C h a p t e rC h a p t e r C h a p t e rC h a p t e r 16 Applications of Aqueous Equilibria Chemistry 4th Edition McMurry/Fay Chemistry 4th Edition McMurry/Fay

Slide 2 Buffer Solutions01 A Buffer Solution : is a solution of a weak acid and a weak base (usually conjugate pair); both components must be present. A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Buffers are very important to biological systems!

Slide 3 Buffer Solutions02 Base is neutralized by the weak acid. Acid is neutralized by the weak base.

Slide 4 Buffer Solutions03 Buffer solutions must contain relatively high concentrations of weak acid and weak base components to provide a high “buffering capacity”. The acid and base components must not neutralize each other. The simplest buffer is prepared from equal concentrations of an acid and its conjugate base.

Slide 5 Buffer Solutions04 a) Calculate the pH of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COONa. b) What is the pH of the system after the addition of 0.10 mole of HCl to 1.0 L of buffer solution? a) pH of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COO – Na + is: 4.74 (see earlier slide)

Slide 6 Buffer Solutions04 b) What is the pH of the system after the addition of 0.10 mole of HCl to 1.0 L of buffer solution? CH 3 CO 2 H (aq) + H 2 O (aq) CH 3 CO 2 – (aq) + H 3 O + (aq) I add 0.10 mol HCl – 0.10 C -x -x +x E 1.1 – x x x K a = 1.8 x = = (0.9+x)(x) (0.9)x (1.1 – x) (1.1) x = ( 1.1 / 0.9 ) 1.8 x x = 2.2 x pH = 4.66 reacts w/

Slide 7 Water – No Buffer What is the pH after the addition of 0.10 mole of HCl to 1.0 L of pure water? HCl – strong acid, completely ionized. H + concentration will be 0.10 molar. pH will be –log(0.10) = 1.0 Adding acid to water: Δ pH = = 6.0 pH units Adding acid to buffer: Δ pH = = 0.09 pH units!! The power of buffers!

Slide 8 Titration: a procedure for determining the concentration of a solution using another solution of known concentration. Titrations involving strong acids or strong bases are straightforward, and give clear endpoints. Titration of a weak acid and a weak base may be difficult and give endpoints that are less well defined. Acid–Base Titrations01

Slide 9 The equivalence point of a titration is the point at which equimolar amounts of acid and base have reacted. (The acid and base have neutralized each other.) For a strong acid/strong base titration, the equivalence point should be at pH 7. Acid–Base Titrations02 H + (aq) + OH – (aq) → H 2 O (l) neutral

Slide 10 The equivalence point of a titration is the point at which equimolar amounts of acid and base have reacted. (The acid and base have neutralized each other.) Titration of a weak acid with a strong base gives an equivalence point with pH > 7. Acid–Base Titrations02 HA (aq) + OH – (aq) → H 2 O (l) + A – (aq) basic

Slide 11 Acid–Base Titration Examples03 Titration curve for strong acid–strong base: Note the very sharp endpoint (vertical line) seen with strong acid – strong base titrations. The pH is changing very rapidly in this region. add one drop of base: get a BIG change in pH

Slide 12 Titration of 0.10 M HCl03 Volume of Added NaOH 1. Starting pH (no NaOH added) mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH mL (total) of 0.10 M NaOH12.05 pH

Slide 13 Acid–Base Titrations04 pH at the equivalence point will always be >7 w/ weak acid/strong base Titration curve for weak acid–strong base: The endpoint (vertical line) is less sharp with weak acid – strong base titrations. strong acid weak acid strong acid equivalence point weak acid equivalence point

Slide 14 Acid–Base Titration Curves weak acid very weak acid With a very weak acid, the endpoint may be difficult to detect.

Slide 15 Acid–Base Titrations09 Strong Acid–Weak Base: The (conjugate) acid hydrolyzes to form weak base and H 3 O +. At equivalence point only the (conjugate) acid is present. pH at equivalence point will always be <7.

Slide 16 Aqueous Solubility Rules for Ionic Compounds A compound is probably soluble if it contains the cations: a. Li +, Na +, K +, Rb + (Group 1A on periodic table) b. NH 4 + A compound is probably soluble if it contains the anions: a. NO 3– (nitrate), CH 3 CO 2 – (acetate, also written C 2 H 3 O 2 – ) b. Cl –, Br –, I – (halides) except Ag +, Hg 2 2+, Pb 2+ halides c. SO 4 2– (sulfate) except Ca 2+, Sr 2+, Ba 2+, and Pb 2+ sulfates Other ionic compounds are probably insoluble. Solubility Equilibria01 Old way to analyze solubility. Answer is either “yes” or “no”.

