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1/18/20141 Applications of Aqueous Equilibria. 1/18/20142 Solutions of Acids or Bases Containing a Common Ion.

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1 1/18/20141 Applications of Aqueous Equilibria

2 1/18/20142 Solutions of Acids or Bases Containing a Common Ion

3 1/18/20143 Solution containing weak acid HA and its salt NaA Salt dissolves in water and breaks up completely into its ions-it is a strong electrolyte –NaA(s) Na+(aq) + A-(aq)

4 1/18/20144 Common Ion Effect When AgNO 3 is added to a saturated solution of AgCl, it is often described as a source of a common ion, the Ag + ion. By definition, a common ion is an ion that enters the solution from two different sources. Solutions to which both NaCl and AgCl have been added also contain a common ion; in this case, the Cl - ion. There is an effect of common ions on solubility product equilibria.

5 1/18/20145 The common-ion effect can be understood by considering the following question: What happens to the solubility of AgCl when we dissolve this salt in a solution that is already 0.10 M NaCl? As a rule, we can assume that salts dissociate into their ions when they dissolve. A 0.10 M NaCl solution therefore contains 0.10 moles of the Cl - ion per liter of solution. Because the Cl - ion is one of the products of the solubility equilibrium, LeChatelier's principle leads us to expect that AgCl will be even less soluble in an 0.10 M Cl - solution than it is in pure water.

6 1/18/20146 Calculate the solubility of AgCl in 0.10 M NaCl. In pure water:C s = 1.3 x M In 0.10 M NaCl: C s = 1.8 x M These calculations show how the common-ion effect can be used to make an "insoluble" salt even less soluble in water.

7 1/18/20147 The common ion effect can be applied to other equilibria, as well. Consider what happens when we add a generic acid (HA) to water. We now have two sources of a common ion ? the H 3 O + ion. HA(aq) + H 2 O(l) !H 3 O + (aq) + A - (aq) 2 H 2 O(l)! H 3 O + (aq) + OH - (aq) Thus, it isn't surprising that adding an acid to water decreases the concentration of the OH - ion in much the same way that adding another source of the Ag + ion to a saturated solution of AgCl decreases the concentration of the Cl - ion.

8 1/18/20148 The solubility of a solid is lowered if the solution already contains ions common to the solid –Dissolving silver chloride in a solution containing silver ions –Dissolving silver chloride in a solution containing chloride ions The common-ion effect can also be used to prevent a salt from precipitating from solution. –Instead of adding a source of a common ion, we add a reagent that removes the common ion from solution.

9 1/18/20149 Ksp (Solubility Product Constant, Solubility Product) –Ksp = [Ca 2+ ][F - ] 2 –Experimentally determined solubility of an ionic solid can by used to calculate its Ksp value –The solubility of an ionic solid can be calculated if its Ksp value is known Relative Solubilities –IF the salts being compared produce the same number of ions in solution, Ksp can be used to directly compare solubility NaCl(s), KF(s) Ksp = [cation][anion] = x 2 IF the salts being compared produce different numbers of ions, Ksp cannot be directly compared –Ag 2 S(s) Ksp = [2x] 2 [x] –Bi 2 S 3 (s) = [2x] 2 [3x] 3

10 1/18/ Common Ion-ion produced by both acid and its salt Ion provided in solution by an aqueous acid (or base) as well as a salt –HF(aq) and NaF (F - in common) HF(aq) H + (aq) + F - (aq) –NaF(s) (in water) Na + (aq) + F - (aq) Excess F - added by NaF Equilibrium shifts away from added component. Fewer H + ions present, making solution less acidic pH is higher than expected. –NH 4 OH and NH 4 Cl (NH 4 + in common) NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) –NH 4 Cl(s) (in water) NH 4 + (aq) + Cl - (aq) Equilibrium shifts to the left. pH of solution decreases due to decrease in OH - conc. Common ion effect-shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction

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12 1/18/ Equilibrium Calculations Consider initial concentration of ion from salt when calculating values for H + and OH - Calculate the pH of a solution that contains 0.10 M HC 2 H 3 O 2 and M NaC 2 H 3 O 2. (The pH of 0.10 M HC 2 H 3 O 2 is 2.9. The addition of the common ion would shift the equilibrium to the left. As a result the [H 3 O + ] would decrease and the pH would rise.)

13 1/18/ K a = 1.8 x HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) Substance[HC 2 H 3 O 2 ][H 3 O + ][C 2 H 3 O 2 - ] Initial conc Change in conc. -x+x Equilibrium conc – xx x

14 1/18/ –K a = [H 3 O + ][C 2 H 3 O 2 - ]/[HC 2 H 3 O 2 ] –1.8 x = x( x)/0.10 – x 0.050x/0.10 –x = [H 3 O + ] = 3.6 x M –pH = -log[H 3 O + ] = -log(3.6 x ) = 4.4 –as we predicted, the pH rose from 2.9 to 4.4.

