6H5COOH = 6.5 x 10-5)[C6H5COO-] = 1.294 M C6H5COONa C6H5COO- + Na+Thus, [C6H5COO-] = 1.294 M = [C6H5COONa] n C6H5COONa = 1.294 x 50 x 10-3= 0.065 mol = 9.36 g?? = − log?? + ???[??3???−][??3????]4.6 = − log(6.5 × 10−5) + ???[??3???−](0.5)= 0.065 mol × 144 gmol mass C -16H5COONa
a) 500 mL of 0.10 M hydrazinium chloride solution, N2H5Cl was added to a 500 mL solution of 0. 20 M hydrazine, N2H4. What is the pH of the mixture?(Kbof N2H4= 1.70 x 10–7, Kw= 1.0 x 10–14)N2H5Cl N2H5+ + ClFrom equation: 1 mol N2H5Cl 1 mol N2H5+ 0.05 mol N2H5Cl 0.05 mol N2H5+??? =??1000??2?5?? =0.10 × 5001000??2?4=0.20 × 5001000??? = ??? + ???[????????? ????][???? ????]??? = − log?? + ???[?2?5+][?2?4]= 0.05 mol= 0.10 mol= 0.05M= 0.10M= 6.70 pH = 7.30[?2?5+] =0.05(500 + 5001000 )[?2?4] =0.10(500 + 5001000 )??? = − log 1.0×10−7 + ???(0.05)(0.10)
• When H+is added into basic bufferH+added + N2H4 N2H5+H+ added is removed by the weak base (N2H4). Thus, pH has no significant change.• When OHis added into acidic bufferOHadded + N2H5+ N2H4 + H2OOH- added is removed by the conjugate acid (N2H5+). Thus, pH has no significant change.b) By writing reaction equations, explain what would happen to the pH of the solution when a small amount of a strong acid or a strong base is added to the solution mixture in (a). 36
a) Calculate the pH of 250 mL solution containing 0.20 M CH3COOH and 0.30 M CH3COONa. (Kaof CH3COOH = 1.8 x 10–5)= 4.92?? = − log?? + ???[??3???−][??3????]?? = − log 1.8×10−5 + ???(0.30)(0.20)
b) Calculate the pH of the resulting solution when 1.0 mL of 1.0 M HCl is added.= 0.075 molCH3COONa CH3COO- + Na+= 0.05 mol???3????? =0.30 × 2501000Initial mol:???3???? =0.20 × 2501000HCl H+ + Cl-= ??+When HCl in added:???? ????? =1.0 × 11000= ???3???−= 1 x 10-3 molH+added + CH3COO- CH3COOHH+from acid reacts with CH3COO-(conjugate base).
39n initialn additionn changen final[ ] finalCH3COO−(aq) + H+(aq) → CH3COOH (aq)0.075 0.0501 x 10-3- 1 x 10-3- 1 x 10-3 + 1 x 10-30.075 – 1 x 10-3= 0.0740 0.050 + 1 x 10-3= 0.0510.0740.251= ?. ??? 0 0.0510.251= ?. ???Using Henderson-Hasselbalch equation : pH = pKa + logCH3COO−CH3COOH= − log 1.8 × 10−5 + log0.2950.203= ?. ??
c) Calculate the pH of the resulting solution when 1.0 mL of 1.0 M NaOH is added.CH3COONa CH3COO- + Na+NaOH Na+ + OH-= 1 × 10-3 mol??? = 0.075 mol 3????? =0.30 × 2501000Initial mol:???3???? =0.30 × 2501000= ???3???−When NaOH in added:????? ????? =1.0 × 11000= 0.05 mol= ???−OHadded + CH3COOH CH3COO- + H2OOHfrom base reacts with CH3COOH (weak acid).
41Using Henderson-Hasselbalch equation := − log 1.8 × 10−5 + log0.3030.195= ?. ??n initialn additionn changen final[ ] finalCH3COOH (aq) + OH−(aq) → CH3COO −(aq) + H2O (l)- 1 x 10-3- 1 x 10-3 + 1 x 10-30.050 – 1 x 10-3= 0.0490 0.075 + 1 x 10-3= 0.076pH = pKa + logCH3COO−CH3COOH0.050 0.075 -1 x 10-3 ----0.0490.251= ?. ???0.0760.2510 = ?. ???
Guideline In Solving Problem Related To Buffer Solution• For original buffer solution apply Henderson-Hasselbalch equation.?? = ??? + ??? [????????? ????]???? ???? ??? = ??? + ??? [????????? ????]???? ????STEP 1:• If acid or base is added to buffer solutionTabulate the reaction and related data involve Assume all acid @ base added is completely used up during the reaction (acid @ base acts as limiting reactant)STEP 2: Determine the new pH of buffer solutionUse Henderson-Hasselbalch equation Use Henderson-Hasselbalch equation
7.2 Acid-base Titrations
• In an acid-base titration, titrant is slowly added from a burette to an analyte until the reaction is complete.TITRATIONTitrantAnalyte• An indicator is added to determine the end point~ An indicator is a weak organic acid/base that changes color when the pH changes.
• When the reaction is completed, we have reached the end pointof the titration.~ End point: the point at which an indicator changes colour.• When the no. of moles of the acid equals the no. of moles of the base, the titration has reached its equivalence point.~ Equivalence point: the point at which the stoichiometrically equivalent amounts of acid and base have reacted. TITRATION
A sample of 0.214 g of unknown monoprotic weak acid, HA was dissolved in 25.00 mL of water and titrated with 0.1 M NaOH. The acid required 27.40 mL of the base to reach the equivalent point. (iii) Explain quantitavely the pH of the solution at equivalent point.When reached equivalence point, the solution contains salt solution @ salt and water. Since the salt is from weak acid and strong base, the anion A- undergoes hydrolysis forming OH-@A− + H2O ⇌ HA + OH−Thus, the salt is a basic salt with pH more than 7. Past Year PSPM 2018/2019
• It is a plot of pH vs volume of titrant added during titration process.• Prior to the equivalence point, the analyte is in excess, so the pH is closest to the pH of the analyte. • The inflection point of the curve is the equivalence point of the titration.• The pH of the equivalence point depends on the pH of the salt solutioni. equivalence point of neutral salt, pH = 7 ii. equivalence point of acidic salt, pH < 7 iii. equivalence point of basic salt, pH > 7• Beyond the equivalence point, the titrant is in excess, so the pH approaches the pH of the titrant.TITRATION CURVE
Titration Curve: Titration Between Strong Acid – Strong Base(Titrant: Strong Base; Analyte: Strong Acid)48Before titrant (a strong base) is added, pH is determined by the strong acid alone.After base has been added, but before equivalence point, excess strong acid determines the pH.Neutral salt solution formed, neither cation nor anion undergoes hydrolysis, therefore pH = 7Beyond equivalence point, excess strong base determines the pH.
• If the titration is run so that the strong acid is in the burette/titrant and the base is in the flask/analyte, the titration curve will be the reflection of the one just shown.Strong baseStrong acid Strong baseStrong acid49
• Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region.• The initial pH is determined using the Ka of the weak acid.• The pH after base is added, but before equivalence point, is determined using the Henderson-Hasselbalch equation of an acidic buffer.• The pH at the equivalence point is determined using the Kb of the conjugate base of the weak acid.• The pH after equivalence point is dominated by the excess strong base (the basicity from the conjugate base anion is negligible).Titration Curve: Titration Between Weak Acid – Strong Base50