Chapter 16. The Danger of Antifreeze. Buffers. Aqueous Equilibrium

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1 hapter 16 Aqueous Equilibrium The Danger of Antifreeze Each year, thousands of pets and wildlife die from consuming antifreeze Most brands of antifreeze contain ethylene glycol sweet taste and initial effect is drunkenness Metabolized in the liver to glycolic acid O 2 OO If present in high enough concentration in the bloodstream, it overwhelms the buffering ability of O 3, causing the blood p to drop When the blood p is low, it ability to carry O 2 is compromised acidosis The treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol Buffers Buffers are solutions that resist changes in p when an acid or base is added They act by neutralizing the added acid or base But just like everything else, there is a limit to what they can do, eventually the p changes Many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion 1

2 Making an Acid Buffer ow Acid Buffers Work A (aq) + 2 O (l) A (aq) + 3 O + (aq) Buffers work by applying Le hâtelier s Principle to weak acid equilibrium Buffer solutions contain significant amounts of the weak acid molecules, A these molecules react with added base to neutralize it you can also think of the 3 O + combining with the O to make 2 O; the 3 O + is then replaced by the shifting equilibrium The buffer solutions also contain significant amounts of the conjugate base anion, A - these ions combine with added acid to make more A and keep the 3 O + constant ommon Ion Effect A (aq) + 2 O (l) A (aq) + 3 O + (aq) Adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left This causes the p to be higher than the p of the acid solution lowering the 3 O + ion concentration 2

3 ommon Ion Effect Practice What is the p of a buffer that is M 2 3 O 2 and M Na 2 3 O 2? Practice What is the p of a buffer that is 0.14 M F (pk a = 3.15) and M KF? 3

4 enderson-asselbalch Equation alculating the p of a buffer solution can be simplified by using an equation derived from the K a expression called the enderson- asselbalch Equation The equation calculates the p of a buffer from the K a and initial concentrations of the weak acid and salt of the conjugate base as long as the x is small approximation is valid enderson-asselbalch Equation p = p K a & - # $ [A ] + log! %[A]" Example What is the p of a buffer that is M 7 5 O 2 and M Na 7 5 O 2? 4

5 Example What is the p of a buffer that is 0.14 M F (pk a = 3.15) and M KF? Using the enderson- asselbalch Equation The enderson-asselbalch equation is generally good enough when the x is small approximation is applicable Generally, the x is small approximation will work when both of the following are true: a)the initial concentrations of acid and salt are not very dilute b)the K a is fairly small For most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of K a Example What is the p of a buffer that has mol 2 3 O 2 and mol Na 2 3 O 2 in 1.00 L that has mol NaO added to it 5

6 Basic Buffers B: (aq) + 2 O (l) :B + (aq) + O (aq) buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, :B + l Example What is the p of a buffer that is 0.50 M N 3 (pk b = 4.75) and 0.20 M N 4 l? Buffering Effectiveness A good buffer should be able to neutralize moderate amounts of added acid or base there is a limit to how much can be added before the p changes significantly The buffering capacity is the amount of acid or base a buffer can neutralize The buffering range is the p range the buffer can be effective The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base (2) the absolute concentrations of acid and base 6

7 Effect of Relative Amounts of Acid and onjugate Base A buffer will be most effective when the [base]:[acid] = 1 equal concentrations of acid and base Effective when 0.1 < [base]:[acid] < 10 A buffer will be most effective when the [acid] and the [base] are large Buffering Range A buffer will be effective when 0.1 < [base]:[acid] < 10 Substituting into the enderson-asselbalch we can calculate the maximum and minimum p at which the buffer will be effective Lowest p p = pk + log 0.10 p = pk a a! 1 ( ) ighest p p = pk + log 10 p = pk a a + 1 ( ) therefore, the effective p range of a buffer is pk a ± 1 when choosing an acid to make a buffer, choose one whose is pk a is closest to the p of the buffer Example Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with p 4.25? hlorous Acid, lo 2 pk a = 1.95 Nitrous Acid, NO 2 pk a = 3.34 Formic Acid, O 2 pk a = 3.74 ypochlorous Acid, lo pk a =

