Acid Base Equilibria, Buffers, Titration

Acid Base Equilibria, Buffers, TitrationAP Chem

Weak Acids & Acid Ionization ConstantMajority of acids are weak. Consider a weak monoprotic acid, HA:

The equilibrium constant for the ionization would be:

or

Acid Ionization ConstantThe magnitude of Ka for an acid determines its strength

Determining pH from KaCalculate the pH of a 0.50 M HF solution at 25 C. The ionization of HF is given by:

Determining pH from Ka

Initial (M)0.500.000.00Change (M)-x+x+xEquilibrium (M)0.50 xxx

Determining pH from KaThis leads to a quadratic equation, so lets simplify:0.50 x 0.50The expression becomes:

Determining pH from KaAt equilibrium, [HF] = (0.50-0.019) = 0.48 M[H+] = 0.019 M[F-] = 0.019 M

This only works if x is less than 5% of 0.50Why? Ka are generally only accurate to 5%

Was the approximation good?Consider if initial concentration of HF is 0.050 M. Same process above, but x = 6.0 x 10-3 M. This would not be valid.

VALID!

NOT VALID!

Weak Bases & Base Ionization ConstantsSame treatment as for acids. Given ammonia in water:

Reminder: we solve using [OH-], not [H+]

Relationship of Ka to their Conjugate BaseKw = Ka Kb

This leads us to conclude the stronger the acid (larger Ka), the weaker its conjugate base (smaller Kb) Mole Buck Opportunity!Diprotic & Polyprotic AcidsA more involved process due to stepwise dissociation of hydrogen ionEach step is like a monoprotic acidBe sure to think about what is present at each step!Note the conjugate base is used as the acid in the next step

Diprotic Acid CalculationCalculate all species present at equilibrium in a 0.10 M solution of oxalic acid (H2C2O4)

Diprotic Acid Calculation

Initial (M)0.100.000.00Change (M)-x+x+xEquilibrium (M)0.10 xxxDiprotic Acid CalculationMaking the approximation 0.10 x 0.10, we get:

STOPApprox. good?

NOT VALID!Quadratic!Diprotic Acid Calculationx = 0.054 MAfter the first stage, we have:[H+] = 0.054 M[HC2O4-] = 0.054 M[H2C2O4] = (0.10 0.054) M = 0.046 MNext step: Treat conj. base as acid for 2nd step

Diprotic Acid CalculationSecond dissociation would be:

Initial (M)0.0540.0540.00Change (M)-y+y+yEquilibrium (M)0.054 y0.054 + yyDiprotic Acid CalculationApplying the approximation (for both) we obtain:

STOPApprox. good?

VALID!Diprotic Acid Calculation!Finally, at equilibrium:[H2C2O4] = 0.046 M[HC2O4-] = (0.054 6.1 x 10-5) = 0.054 M[H+] = (0.054 + 6.1 x 10-5) = 0.054 M[C2O42-] = 6.1 x 10-5 M

Conclusions on Polyprotic AcidsThe past example shows that for diprotic acids, Ka1 >> Ka2From this, we can assume the majority of H+ ions are produced in the first stage of ionizationSecondly, concentration of conj. base is numerically equal to Ka2Molecular Structure and Strength of AcidsStrength of hydrohalic acids (HX) depends on 2 factors:Bond Strength (called bond enthalpy)Polarity of BondStrength depends on ease of ionizationStronger the bond, more difficult to ionizeHigher the polarity, better chance of ionizingStrength of Binary AcidsHF HBrO3Strength of OxoacidsOxoacids have same central atom but different number of attached groupsIn this group, acid strength increases as oxidation number of central atom increasesi.e. the more oxygen, the merrier!

