Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107
Chapter 15 Applications of Aqueous Equilibria (mainly acid/base & solubility) The Common Ion Effect: Problem: What is the common ion when acetic acid and sodium acetate are added together? Predict what will happen to ph when sodium acetate is added to a solution of acetic acid. Write the equation of ammonia added to water. What will happen to ph of NH 4 Cl is added? The common ion effect is important with polyprotic acids. The production of H + ions in the first step greatly inhibits the production of H + in succeeding steps.
Problem: Calculate the ph of a) solution containing 0.20 M HC 2 H 3 O 2 Ka = 1.8 x 10-5 b) solution containing 0.20 M HC 2 H 3 O 2 and 0.30 M NaC 2 H 3 O 2 c) Compare the percent ionization of acetic acid in a) and b)
Problem: Determine the ph of a solution containing 0.20 M NH 3 and 0.30 M NH 4 Cl. K b = 1.8 x 10-5 Henderson-Hasselbalch Equation (H.H. Equation) used to calculate the ph where there is a common ion. (acid/base pair) ph = pka + log [Conjugate Base] [Acid] where pka = -log Ka Solve the above problem using the H.H. Equation.
Buffered solution 1) solution that resists a change in ph when an acid or a base is added to it. and 2) Contains a weak acid plus the salt (conjugate base) of that acid. or Contains a weak base plus the salt (conjugate acid) of that base. Idea behind a buffer: an acid and its conjugate base will not neutralize each other, but they will neutralize all other acids and bases. Problem: Identify which of the following are buffer systems: A) H 3 PO 4 /KH 2 PO 4 B) HClO 4 /NaClO 4 C) NaHCO 3 /Na 2 CO 3 D) NH 3 /NH 4 Cl E) C 5 H 5 N/C 5 H 5 NHCl
Problem: A) A 1 Liter solution contains.75 M lactic acid (HC 3 H 5 O 3 ) w/ Ka = 1.4 x 10-4 and 0.25 M sodium lactate. Is this a buffered solution? B) Calculate the ph of this solution. C) Calculate the ph of this solution after 0.10 mole HCl (negligible volume change) is added to this buffer.
Student Problem: A) A 1 Liter solution contains.75 M lactic acid (HC 3 H 5 O 3 ) w/ Ka = 1.4 x 10-4 and 0.25 M sodium lactate. Is this a buffered solution? B) Calculate the ph of this solution. C) Calculate the ph of the solution of 0.10 mole HCl (negligible volume change) is added to this buffer. D) Calculate the ph if.10 mole NaOH (negligible volume change) is added to the original buffer solution. E) What would the ph be if.10 mole HCl were added to 1L pure water, rather than a buffer? F) What would the ph be if.10 mole NaOH were added to 1L pure water, rather than a buffer?
Buffer Capacity or Range A buffer is only effective within a certain range of ph ph range = pka +/- 1 Problem: What ph range would an acetic acid buffer be useful? Problem: What ph range would a methylamine buffer be useful? K b = 4.38 x 10-4 Outside this range, the ph is susceptible to large changes in ph because 1 component of the buffer is too low to neutralize an added acid or base.
Preparing a buffer with a specific ph: Problem: How would you prepare a phosphate buffer with a ph = 7.4 H 3 PO 4 H + + H 2 PO 4 K a1 = 7.5 x 10-3 pka = 2.12 H 2 PO 4 - H+ HPO 4-2 K a2 = 6.2 x 10-8 pka = 7.21 HPO 4-2 H + + PO 4-3 K a3 = 4.8 x 10-13 pka = 12.32 Problem: Could a good phosphate buffer of ph 5.0 be made?
Problem: Our blood is buffered to a ph = 7.40 using H 2 CO 3 /NaHCO 3 Determine the ratio of each component where Ka H 2 CO 3 = 4.2 x 10-7 Homework: 21, 23, 27, 29, 39, 43, 48, 49, 51, 53
Titrations and ph curves: Plot conc. of species during a titration. 3 scenarios: 1) Strong acid + strong base At approximately the equivalence point, the ph changes sharply, as the only source of H + + and OH - are from water. Consequently, 1 drop of excess acid or base (especially w/ high molarity) will alter [H + ] concentration drastically. [H + ] Vol. NaOH added Problem: Calculate the ph if 25 ml of.10 M HCl is mixed with a) 10. ml.10 M NaOH
Titrations and ph curves: Plot conc. of species during a titration. 3 scenarios: 1) Strong acid + strong base At approximately the equivalence point, the ph changes sharply, as the only source of H + + and OH - are from water. Consequently, 1 drop of excess acid or base (especially w/ high molarity) will alter [H + ] concentration drastically. [H + ] Vol. NaOH added Problem: Calculate the ph if 25 ml of.10 M HCl is mixed with a) 10. ml.10 M NaOH b) 25 ml.10 M NaOH c) 35 ml.10 M NaOH
2. Titration of a weak acid and a strong base - more complex because they involve buffers. Note: Buffer region Equivalence Point Region where ph changes drastically
Problem: Determine the ph if 25 ml.10 HC 2 H 3 O 2 are titrated with: a) 10 ml of.10 M NaOH
Problem: Determine the ph if 25 ml.10 HC 2 H 3 O 2 are titrated with: a) 10 ml of.10 M NaOH b) 25 ml of.10 M NaOH
Problem: Determine the ph if 25 ml.10 HC 2 H 3 O 2 are titrated with: a) 10 ml of.10 M NaOH b) 25 ml of.10 M NaOH c) 35 ml of.10 M NaOH
A couple final points: 1. Equivalence point is when all reactants are consumed, not by ph 2. Strong acid of same molarity as weak acid start at a lower point on titration curve. Why? 3. Weak acid, when titrated, changes ph rapidly at first, then slows down. Why?
