Chapter 17 Acids, Bases and Buffers

Overview strong acid : strong base strong acid : weak base weak acid : strong base weak acid : weak base common ion effect buffers Henderson-Hasselbalch Equation titration curves

strong acid : strong base at the equivalence point (OH - = H + ) the pH = 7 strong acid : weak base at the equivalence point, pH < 7 weak acid : strong base at the equivalence point, pH > 7 weak acid : weak base at the equivalence point, pH depends on the relative value of the K a and the K b of the acid and base

example: weak acid : strong base solution H 2 O + HCN H 3 O + + CN - K a H 3 O + + OH - 2H 2 O 1/K w HCN + OH - H 2 O + CN - K a /K w K a /K w is large so rxn. will be, essentially, complete WASB

example: strong acid : weak base solution H 2 O + CN - HCN + OH - K b H 3 O + + OH - 2H 2 O 1/K w CN - + H 3 O + H 2 O + HCN K b /K w K b /K w is large so rxn. will be, essentially, complete SAWB

example: weak acid : weak base solution H 2 O + HCN H 3 O + + CN - K a H 2 O + NH 3 OH - + NH 4 + K b pH depends upon relative magnitude of K a and K b

Common Ion Effect: HCN + H 2 O H 3 O + + CN - Initial addition of CN - (as NaCN) shifts equilibrium, decreasing H 3 O + thereby increasing the pH

How does this affect the pH quantitatively? HCN + H 2 O H 3 O + + CN M M initail -x+ x + x change x x x equil. 4.9 x = (x)( x) x x = 9.8 x M = [H 3 O + ] = pH = 9.0 (vs. 9.9 x M = pH = 5.0 with no added NaCN)

Buffers: Resist change in pH on addition of small amounts of strong acid or base Composed of : A weak acid and the salt of its conj. base or A weak base and the salt of its conj. acid Most effective when pH is  1 of the pK a

HX H + + X - [H + ] = K a [HX] [X - ] adding OH - causes inc. in X - dec. in HX OH - + HX H 2 O + X - adding H + causes dec. in X - inc. in HX H + + X - HX as long as amt. of OH - or H + is small compared to HX & X -, ratio changes little

[H + ] = K a [HX] [X - ] -log [H + ] = - log K a -log [HX] [X - ] pH = pK a - log [HX] [X - ] pH = pK a + log [X - ] [HX] Henderson-Hasselbalch Equation

Henderson-Hasselbalch Equation: Calculates pH of buffer solutions Can be used only for buffer solutions Can be used only when the equilibrium approximation can be used pH = pK a + log ([X - ] / [HX]) where HX is the weak acid and X - is its conjugate base

Titration Curves: strong acid : strong base Calculation of pH after addition of aliquots of base HCl + NaOH  H 2 O + NaCl pH 50 mL mL 0.10 M 1.0 “ 10 mL 1.2 “ 20 mL 1.4 “ 49 mL 3.0 “ 50 mL 7.0 “ 55 mL11.7 “ 80 mL12.4 “ 100 mL12.5 Note: large increase in pH near equivalence pt.

Volume of Base, mL pH Equivalence Pt. 50 mL 0.10 M HCl titrated with 0.10 M NaOH

Titration Curves: Weak acid : strong base Calculation of pH after addition of aliquots of base HC 2 H 3 O 2 + NaOH  H 2 O + NaC 2 H 3 O 2 pH 50 mL mL 0.10 M 2.9 “ 10 mL 4.1 “ 20 mL 4.6 “ 25 mL half equivalence point [ pH = pK a ] 4.7 “ 49 mL 6.4 “ 50 mL equivalence point 8.7 “ 55 mL11.7 “ 80 mL12.4 “ 100 mL12.5 Note: large increase in pH near equivalence pt.

Volume of Strong Base, mL pH Equivalence Pt. 50 mL 0.10 M HC 2 H 3 O 2 titrated with 0.10 M NaOH half equivalence point = pK a 25

Strong Base Titrated with Strong Acid

Volume of Strong Acid, mL pH Equivalence Pt. 50 mL 0.10 M NH 3 titrated with 0.10 M HCl half equivalence point = pK b 25

Volume of Base, mL pH second equivalence pt. Diprotic acid, H 2 C 2 O 4, titrated with 0.10 M NaOH first equivalence point = pK a 100