8 Example What ratio of NaO 2 : O 2 would be required to make a buffer with p 4.25? Buffering apacity Buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness The buffering capacity increases with increasing absolute concentration of the buffer components As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves Buffers that need to work mainly with added acid generally have [base] > [acid] Buffers that need to work mainly with added base generally have [acid] > [base] Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete when the reaction is complete we have reached the endpoint of the titration An indicator may be added to determine the endpoint an indicator is a chemical that changes color when the p changes When the moles of 3 O + = moles of O, the titration has reached its equivalence point 8

9 Titration Titration Titration urve A plot of p vs. amount of added titrant The inflection point of the curve is the equivalence point of the titration Prior to the equivalence point, the known solution in the flask is in excess, so the p is closest to its p The p of the equivalence point depends on the p of the salt solution equivalence point of neutral salt, p = 7 equivalence point of acidic salt, p < 7 equivalence point of basic salt, p > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the p approaches its p 9

10 Titration urve: Strong Base Added to Strong Acid Example Show the titration of 25 ml of M l with M NaO Titrating Weak Acid with a Strong Base: Overview 1. Before any titrant (p of weak acid) 2. After a small amount of titrant (~5mL) 3. At alf titration point (p = pka) 4. At the equivalence point (mol acid = mol base) 5. A drop beyond the endpoint (~0.5mL) 10

11 Titrating Weak Acid with a Strong Base 1. The initial p is that of the weak acid solution calculate like a weak acid equilibrium problem 2. Before the equivalence point, the solution becomes a buffer calculate mol A init and mol A init using reaction stoichiometry calculate p with enderson-asselbalch using mol A init and mol A init 3. alf-neutralization p = pk a Titrating Weak Acid with a Strong Base 4. At the equivalence point, the mole A = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established mol A = original mole A calculate the volume of added base like Ex 4.8 [A ] init = mol A /total liters calculate like a weak base equilibrium problem 5. Beyond equivalence point, the O is in excess [O ] = mol MO xs/total liters [ 3 O + ][O ]=1 x Example A 40.0 ml sample of M NO 2 is titrated with M KO. alculate all 5 points. 11

12 Titration urve of a Weak Base with a Strong Acid Titration of a Polyprotic Acid titration of 25.0 ml of M 2 SO 3 with M NaO Monitoring p During a Titration The general method for monitoring the p during the course of a titration is to measure the conductivity of the solution due to the [ 3 O + ] using a probe that specifically measures just 3 O + The endpoint of the titration is reached at the equivalence point in the titration at the inflection point of the titration curve If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator 12

13 Monitoring p During a Titration Phenolphthalein Methyl Red ( 3 ) 2 N N N N 3 O+ O- NaOO ( 3 ) 2 N N N N NaOO 13

14 Monitoring a Titration with an Indicator For most titrations, the titration curve shows a very large change in p for very small additions of base near the equivalence point An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in p pk a of Ind p at equivalence point Indicators Solubility Equilibria All ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble The concepts of equilibrium can be applied to salts dissolving, and then one can use the equilibrium constant for the process to measure relative solubilities in water 14

15 Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s) nm m+ (aq) + mx n (aq) The solubility product would be K sp = [M m+ ] n [X n ] m For example, the dissociation reaction for Pbl 2 is Pbl 2 (s) Pb 2+ (aq) + 2 l (aq) Its equilibrium constant is K sp = [Pb 2+ ][l ] 2 Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature The molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution For the general reaction M n X m (s) nm m+ (aq) + mx n (aq) ( n m) Ksp molar solubility = + n m n m 15