HClO4 > HClO3 > HClO2 > HClOAcid-Base Properties of SaltsSalt hydrolyis is the reaction of an anion or a cation of a salt (or both) with waterSalts can be neutral, basic, or acidic and follow certain trends

Acids mixed with bases forms salt + water!BEWARE!Salts that Produce Neutral SolnsAlkali metal ion or alkaline earth metal ion (except Be) do not undergo hydrolysisConjugate bases of strong acids (or bases) do not undergo hydrolysisi.e. Cl-, Br-, NO3-So NaNO3 forms a neutral solutionSalts that Produce Basic SolnsConjugate bases of a weak acid will react to form OH- ionsFor example, NaCH3COO forms Na+ and CH3COO- in solution. Acetate ion is the conjugate base of acetic acid, and undergoes hydrolysis:

Basic Salt Hydrolysis CalculationCalculate the pH of a 0.15 M solution of sodium acetate (CH3COONa)

Since the dissocation is 1:1 mole ratio, the concentration of the ions is the same

Basic Salt Hydrolysis Calc.Because acetate ion is the conj. base of a weak acid, it hydrolyzes as:

Initial (M)0.150.000.00Change (M)-x+x+xEquilibrium (M)0.15 xxxBasic Salt Hydrolysis Calc.Applying the approximation,

STOPApprox. good?

Also known as percent hydrolysis!Acidic Salt HydrolysisConjugate acid of a weak base will react to form H3O+ ionsSmall, highly charged cation will also produce acidic solutions when hydratedTypically transition metals (Al3+, Cr3+, Fe3+)

Common Ion Effect RevisitedRecall that the addition of a common ion causes an equilibrium to shiftEarlier we related this to the solubility of a saltThe idea is just an application of Le Chateliers PrinciplepH changes due to Common Ion EffectConsider adding sodium acetate (NaCH3COO) to a solution of acetic acid:

The addition of a common ion here (CH3COO-) will increase the pHBy consuming H+ ions

Henderson-HasselbalchConsider:

Rearranging Ka for [H+]we get:

Henderson-HasselbalchTake negative log of both sides:

or:

Finding pH with common ion presentCalculate the pH of a solution containing both 0.20 M CH3COOH and 0.30 M CH3COONa? The Ka of CH3COOH is 1.8x10-5.Sodium acetate fully dissociates in solution

Finding pH with common ion presentCan use I.C.E. table, or Henderson-Hasselbach

Effect of Common Ion on pHConsider calculating the pH for a 0.20 M acetic acid solutionpH = 2.72From our last example, its obvious the pH has increased due to the common ionBuffer SolutionsBuffers are a solution of (1) weak acid or weak base and (2) its saltBuffers resist changes in pH upon addition of acid or baseBuffer Capacity: refers to the amount of acid or base a buffer can neutralizeBuffer SolutionsConsider a solution of acetic acid and sodium acetateUpon addition of an acid, H+ is consumed:

Upon addition of a base, OH- is consumed:

Buffer Animationhttp://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swfBuffer Problem(a) Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0 M CH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume volume of soln does not change.Buffer ProblemTo calculate pH of buffer, I.C.E. or H.H.

With addition of HCl, we are adding 0.10 M H+, which will react completely with acetate ion

Buffer ProblemNow acetic acid will still dissociate and amount of H+ formed is the pH of soln

Initial (mol)1.00.101.0Change (mol)-0.10-0.10+0.10Equilibrium (mol)0.9001.10Buffer ProblemUsing H.H. (or I.C.E.)

Finding Buffers of Specific pHIf concentrations of both species are equal this means:

Using H.H., to find specific pH, search for pKa pH

Finding Buffers of Specific pHDescribe how you would prepare a phosphate buffer with a pH of about 7.40

Finding Buffers of Specific pHUsing the HPO42-/H3PO4- buffer:

This means a mole ratio of 1.5 moles disodium hydrogen phosphate : 1.0 mole Monosodium dihydrogen phosphate will result in a buffer solution with a 7.4 pH Finding Buffers of Specific pHTo obtain this solution, disodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) is in 1.5:1.0 ratioMeaning it is 1.5 M Na2HPO4 and 1.0 M NaH2PO4 are combined per liter of solutionAcid-Base TitrationsThree situations will be considered:Strong Acid/ Strong BaseWeak Acid/ Strong BaseStrong Acid/ Weak BaseTitrations of weak acid/base are complicated by hydrolysisStrong Acid-Strong Base TitrationReacting HCl and NaOH, the net ionic equation would be:

For our example, 0.100 M HCl is being titrated by 0.100 M NaOHCalculating pH changes depends on the stage of the titration

Strong Acid-Strong Base TitrationScenario #1: After addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.Total volume = 35.0 mL

Strong Acid-Strong Base TitrationScenario #1

Strong Acid-Strong Base TitrationScenario #2: After addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl (aka equivalence point)Because equivalent moles of acid/base

Strong Acid-Strong Base TitrationScenario #2Since no hydrolysis occurs (both are strong) equivalence point of all strong acid/base titrations has a pH = 7Keep in mind, equivalence point means [OH-]=[H+]Strong Acid-Strong Base TitrationScenario #3: After addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HClTotal volume = 60.0 mL

Strong Acid-Strong Base TitrationScenario #3

Weak Acid-Strong Base TitrationConsider the neutralization between acetic acid and sodium hydroxide:

The net ionic is:

Lets calculate the pH for this reaction at different stages

Weak Acid-Strong Base TitrationScenario #1: 25.0 mL of 0.100 M acetic acid is titrated with 10.0 mL of 0.100 M NaOHTotal volume = 35.0 mL

Weak Acid-Strong Base TitrationKey things to note:Work in moles, not molarity in I.C.F. tableIf volume is the same for all species, ratio of moles is equal to ratio of molar concentrationThis means for Ka, no need to convert back to molarityEquivalance point is not 7Hydrolysis of salt causes this shift

Weak Acid-Strong Base TitrationScenario #1Buffer system exists here (CH3COONa/CH3COOH)Calculate moles of each

Weak Acid-Strong Base Titration

Initial (mol)2.5x10-31.00x10-30Change (mol)-1.00x10-3-1.00x10-3+1.00x10-3Equilibrium (mol)1.50x10-301.00x10-3Buffer System Wooo!!Weak Acid-Strong Base Titration

Weak Acid-Strong Base TitrationScenario #2: Adding 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M acetic acidEquivalence point is not at 7!

Weak Acid-Strong Base Titration

Initial (mol)2.5x10-32.50x10-30Change (mol)-2.50x10-3-2.50x10-3+2.50x10-3Equilibrium (mol)002.50x10-3Weak Acid-Strong Base TitrationAcid/Base concentration being zero at the equivalence point, pH is determined by hydrolysis of salt

Weak Acid-Strong Base TitrationI.C.E. it! Equation becomes:

Weak Acid-Strong Base TitrationScenario #3: Adding 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M acetic acid

Weak Acid-Strong Base Titration

Initial (mol)2.5x10-33.50x10-30Change (mol)-2.50x10-3-2.50x10-3+2.50x10-3Equilibrium (mol)01.00x10-32.50x10-3Weak Acid-Strong Base TitrationAt this point, both OH- and hydrolysis of CH3COO- affects pHSince OH- is much stronger, we can ignore the impact of the hydrolysis

Strong Acid-Weak Base TitrationConsider the titration of HCl with NH3:

Or

Lets calculate the pH for this titration at different stages

Strong Acid-Weak Base TitrationKey main differences from Weak Acid/ Strong Base:Equivalence point is less then 7 due to salt hydrolysisCalculation of pH after equivalence point is based on remaining [H+] onlyEven though salt still exists, effect on pH minimalFirst part of titration is the same (buffer!)Strong Acid-Weak Base TitrationCalculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is titrated by a 0.100 M HCl solution

Strong Acid-Weak Base TitrationAt equivalence point, moles of acid equal moles of base

Strong Acid-Weak Base Titration

Initial (mol)2.5x10-32.50x10-30Change (mol)-2.50x10-3-2.50x10-3+2.50x10-3Equilibrium (mol)002.50x10-3Strong Acid-Weak Base Titration

Initial (mol)0.050000Change (mol)-x+x+xEquilibrium (mol)0.0500 - xxxStrong Acid-Weak Base TitrationApprox. good (0.01%)