Chapter 17 HW problems: 3, 5, 14, 15, 16, 23, 24, 27a, 28a, 31, 37, 43, 45, 51, 57 Aqueous Equilibria
The presence of a common ion suppresses the ionization of a weak acid or a weak base. CH3COONa (s) Na+(aq) + CH3COO-(aq) common ion CH3COOH (aq) H+(aq) + CH3COO-(aq) The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
Consider an equal molar mixture of CH3COOH and CH3COONa • A buffer solution is a solution of: • A weak acid or a weak base and • The salt of the weak acid or weak base • Both must be present! (and should NOT neutralize EACH OTHER!!!!) A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. CH3COOH (aq) H+(aq) + CH3COO-(aq) Adding more acid creates a shift left IF enough acetate ions are present
Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na2CO3/NaHCO3 (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HCl is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is it conjugate acid buffer solution
HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Ka for HCOOH = 1.8 x 10 -4 x = 1.038 X 10 -4 [H+] [HCOO-]Ka = [HCOOH] pH = 3.98
OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA. NaA (s) Na+(aq) + A-(aq) HA (aq) H+(aq) + A-(aq) Ka [HA] [H+][A-] Ka = [H+] = [HA] [A-] -log [H+] = -log Ka - log [HA] [A-] [A-] -log [H+] = -log Ka + log [HA] [HA] [A-] pH = pKa + log [conjugate base] pH = pKa + log [acid] pKa = -log Ka Henderson-Hasselbach equation
HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) [HCOO-] pH = pKa + log [HCOOH] [0.52] pH = 3.74 + log [0.30] What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Common ion effect 0.30 – x 0.30 = 3.98 0.52 + x 0.52 HCOOH pKa = 3.77
HCl H+ + Cl- HCl + CH3COO- CH3COOH + Cl-
NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? [NH4+] [OH-] [NH3] Kb = = 1.8 X 10-5 0.36 0 Initial 0.30 + x + x Change - x End 0.30 - x 0.36 + x x (.36 + x)(x) (.30 – x) 1.8 X 10-5 = 0.36x 0.30 1.8 X 10-5 x = 1.5 X 10-5 pOH = 4.82 pH= 9.18
Equilibrium Calculation NH4+(aq) H+(aq) + NH3(aq) NH4+(aq) + OH-(aq) H2O (l) + NH3(aq) What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? final volume = 80.0 mL + 20.0 mL = 100 mL NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M 0.01 0.24 start (M) 0.29 end (M) 0.28 0.0 0.25 [H+] [NH3] [NH4+] [H+] 0.25 0.28 = 5.6 X 10-10 Ka= = 5.6 X 10-10 [H+] = 6.27 X 10 -10 pH = 9.20
NH4+(aq) H+(aq) + NH3(aq) pH = 9.25 + log [0.30] [0.25] [NH3] pH = pKa + log [NH4+] [0.36] [0.28] NH4+(aq) + OH-(aq) H2O (l) + NH3(aq) pH = 9.25 + log Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? pKa= 9.25 = 9.17 final volume = 80.0 mL + 20.0 mL = 100 mL 0.01 0.24 start (M) 0.29 end (M) 0.28 0.0 0.25 = 9.20
Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at the endpoint (hopefully close to the equivalence point) Slowly add base to unknown acid UNTIL The indicator changes color (pink)
Titration Curve A titration curve is a plot of pH vs. the amount of titrant added. Typically the titrant is a strong (completely) dissociated acid or base. Such curves are useful for determining endpoints and dissociation constants of weak acids or bases.
Features of the Strong Acid-Strong Base Titration Curve • The pH starts out low, reflecting the high [H3O+] of the strong acid and increases gradually as acid is neutralized by the added base. • Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point. • Beyond this steep portion, the pH increases slowly as more base is added.
Sample Calculation: Strong Acid-Strong Base Titration Curve Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH. Region 1. Before the equivalence point, after adding 20.0 mL of 0.100 M NaOH. (Half way to the equivalence point.) Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 mol H3O+ - Moles of OH- added = 0.0200 L x 0.100 M =0.00200 mol OH- pH=1.477
Sample Calculation: Strong Acid-Strong Base Titration Curve (Cont. I) Region 2. At the equivalence point, after adding 40.0 mL of 0.100 M NaOH. Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 mol H3O+ - Moles of OH- added = 0.0400 L x 0.100 M =0.00400 mol OH- An equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt (NaCl). The pH=7.00 because the cation of the strong base and the anion of the strong acid have no effect on pH.
Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. II) Region 3. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate excess OH-) Total moles of OH- = 0.0500 L x 0.100 M = 0.00500 mol OH- -Moles of H3O+ consumed = 0.0400 L x 0.100 M =0.00400 mol pOH=1.954 pH=12.046
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH-(aq) CH3COO-(aq) + H2O (l) CH3COO-(aq) + H2O (l) OH-(aq) + CH3COOH (aq) Weak Acid-Strong Base Titrations At equivalence point (pH > 7):
The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve • The initial pH is higher. • A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. • The pH at the equivalence point is greater than 7.00. • The steep rise interval is less pronounced.
HCl (aq) + NH3(aq) NH4Cl (aq) H+(aq) + NH3(aq) NH4Cl (aq) NH4+(aq) + H2O (l) NH3(aq) + H+(aq) Strong Acid-Weak Base Titrations At equivalence point (pH < 7):
The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve • The initial pH is above 7.00. • A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point. • The pH at the equivalence point is less than 7.00. • Thereafter, the pH decreases slowly as excess strong acid is added.
Features of the Titration of a Polyprotic Acid with a Strong Base • The loss of each mole of H+ shows up as separate equivalence point (but only if the two pKas are separated by more than 3 pK units). • The pH at the midpoint of the buffer region is equal to the pKa of that acid species. • The same volume of added base is required to remove each mole of H+.
Which indicator(s) would you use for a titration of HNO2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein
Finding the Equivalence Point(calculation method) • Strong Acid vs. Strong Base • 100 % ionized! pH = 7 No equilibrium! • Weak Acid vs. Strong Base • Acid is neutralized; Need Kb for conjugate base equilibrium • Strong Acid vs. Weak Base • Base is neutralized; Need Ka for conjugate acid equilibrium • Weak Acid vs. Weak Base • Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7
Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO2(aq) + OH-(aq) NO2-(aq) + H2O (l) Initial (M) 0.01 Change (M) [NO2-] = = 0.05 M Equilibrium Calculation NO2-(aq) + H2O (l) OH-(aq) + HNO2(aq) 0.200 Equilibrium (M) x2 [OH-][HNO2] = Kb = 0.05-x [NO2-] start (moles) 0.01 0.01 end (moles) 0.0 0.0 0.01 Final volume = 200 mL 0.05 0.00 0.00 -x +x +x 0.05 - x x x pOH = 5.98 = 2.2 x 10-11 pH = 14 – pOH = 8.02 x 1.05 x 10-6 = [OH-]
Solubility Equilibria • Looking at dissolution or precipitation reactions of ionic compounds • Ksp(solubility product) is the equilibrium constant for the equilibrium reaction between a solid and its saturated ions • Indicates how soluble the solid is when in water • Solids, liquids and solvents do not appear in the equilibrium equation (just like before)
Solubility and Ksp • Solubility is how much actually dissolves while Ksp is the equilibrium constant. • pH, temperature, and the presence of other ions in solution affect solubility • Only temperature affects Ksp • Can use Ksp to calculate solubility but need to be careful…not always correct!
Factors that Affect Solubility(4) • Common Ion Effect • presence of other ions reduces solubility • presence of 2nd solute that gives a common ion reduces solubility • pH • dissolving increases when pH is made more acidic (more of the basic ion X- will be made to react with extra H+ ions) • the more basic the anion, the more its solubility is affected by pH • Formation of a Complex Ion • Amphoterism
2- Co2+(aq) + 4Cl-(aq) CoCl4(aq) 2+ Co(H2O)6 2- CoCl4 Complex Ion Equilibria and Solubility Where they “smush “ it all together A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. • Metal ions can act as Lewis acids • Can react with water • Can react with other Lewis bases (must be able interact more than water can) • Keq becomes Kf(formation constant) • Kf’s size shows stability of complex ion in solution
Complex Ion Formation • These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). • As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)42+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2+. • Memorize the common ligands.
Names • Names: ligand first, then cation Examples: • tetraamminecopper(II) ion: Cu(NH3)42+ • diamminesilver(I) ion: Ag(NH3)2+. • tetrahydroxyzinc(II) ion: Zn(OH)4 2- • The charge is the sum of the parts (2+) + 4(-1)= -2.
When Complexes Form • Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H2O)63+ • Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. • The odd complex ion, FeSCN2+, shows up once in a while • Acid-base reactions may change NH3 into NH4+ (or vice versa) which will alter its ability to act as a ligand. • Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)42+, forms.
Factors that Affect Solubility (cont.) • Amphoterism • Some insoluble, metal hydroxides will dissolve in strong acids or strong bases • will dissolve in acid because of basic anion • Will dissolve in base because a complex ion forms • Solubility varies with the metal bonded
Precipitation & Separation of Ions Q > Ksp get precipitation until Q = Ksp Q = Ksp at equilibrium Q < Ksp solid dissolves until Q = Ksp • Selective Precipitation of ions • Separate ions in an aqueous solution using a reagent that forms a precipitate 1 or few ions in solution • Ex. Ag+ + Cu2+ + HCl AgCl + Cu2+ + H+