Indicators: Indicators are generally weak acids HInd + H 2 O H 3 O + + Ind - K a Ind = [H 3 O + ][Ind - ] [HInd] [H 3 O + ] = [HInd] this ratio controls color K a [Ind - ]

Solubility Equilibria Solubility and Solubility Product Precipitation of Insoluble Salts K sp and Q Common Ion Effect and Solubility Simultaneous Equilibria Solubility and pH Solubility and Complex Ions Separations and Qualitative Analysis

Solubility Product Constant: The equilibrium constant expression for the solution of a solid AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] All rules for equilibrium constants and expressions apply Solubility Product

Examples: CaF 2(s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][F - ] 2 Ag 3 PO 4(s) 3Ag + (aq) + PO 4 3- (aq) K sp = [Ag + ] 3 [PO 4 3- ]

Ca 2+ F - Ca 2+ solid CaF 2(s) dissolved Ca 2+ (aq) & F - (aq) ions CaF 2(s) Ca 2+ (aq) + 2F - (aq) The quantity of CaF 2(s) dissolved is reflected by the quantity of Ca 2+ (aq) ions in solution The Molar Solubility of CaF 2(s) is equal to the [Ca 2+ ] at the eq. point = a saturated solution

Calculate the K sp from experimental data: Prob: [Ba 2+ ] = 7.5 x M in saturated BaF 2. Calculate K sp. BaF 2(s) Ba 2+ (aq) + 2F - (aq) x 2x 7.5 x (7.5 x ) K sp = (7.5 x )(15.0 x ) 2 = 1.6 x 10 -6

Estimating Solubility: Much like doing equilibrium problems: Determine the molar solubility of CaCO 3 if the K sp is 3.8 x at 25  C CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) initial 0 0 change +x+x equil. x x x 2 = 3.8 x x = 6.2 x M = [Ca 2+ ] = [CO 3 2- ]

Prob: Calculate the molar solubility of Mg(OH) 2 if the K sp is 1.5 x Mg(OH) 2(s) Mg 2+ (aq) + 2OH - (aq) initial0 0 change+x +2x equil. x 2x (x)(2x) 2 = 4x 3 = 1.5 x x = 1.6 x M [Mg 2+ ] = 1.6 x M [OH - ] = 3.2 x M K sp = [Mg 2+ ][OH - ] 2 (x) (2x)

Which is more soluble in water? AgClor Ag 2 CrO 4 AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.8 x Ag 2 CrO 4(s) 2Ag + (aq) + CrO 4 2- (aq) K sp = 9.0 x molar solubility = 1.3 x M molar solubility = 1.3 x M Direct comparisons of K sp ’s can only be used if the ion ratios are the same

Which is more soluble in water? AgClK sp = 1.8 x or AgCNK sp = 1.2 x Mg(OH) 2 K sp = 1.5 x or Ca(OH) 2 K sp = 7.9 x 10 -9

Q, Reaction Quotient Q = K sp system is at equilibrium Q > K sp system not at equilibrium solid forms (rxn. shifts ) Q < K sp system not at equilibrium solid dissolves (rxn. shifts ) Prob: PbI 2(s) (K sp = 8.7 x ) placed in solution where [Pb 2+ ] = 1.1 x M. Is the solution saturated? PbI 2(s) Pb 2+ (aq) + 2I - (aq) Q = 5.3 x < K sp No, more will dissolve

Concentrations required for precipitation: Prob: What is the minimum conc. of I - that can cause precipitation of PbI 2 from a M solution of Pb(NO 3 ) 2 ? K sp (PbI 2 ) = 8.7 x K sp = [Pb 2+ ][I - ] x = [I - ] [I - ] = 4.2 x M How much Pb 2+ remains when [I - ] = M [Pb 2+ ] = 8.7 x = 3.9 x M (0.0015) 2

K sp and Precipitation: Prob: You have mL of M AgNO 3. Does AgCl precipitate if you add 5.0 mL of HCl? AgCl (s) Ag + (aq) + Cl - (aq) AgNO 3(aq) Ag + (aq) + NO 3 - (aq) HCl (aq) + H 2 O H 3 O + (aq) + Cl - (aq) 1.0 x mol 1.25 x mol L L 9.5 x M 1.2 x M Q = (9.5 x )(1.2 x ) Q = 1.1 x > K sp will precipitate

Solubility and Common Ion Effect: CaF 2(s) Ca 2+ (aq) + 2F - (aq) Adding extra Ca 2+ or F - shifts equilibrium causing a decrease in solubility of CaF 2(s) shift toward solid CaF 2