Slide 17 Solubility Equilibria02 measure New method to measure solubility: Consider solution formation an equilibrium process: MCl 2 (s)  M 2+ (aq) + 2 Cl – (aq) Give equilibrium expression K c for this equation: K C = [M 2+ ][Cl – ] 2 K sp This type of equilibrium constant K c that measures solubility is called: K sp

Slide 18 Solubility Equilibria03 Solubility Product: is the product of the molar concentrations of the ions and provides a measure of a compound’s solubility. MX 2 (s)  M 2+ (aq) + 2 X – (aq) K sp = [M 2+ ][X – ] 2 “Solubility Product Constant”

Slide 19 Al(OH) x 10 –33 BaCO x 10 –9 BaF x 10 –6 BaSO x 10 –10 Bi 2 S x 10 –72 CdS8.0 x 10 –28 CaCO x 10 –9 CaF x 10 –11 Ca(OH) x 10 –6 Ca 3 (PO 4 ) x 10 –26 Cr(OH) x 10 –29 CoS4.0 x 10 –21 CuBr4.2 x 10 –8 CuI5.1 x 10 –12 Cu(OH) x 10 –20 CuS6.0 x 10 –37 Fe(OH) x 10 –14 Fe(OH) x 10 –36 FeS6.0 x 10 –19 PbCO x 10 –14 PbCl x 10 –4 PbCrO x 10 –14 PbF x 10 –8 PbI x 10 –8 PbS3.4 x 10 –28 MgCO x 10 –5 Mg(OH) x 10 –11 MnS3.0 x 10 –14 Hg 2 Cl x 10 –18 HgS 4.0 x 10 –54 NiS 1.4 x 10 –24 AgBr 7.7 x 10 –13 Ag 2 CO x 10 –12 AgCl 1.6 x 10 –10 Ag 2 SO x 10 –5 Ag 2 S 6.0 x 10 –51 SrCO x 10 –9 SrSO x 10 –7 SnS 1.0 x 10 –26 Zn(OH) x 10 –14 ZnS 3.0 x 10 –23 Solubility Equilibria - K sp Values04

Slide 20 Solubility Equilibria05 The solubility of calcium sulfate (CaSO 4 ) is found experimentally to be 0.67 g/L. Calculate the value of K sp for calcium sulfate. The solubility of lead chromate (PbCrO 4 ) is 4.5 x 10 –5 g/L. Calculate the solubility product of this compound. Calculate the solubility of copper(II) hydroxide, Cu(OH) 2, in g/L.

Slide 21 Equilibrium Constants - Review The reaction quotient (Q c ) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction. Q c K c System forms more reactants (left)

Slide 22 Equilibrium Constants - Q c Predicting the direction of a reaction. Q c < K c Qc > KcQc > KcQc > KcQc > Kc

Slide 23 Solubility Equilibria06 We use the reaction quotient ( Q c ) to determine if a chemical reaction is at equilibrium: compare Q c and K c K sp values are also a type of equilibrium constant, but are valid for saturated solutions only. We can use “ion product” (IP) to determine whether a precipitate will form: compare IP and K sp

Slide 24 Solubility Equilibria06 Ion Product (IP): solubility equivalent of reaction quotient ( Q c ). It is used to determine whether a precipitate will form. IP < K sp IP = K sp IP > K sp Unsaturated (more solute can dissolve) Saturated solution Supersaturated; precipitate forms.

Slide 25 Solubility Equilibria07 A BaCl 2 solution (200 mL of M) is added to 600 mL of M K 2 SO 4. Will precipitate form? (K sp for BaSO 4 is 1.1 x ) L x mol/L = moles Ba 2+ [Ba 2+ ] = mol  L = M L x mol/L = moles SO 4 2– [SO 4 2– ] = mol  L = M [ Ba 2+ ] [ SO 4 2– ]

Slide 26 Solubility Equilibria07 IP= [ Ba 2+ ] 1 x [ SO 4 2– ] 1 = (0.0010) x (0.0060) = 6.00 x A BaCl 2 solution (200 mL of M) is added to 600 mL of M K 2 SO 4. Will precipitate form? (K sp for BaSO 4 is 1.1 x ) IP> K sp, so ppt forms

Slide 27 Solubility Equilibria07 Exactly 200 mL of M BaCl 2 are added to exactly 600 mL of M K 2 SO 4. Will a precipitate form? If 2.00 mL of M NaOH are added to 1.00 L of M CaCl 2, will precipitation occur?

Slide 28 The solubility product (K sp ) is an equilibrium constant; precipitation will occur when the ion product (IP) exceeds the K sp for a compound. If AgNO 3 is added to saturated AgCl, the increase in [Ag + ] will cause AgCl to precipitate. IP = [Ag + ] 0 [Cl – ] 0 > K sp The Common-Ion Effect and Solubility

Slide 29 The Common-Ion Effect and Solubility MgF 2 becomes less soluble as F - conc. increses

Slide 30 The Common-Ion Effect and Solubility CaCO 3 is more soluble at low pH.

Slide 31 Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10 –3 M silver chloride solution. Calculate the solubility of AgBr (in g/L) in: (a) pure water (b) M NaBr The Common-Ion Effect and Solubility