3. Titration of a weak base and a strong acid also includes buffers Vol.10 M NH 3 added Note: Equivalence Point Buffer region Region where ph changes drastically What are the components of the buffe
Problem: Determine the ph when 25 ml 0.100 M HCl is titrated with a) 10. ml 0.10 M NH 3
Problem: Determine the ph when 25 ml 0.100 M HCl is titrated with a) 10. ml 0.10 M NH 3 b) 25 ml 0.10 M NH 3
Problem: Determine the ph when 25 ml 0.100 M HCl is titrated with a) 10. ml 0.10 M NH 3 b) 25 ml 0.10 M NH 3 c) 35 ml 0.10 M NH 3 a) ph = 1.37 b) ph = 5.28 c) ph = 8.86
Problem: Determine the Ka of a monoprotic acid..0020 mole are dissolved in 100 ml solution. 20.0 ml of.0500 M NaOH is added and the ph is measured to be 6.00. (This is the same problem as the lab you will do.)
Shape of phosphoric acid, a polyprotic weak acid titrated with a strong base. 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Equivalence point for titration Of HPO 4- Equivalence point for titration of H 2 PO 4- Equivalence point for titration of H 3 PO 4 Volume NaOH added Homework: 55, 57, 59, 67
To determine the equivalence point, 2 methods are available. 1) Use a ph meter, plot a titration curve, where ph changes sharply, this is the equivalence point. (inflection point) 2) Use an acid/base indicator, which is a chemical that changes color at a certain ph. The color change is called the End Point. If the endpoint occurs at the equivalence point, this is a good thing.
Which indicator would work for these titration curves?
Indicators themselves are weak acids or bases where: Phenolphthalein HIn (aq) H + (aq) Red Ka = [H + ][In - ] [HIn] + In - (aq) Yellow If the ratio of In - : HIn were 100 : 1 we would just see yellow. Our eyes can begin to see a color change at 10:1, so at this ratio, we would see a slight change toward orange. (Yellow) In - : HIn (Red) at 10:1 we see yellow at 1:1 we see orange at 1:10 we see red only Problem: If Ka of an indicator = 1 x 10-6, what ph range would we see a color change? Homework: 69, 71, 77
Solubility Equilibria Solubility measurement of the maximum number of moles of solute that will dissolve in a solvent, usually in moles/liter. For CaF 2(s) Ca +2 (aq) + 2F - (aq) K = [Ca +2 ] [F - ] 2 = K sp (the solubility product constant.) Equilibrium also means saturated. Problem: Write a dissociation equation and Ksp expression for Ag 2 CrO 4(s) 2 General types of solubility problems that are easily confused: 1. Given the solubility, determine K sp Problem: The solubility of silver sulfate is.015 mole/l. Determine K sp 2. Given K sp, Determine Solubility Problem: The K sp of copper (II) iodate = 1.4 x 10-7. Determine its solubility and concentration of [IO 3- ] when it is saturated.
Problem: Ksp of Ca 3 (PO 4 ) 2 = 1.2 x 10-26. Determine its solubility and the [Ca +2 ] when the solution is saturated.
Relative Solubilities: Which is more soluble? AgI Ksp = 1.5 x 10-16 CuI Ksp = 5.0 x 10-12 Ag 2 S Ksp = 1.6 x 10-49 Bi 2 S 3 Ksp = 1.1 x 10-73 Common ion effect: Problem: Determine the solubility of A) CaF 2 Ksp = 4.0 x 10-11 B) CaF 2 in.025 M NaF(aq)
ph and Solubility the solubility of many substances is drastically affected by ph Mg(OH) 2(s) Mg +2 (aq) + 2OH - (aq) Add HCl Add NaOH Would Ag 3 PO 4 dissolve better in an acidic solution? Ag 3 PO 4(s) 3Ag + (aq) + PO -3 4(aq) where PO -3 4 is a fairly strong base. Add HCl: All ionic substances w/ : OH -, S -2, CO 3-2, C 2 O 4-2, CrO 4-2 will dissolve better in acidic solutions, as all of these anions are bases, and will react with H + to increase the ionic salts solubility.
