Unit 8: Acids And Bases

Unit 8: Acids and BasesChapter 15Applications of Aqueous Equilibria

March 4th, 2020Percent ionizationSalts that produce basic solutions and acidic solutions 14.8 (find your notes)Alka Seltzer- labette15.2 Buffer SolutionsBuffering capacity2017 FRQ Qu 3.

Percent ionization questionHNO2(aq) H (αq) NO2 (aq)Kα 4.0 10 4On the basis of the information above, what is the approximatepercent ionization of HNO2 in a 1.0 M HNO2(aq) solution?0.00040%0.020%0.040%2.0%

Percent ionization questionHNO2(aq) H (αq) NO2 (aq)1.0M00-x x x1.0 - xxxKα 4.0 10 4Kα 4.0 10 4 [H ] [NO2-.] x2[HNO2]1.0-xapprox x21.0

Percent ionization questionHNO2(aq) H (αq) NO2 (aq)Kα 4.0 10 4Kα 4.0 10 4 [H ] [NO2-.] x2[HNO2] 4.0 10 4 xx1.0-x1.0x [H ] 2.0 x 10-2 Mx 100 2.0 x 10-2 x 100 2%[HNO2]approx x21.05% approx. is valid

Percent ionization questionHNO2(aq) H (αq) NO2 (aq)Kα 4.0 10 4Kα 4.0 10 4 [H ] [NO2-.] x2[HNO2] 4.0 10 4 xx[HNO2]approx x21.0-x1.0x [H ] 2.0 x 10-2 Mx 100 2.0 x 10-2 x 100 2% , percent of acid dissociated1.0

Percent ionization question 4 is theNoticethisKα 4.0 10same step asKα 4.0 10 4 [H ] [NO2-.] x2checkingapprox 5%x2 ruletheis a valid 1.0[HNO2]1.0-xapproximation.HNO2(aq) H (αq) NO2 (aq) 4.0 10 4 xx[HNO2]x [H ] 2.0 x 10-2 Mx 100 2.0 x 10-2 x 100 2% , percent of acid dissociated1.0

Percent ionization questionHNO2(aq) H (αq) NO2 (aq)Kα 4.0 10 4On the basis of the information above, what is the approximatepercent ionization of HNO2 in a 1.0 M HNO2(aq) solution?0.00040%0.020%0.040%2.0%

14.8 Salts that produce basic solutions - find yournotesThe ionization constant for acetic acid is 1.8 x 10-5; that forhydrocyanic acid is 4 x 10-10. In 0.1 M solutions of sodium acetateand sodium cyanide, it is true thatA. [H ] equals [OH-] in each solutionB. [H ] exceeds [OH-] in each solutionC. [H ] of the sodium acetate solution is less than that of thesodium cyanide solutionD. [OH-] of the sodium acetate solution is less than that of thesodium cyanide solutionE. [OH-] for the two solutions is the same

14.8 Salts that produce basic solutions - find yournotesThe ionization constant for acetic acid is 1.8 x 10-5; that forhydrocyanic acid is 4 x 10-10. In 0.1 M solutions of sodium acetateand sodium cyanide, it is true that:Sodium salts are very soluble so equilibria to the right:C2H3O2Na(aq) Na (aq)NaCN(aq) C2H3O2-(aq) Na (aq) CN-(aq)Na ions have neither acid nor base properties

14.8 Salts that produce basic solutions - find yournotesThe ionization constant for acetic acid is 1.8 x 10-5; that for hydrocyanicacid is 4 x 10-10. In 0.1 M solutions of sodium acetate and sodiumcyanide, it is true that:Sodium salts are very soluble so equilibria to the right:C2H3O2Na(aq) Na (aq)NaCN(aq) C2H3O2-(aq) Na (aq) CN-(aq)BUT, the ionization constants (or Ka) values for acetic acid andhydrocyanic acid tell us that they are both weak acids (not very ionized)and that the hydrocyanic acid is a weaker acid than the acetic.

14.8 Salts that produce basic solutions - find yournotesThe conjugate bases of weak acids, will therefore be strong bases with asignificant affinity for a proton (available from water molecules).C2H3O2-(aq) H2O(l) C2H3O2H (aq) OH-(aq)CN-(aq) H2O(l) HCN OH-(aq)Since the HCN was the weakest acid, it will have the strongestconjugate base, and so the second equilibrium will have an equilibriumfavoring the RHS most; so highest [OH-] at equilibrium.