Molar Solubility of CaF 2 (no added F - ): CaF 2(s) Ca 2+ (aq) + 2F - (aq) initial 0 0 change +x +2x equil. x 2x x = 2.1 x M = [Ca 2+ ] K sp = 3.9 x = (x)(2x) 2 molar solubility is 2.1 x moles CaF 2 / L

Molar Solubility of CaF 2 (with M NaF): CaF 2(s) Ca 2+ (aq) + 2F - (aq) initial change +x +2x equil. x x K sp = 3.9 x = (x)( x) 2 x = 3.9 x M = [Ca 2+ ] molar solubility is 3.9 x moles CaF 2 / L with added NaF which suppresses solubility

What happens if acid is added to CaF 2 ? CaF 2(s) Ca 2+ (aq) + 2F - (aq) H 3 O + + F - HF + H 2 O SAWB The second rxn. has the effect of removing F - from the first equilibrium, affecting the solubility of CaF 2 H + addition shifts rxn, increasing solubility complete reaction

Which of the following would be more soluble in acid solution? PbCl 2 CaCO 3 Mg(OH) 2 the stronger the conj. base the more soluble if the anion is a hydrolyzing conjugate base, the stronger base it is, the more soluble the salt is in acid solution

Cl - + H 2 O HCl + OH - K very small CO H 2 O HCO OH - K b 2.1 x OH - + H 2 O H 2 O + OH - K very large

For example: CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) K sp 3.8 x CO 3 2- (aq) + H 2 O HCO OH - K b 2.1 x OH - + H 3 O + 2H 2 O K w x CaCO 3(s) + H 3 O + Ca 2+ + HCO H 2 O K = 79.8

Which of the following would be more soluble in acid solution? Ca 10 (PO 4 ) 6 (OH) 2 Ca 10 (PO 4 ) 6 F 2 hydroxy apatite - tooth enamel fluoro apatite - fluoridated tooth enamel

Formation of Complex Ions: Complex ions are large, polyatomic ions Most complex ions have large K values Formation of complex ions can affect solubility of some salts K f is the formation constant Most transition metals form stable complex ions- the transition metal is a Lewis Acid

Ag + + 2NH 3(aq) Ag(NH 3 ) 2 + K f 1.7 x 10 7 x M0.20 M ( x) M K = 1.7 x 10 7 = [Ag(NH 3 ) 2 + ] = [Ag + ][NH 3 ] 2 (x)(0.20) 2 For example: Ex Calc. [Ag + ] present at eq. when conc. NH 3 added to M AgNO 3 to give eq. [NH 3 ] = 0.20 M. Neglect vol. change. x = 1.5 x M = [Ag + ]

Selective Precipitation: Prob: M Mg 2+ & M Cu 2+. Which will ppt first as OH - is added? Mg(OH) 2 Mg OH - K sp = 1.8 x Cu(OH) 2 Cu OH - K sp = 2.2 x [OH - ] = (K sp / [M 2+ ]) 1/2 Cu(OH) 2

What concentration of OH - is necessary? [OH - ] = (K sp / [Cu 2+ ]) 1/2 = (2.2 x / 0.020) 1/2 = 1.0 x M [OH - ] = (K sp / [Mg 2+ ]) 1/2 = (1.8 x / 0.050) 1/2 = 1.9 x M takes less OH -

AgCl (s) Ag + (aq) + Cl - (aq) K sp 1.8 x Ag + + 2NH 3(aq) Ag(NH 3 ) 2 + K f 1.7 x 10 7 AgCl (s) + 2NH 3(aq) Ag(NH 3 ) Cl - K = 3.1 x = [Ag(NH 3 ) 2 + ][Cl - ] [NH 3 ] 2 formation For example:

Prob: Does 100 mL of 4.0 M aqueous ammonia completely dissolve mol of AgCl suspended in 1.0 L of solution? AgCl (s) + 2NH 3(aq) Ag(NH 3 ) Cl - [NH 3 ] 2 = (1.0 x )(1.0 x ) = (0.032) 1/ 2 = 0.18 M (3.1 x ) we have 0.4 moles of NH 3 available which is plenty to provide the mol necessary to form the complex plus 0.16 mol to achieve an equilibrium concentration of 0.18 M [NH 3 ] K = 3.1 x = [Ag(NH 3 ) 2 + ] [Cl - ] [NH 3 ] moles0.020 moles0.010 moles