Predicting whether a precipitate will form using Q, the ion product. A solution contains 200. ml of 0.0040 M BaCl 2 K 2 SO 4. Will a ppt. occur? + 600 ml 0.0080 M
Problem: 150. ml 0.0100 M Mg(NO 3 ) 2 are added to 250 ml.100 M NaF. Ksp of MgF 2 = 6.4 x 10-9. Determine the [Mg +2 ] and [F - ] concentration at equilibrium.
Fractional or selective precipitation: Problem: A solution contains K + (aq) them? and Ba+2 (aq). How could you separate Problem: A solution contains.020 M Cl- and.020 M Br- Ksp AgCl = 1.6 x 10-10 AgBr = 7.7 x 10-13 a) Describe how to separate the Br - from the Cl - b) Calculate the [Ag + ] conc. needed to ppt. AgBr c) Calculate the [Ag + ] conc. needed to ppt. AgCl d) What is the concentration and remaining % of Br - when Cl - begins to ppt.? Homework: 81, 85, 91, 93, 97, 99
Qualitative Analysis A mixture of cations or anions are separated and identified. H 2 S 2H + + S -2 + HCl makes a very low [S -2 ] concentration. Only extremely insoluble sulfides precipitate. Add base to neutralize HCl, now more S -2 is available to make sulfide precipitates.
Flame Tests are done on group 5 to determine identity of alkali ions, which don t precipitate. Potassium flame is pink/purple Sodium flame is yellow/orange
Complex Ion Equilibrium Some metal ions, especially transition metals, form complex ions, which consist of a central metal cation bonded to 1 or more molecules or ions, called ligands. The number of ligands/metal cation is called the coordination number.
The hybrid orbitals required for tetrahedral, square planar, and linear complex ions.
The formation of this ligand is a lewis acid/base reaction. The ligand is always the base, providing the pair of electrons.
(From left to right) Metals containing the metal ions Co 2+ Mn 2 + Cr 3+ Fe 3 + Ni 2+
Other complexes to be aware of: Ni(NH 3 ) +2 6, CoCl -2 4, Cu(NH 3 ) +2 4, Ag(NH 3 ) + 2 Typical ligands are: H 2 O, NH 3, Cl -, CN -, S 2 O 3-, SCN - Complex ions form in steps: K f values are generally large, which means the reaction tends to go to completion. Write the K f expression for the silver ammonia complex reaction. (The coordination number of silver is 2)
What would happen if NH 3 were slowly added to a precipitate of AgCl? K f Ag(NH 3 ) 2 + (aq) = 1.7 x 10 7
What would happen if NH 3 were slowly added to CuSO 4(aq)? (pale blue) CuSO 4 (s) Cu +2 (aq) + SO 4-2 (aq) NH 3(aq) + H 2 O NH 4 + (aq) + OH - (aq) Cu +2 + 2OH - Cu(OH) 2(s) will occur as long as the [NH 3 ] is low. As more NH 3 is added: (dark blue solution) Cu(OH) 2(s) + 4NH 3 (aq) Cu(NH 3 ) 4 +2 (aq) + 2OH - (aq) and nearly all the precipitate re-dissolves to make the copper ammonia complex.
Problem: 0.20 mole CuSO 4 is added to 1.00 Liter of 1.20 M NH 3. Determine the Cu +2 (aq) concentration at equilibrium.
Skip this problem. Problem: Calculate [Ag + ], [AgS 2 O 3- ] and Ag(S 2 O 3 ) -3 2 in a solution prepared by mixing 150. ml 0.0010 M AgNO 3 + 200. ml 5.00 M Na 2 S 2 O 3 where: Ag + (aq) + S 2 O 3-2 (aq) Ag(S 2 O 3 ) - (aq) K 1 = 7.4 x 10 8 Ag(S 2 O 3 ) - (aq) + S 2 O 3 - (aq) Ag(S 2 O 3 ) 2-3 (aq) K 2 = 3.9 x 10 4 Homework: 14, 16, 104b, 105, 107,
Application of complex ions to Qualitative Analysis: Why is NH 3 added?
Caves: CaCO 3(s) + CO 2(aq) + H 2 O Ca +2 (aq) + 2HCO 3 - (aq)
Teeth Chemistry Tooth Enamel Ca 5 (PO 4 ) 3 OH (s) (calcium hydroxy apatite)