14.8 Salts that produce basic solutions - find yournotesThe ionization constant for acetic acid is 1.8 x 10-5; that forhydrocyanic acid is 4 x 10-10. In 0.1 M solutions of sodium acetateand sodium cyanide, it is true thatA. [H ] equals [OH-] in each solutionB. [H ] exceeds [OH-] in each solutionC. [H ] of the sodium acetate solution is less than that of thesodium cyanide solutionD. [OH-] of the sodium acetate solution is less than that of thesodium cyanide solutionE. [OH-] for the two solutions is the same

14.8 Salts that produce basic solutions - find yournotesThe ionization constant for acetic acid is 1.8 x 10-5; that forhydrocyanic acid is 4 x 10-10. In 0.1 M solutions of sodium acetateand sodium cyanide, it is true thatA. [H ] equals [OH-] in each solutionB. [H ] exceeds [OH-] in each solutionC. [H ] of the sodium acetate solution is less than that of thesodium cyanide solutionD. [OH-] of the sodium acetate solution is less than that of thesodium cyanide solutionE. [OH-] for the two solutions is the same

14.8 Salts that produce acidic solutions - find yournotesSamples of NaF(s) and NH4Cl(s) are dissolved in separate beakersthat each contain 100mL of water. One of the salts produces a slightlyacidic solution.Which one and why?“In general, salts in which the anion (-ve) is NOT a base and thecation ( ve) is the conjugate acid of a weak base produce acidicsolutions.”

14.8 Salts that produce acidic solutions - find yournotesSamples of NaF(s) and NH4Cl(s) are dissolved in separate beakersthat each contain 100mL of water. One of the salts produces a slightlyacidic solution.NH4 (aq) H2O(l) NH3(aq) H3O (aq)“In general, salts in which the anion (-ve) is NOT a base and thecation ( ve) is the conjugate acid of a weak base produce acidicsolutions.”

14.8 Salts that produce acidic solutions - find yournotesAl(H2O)63 (aq) H2O(l) Al(OH)(H2O)52 H3O (aq)Also look out for salts with a highly charged metal ion as above. The3 charge on the aluminum ion polarizes (distorts in one direction theelectrons in the bonds) the O-H bonds in the attached watermolecules, making the hydrogens in the water molecules more acidicthan those in free water molecules. Typically, the higher the charge onthe metal ion, the stronger the acidity of the hydrated ion.

Alka-seltzer labette 20 mins

15.2 Buffered SolutionsA buffered solution is one that resists a changein the pH when either hydroxide ions or protonsare added. (Alka-seltzer labette)Eg. Our blood - can absorb acids and basesproduced in biological reactions withoutchanging pH. Constant pH necessary becausecells can only survive in a narrow pH range.

Types of buffered solutionsWeak acid and its salt (eg. HF and NaF)Weak base and its salt (eg. NH3 and NH4Cl)

pH of a buffered solution (1)0.5 M acetic acid (HC2H3O2 , Ka 1.8 x 10-5)0.5 M sodium acetate (NaC2H3O2)Major species in the solution are:HC2H3O2(aq) Na (aq)Weak acidC2H3O2-(aq)neither acid nor baseBaseand H2Overy weakacid or base(quickly draw the structural formula for acetic acid - see wherethe acidic proton is - attached to an O)

HC2H3O2(aq) H (aq)Initial0.50Change-x C2H3O2-(aq)0Equilibrium 0.50 - x0.50 (from salt)xxx0.50 xKa 1.8 x 10-5 [H ][C2H3O2-][ HC2H3O2 ] (x)(0.5 x)0.5 - xx(0.5)0.50(assuming the x is small compared with [HA]0,, valid by 5% ruleThusx1.8 x 10-5

So [H ] x 1.8 x 10-5 MpH -log(1.8 x 10-5)pH 4.74

So [H ] x 1.8 x 10-5 MpH -log(1.8 x 10-5)pH 4.74Now calculate the change in pH that occurs when 0.010 mole ofsolid NaOH is added to 1.0L of the buffered solution above.NaOH will be completely dissociated, so the major species will be:HC2H3O2(aq) Na (aq) OH-(aq)C2H3O2-(aq) and H2O

So [H ] x 1.8 x 10-5 MpH -log(1.8 x 10-5)pH 4.74Now calculate the change in pH that occurs when 0.010 mole ofsolid NaOH is added to 1.0L of the buffered solution above.NaOH will be completely dissociated, so the major species will be:HC2H3O2(aq) Na (aq) OH-(aq)C2H3O2-(aq) and H2OSource of protonsstrong base

HC2H3O2(aq) OH-(aq) H2O(l) C2H3O2-(aq)The hydroxide ion is such a strong base (proton acceptor) thatthe reaction will proceed essentially to completion - or until allthe hydroxide ions are consumed.So assume this reaction goes to completion and deal with itfirst and then carry out the equilibrium calculations.