16 Example alculate the molar solubility of Pbl 2 in pure water at 25 Example Determine the K sp of PbBr 2 if its molar solubility in water at 25 is 1.05 x 10-2 M K sp and Relative Solubility Molar solubility is related to K sp But you cannot always compare solubilities of compounds by comparing their K sp s In order to compare K sp s, the compounds must have the same dissociation stoichiometry 16

17 The Effect of ommon Ion on Solubility Addition of a soluble salt that contains one of the ions of the insoluble salt, decreases the solubility of the insoluble salt For example, addition of Nal to the solubility equilibrium of solid Pbl 2 decreases the solubility of Pbl 2 Pbl 2 (s) Pb 2+ (aq) + 2 l (aq) addition of l shifts the equilibrium to the left Example alculate the molar solubility of af 2 in M NaF at 25 The Effect of p on Solubility For insoluble ionic hydroxides, the higher the p, the lower the solubility of the ionic hydroxide and the lower the p, the higher the solubility higher p = increased [O ] M(O) n (s) M n+ (aq) + no (aq) For insoluble ionic compounds that contain anions of weak acids, the lower the p, the higher the solubility M 2 (O 3 ) n (s) 2 M n+ (aq) + no 3 2 (aq) 3 O + (aq) + O 3 2 (aq) O 3 (aq) + 2 O(l) 17

18 Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur Q = K sp, the solution is saturated, no precipitation Q < K sp, the solution is unsaturated, no precipitation Q > K sp, the solution would be above saturation, the salt above saturation will precipitate Some solutions with Q > K sp will not precipitate unless disturbed these are called supersaturated solutions precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added Selective Precipitation A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others A successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different 18

19 Example What is the minimum [O ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? Qualitative Analysis An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry A sample containing several ions is subjected to the addition of several precipitating agents Addition of each reagent causes one of the ions present to precipitate out Qualitative Analysis 19

20 Group 1 Group one cations are Ag +, Pb 2+, and g 2 2+ All these cations form compounds with l that are insoluble in water as long as the concentration is large enough Pbl 2 may be borderline molar solubility of Pbl 2 = 1.43 x 10-2 M Precipitated by the addition of l Group 2 Group two cations are d 2+, u 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and g2+ All these cations form compounds with S and S 2 that are insoluble in water at low p Precipitated by the addition of 2 S in l 20

21 Group 3 Group three cations are Fe 2+, o 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as r 3+, Fe 3+, and Al3+ precipitated as hydroxides All these cations form compounds with S 2 that are insoluble in water at high p Precipitated by the addition of 2 S in NaO Group 4 Group four cations are Mg 2+, a 2+, Ba 2+ All these cations form compounds with PO 4 3 that are insoluble in water at high p Precipitated by the addition of (N 4 ) 2 PO 4 Group 5 Group five cations are Na +, K +, N 4 + All these cations form compounds that are soluble in water they do not precipitate Identified by the color of their flame 21

22 omplex Ion Formation Transition metals tend to be good Lewis acids They often bond to one or more 2 O molecules to form a hydrated ion 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) O(l) Ag( 2 O) 2 + (aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag( 2 O) 2 + The attached ions or molecules are called ligands e.g., 2 O omplex Ion Equilibria If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag( 2 O) 2 + (aq) + 2 N 3(aq) Ag(N 3 ) 2 + (aq) O (l) Generally 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 N 3(aq) Ag(N 3 ) 2 + (aq) omplex Ion Formation The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 N 3(aq) Ag(N 3 ) 2 + (aq) The equilibrium constant for the formation reaction is called the formation constant, K f 22

23 Example ml of 1.5 x 10-3 M u(no 3 ) 2 is mixed with ml of 0.20 M N 3. What is the [u 2+ ] at equilibrium? The Effect of omplex Ion Formation on Solubility The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands Agl (s) Ag + (aq) + l (aq) K sp = 1.77 x Ag + (aq) + 2 N 3(aq) Ag(N 3 ) 2 + (aq) K f = 1.7 x 10 7 Adding N 3 to a solution in equilibrium with Agl (s) increases the solubility of Ag + 23

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