Stoichiometry problemHC2H3O2(aq) OH-(aq) H2O(l)1.0 L x 0.50mol/L 0.5mol C2H3O2-(aq)

Stoichiometry problemHC2H3O2(aq) OH-(aq) H2O(l)1.0 L x 0.50M 0.5mol0.01 mol C2H3O2-(aq)1.0 L x 0.50M 0.5mol

Stoichiometry problemHC2H3O2(aq) OH-(aq) H2O(l)1.0 L x 0.50M 0.5mol C2H3O2-(aq)1.0 L x 0.50M 0.5mol0.01 molAfter reaction0.5 mol - 0.01 mol0.49mol(OH- all gone)0.5 mol 0.01mol0.51 mol

Now treat the Equilibrium part of the problemHC2H3O2(aq) OH-(aq) H2O(l) C2H3O2-(aq)After this reaction is completeThe major species in solution areHC2H3O2Na C2H3O2-and H2ODominant equilibrium involves dissociation of acetic acid

HC2H3O2(aq) H (aq)Initial0.49Change-x0Equilibrium 0.49 - x C2H3O2-(aq)0.51xxx0.51 xKa 1.8 x 10-5 [H ][C2H3O2-][ HC2H3O2 ] (x)(0.51 x)0.49 - xx(0.51)0.49(assuming the x is small compared with [HA]0,, valid by 5% ruleThusx1.7 x 10-5

So [H ] x 1.7 x 10-5 MpH -log(1.7 x 10-5)pH 4.76The change in pH produced by the addition of 0.01 mole of OH- tothis buffered solution is then4.76 - 4.74 0.02The pH has increased by only 0.02 pH units.Now compare that to what happens when 0.01 mole of solidNaOH is added to 1.0L of water to give 0.01M NaOH.

[OH-] 0.01 M[H ] Kw 1.0 x 10-14 1.0 x 10-12[OH-]1.0 x 10-2pH 12.00Change in pH is from pure water to this12.00 - 7.00 5.00pH has gone up by 5 pH units.Notice how well the buffer resists a change inpH compared to water.

Ka [H ][C2H3O2-][ HC2H3O2 ] [H ][A-][ HA ]We often know the concentration of the acid and itsconjugate base so[H ] Ka [ HA ][A-]will be helpful

Then taking negative logs we can generate another useful form- log[H ] - log Ka - log [ HA ][A-]pH pKa log [A-][HA]Believe it or not, this is called the Henderson-HasselbachEquation (Topic 8.9)

Then-log[H ] -log Ka - log [ HA ][A-]pH pKa log [A-][HA]If we know theratio of theconcentrations ofthe Acid and itsconjugate base,then we can findthe pH of thebuffered solution.

Then-log[H ] -log Ka - log [ HA ][A-]pH pKa log [A-][HA]Or stated another way,the pH of the bufferedsolution will depend onthe [A-] to [HA] ratio.When this ratio is leastaffected by addedprotons or hydroxideions, the solution is mostresistant to change inpH.

Then-log[H ] -log Ka - log [ HA ][A-]pH pKa log [A-][HA]For the mosteffective bufferingwe want to avoidlarge changes inthe [A-] to [HA]ratio. This will bewhen theirconcentrationsare equal.

Then-log[H ] -log Ka - log [ HA ][A-]For most effective buffering:pH pKa log (1)pH pKaFor the mosteffective bufferingwe want to avoidlarge changes inthe [A-] to [HA]ratio. This will bewhen theirconcentrationsare equal.

Then-log[H ] -log Ka - log [ HA ]-[A ]For most effective buffering:pH pKa log (1)pH pKaSo the pKa of theweak acid to beused in the buffershould be asclose as possibleto the desired pH.

Note that to usepH pKa log [A-][HA]We assume that the equilibrium concentrations of A- and HAare equal to the initial concentrations (the amount of acid thatdissociates is small compared to the initial concentration of theacid). Since the initial concentrations of HA and A- arerelatively large in a buffered solution, this is generallyacceptable.

15.3 Buffering Capacityrepresents amount of protons or hydroxide ions the buffer canabsorb without a significant change in pH.pH of a buffered solution is determined by the ratio [A-]/[HA].The capacity of a buffered solution is determined by themagnitudes of [HA] and [A-] - increasing the concentration ofthe buffer components (while keeping the ratio of theseconcentrations constant) keeps the pH of the buffer the samebut increases the capacity of the buffer to neutralize addedacid or base.

When a buffer has more conjugate acid thanbase, it has greater buffer capacity for addition ofbase than acid.When a buffer has more conjugate base thanacid, it has greater buffer capacity for addition ofacid than base.

AP Classroom 8.4 Quiz Acids, bases and buffers

FRQ 2017 Qu. 3.N2(g) O2(g)2 NO(g)At high temperatures N2(g) and O2(g) can react toproduce nitrogen monoxide, NO(g), asrepresented by the equation above.a) Write the expression for the equilibriumconstant, Kp, for the forward reaction.

FRQ 2017 Qu. 3.N2(g) O2(g)2 NO(g)a) Write the expression for the equilibriumconstant, Kp, for the forward reaction.Kp (PNO)2(PN )(PO )22