15 1/18/ Buffered Solutions

16 1/18/ Until now, we have considered solutions containing only pure weak acids, weak bases, or their salts dissolved in distilled water A buffer is a solution with a very stable pH You can add acid or base to a buffer solution without greatly affecting the pH of the solution The pH of a buffer will also remain unchanged if the solution is diluted with water or if water is lost through evaporation A mixture that contains a conjugate acid-base pair is known as buffer solution because the pH changes by a relatively small amount if a strong acid or base is added to it –Buffer systems are used to control pH in many biological and chemical reactions

17 1/18/ A buffer is created by placing a large amount of a weak acid or base into a solution along with its conjugate A weak acid and its conjugate base (B-) will remain in solution together without neutralizing each other A weak base and its conjugate acid (A+) will do the same

18 1/18/ When both the acid and the conjugate base are together in the solution, any hydrogen ions that are added will be neutralized by the base while any hydroxide ions that are added will be neutralized by the acid without having much of an effect on the solutions pH A buffer system is a chemical sponge for H 3 O + or OH - ions that may be produced within a solution –When additional H 3 O + ions are added, they react with the conjugate base component of the buffer system to produce H 2 O and a weak acid that is only slightly ionized, so the pH drops only minimally H 3 O + (aq) + conjugate base - (aq) weak acid(aq) + H 2 O(l) –When additional OH - ions are added, the basic OH - ions react with the acid component of the buffer system to produce H 2 O and a weak base, so the pH rises only minimally OH - (aq) + conjugate acid (aq) weak base - (aq) + H 2 O(l)

19 1/18/ These solutions are governed by the same equilibrium law as are weak acids and weak bases Buffered solutions-Equilibrium systems that resist changes in acidity and maintain constant pH when acids or bases are added to them –Most effective pH range for any buffer is at or near the pH where the acid and salt concentrations are equal (pK a ) [H 3 O + ] of weak acid/conjugate base buffer equals the K a of the weak acid [OH - ] of weak base/conjugate acid buffer equals the K b of the weak base In general the most effect pH range of a buffer system is (optimum pH 1.0 pH unit

20 1/18/ –Since K a for a weak acid is small, equilibrium concentrations of the weak acid and its conjugate base are very nearly their initial concentrations All weak acid-conjugate base buffer systems have general expression –[H 3 O + ] = K a x [weak acid] 0 /[conjugate base] 0 –Convert the buffer relationship into its logarithmic form pH for buffer-pH = pK a + log[A - ]/[HA] = pK a + log [conjugatebase] 0 /[weak acid] 0 -obtained from equation for weak acid equilibrium (K a = [H + ][A - ]/[HA]) this relationship is the Henderson-Hasselbalch equation [OH - ] = K b x [weak base] 0 /[conjugate acid] 0 pOH = pK b + log x [conjugate acid] 0 /[weak base] 0 –Buffered solutions contain either: A weak acid and its salt A weak base and its salt

21 1/18/ Preparing a buffer A buffer can be prepared by dissolving mole of HF and mole of NaF in water to make 1.00 liter of solution –Dissociation reaction of HF is HF(aq) F - (aq) + H + (aq) –Equilibrium law is K a = [F - ][H + ]/[HF] = 6.8 x –Set up an equilibrium table using the given information

22 1/18/ ReactionHF F-F- +H+H+ Initial conc Change-x+x Equilibriu m – x x x Solution x 10 -3

23 1/18/ –Substituting the information from the equilibrium line into the equilibrium law yields K a = ( x)(x)/0.200 – x = 6.8 x –Assuming that x is small compared to both and 0.200, we simplify the equation to K a = (0.100)(x)/0.200 = 6.8 x x = 1.36 x –This value of x satisfies the assumptions made, and the last line of the equilibrium table can be completed The last column gives us [H + ] = 1.36 x M The pH of the buffer solution is calculated as 2.87 For almost all buffer solutions the assumptions hold true, and this type of problem is solved by simply entering the given concentrations of the conjugate acid and conjugate base directly into the equilibrium law

24 1/18/ Calculate the pH of the following buffer solutions M acetic acid and M sodium acetate –K a = [C 2 H 3 O 2 - ][H + ]/[HC 2 H 3 O 2 ] –1.8 x = (0.150)[H + ]/0.250 –[H + ] = 3.0 x 10 -3, and pH = 2.52 A solution of 10.0 g each of formic acid and potassium formate dissolved in 1.00 L of H 2 O –K a = [CHO 2 - ][H + ]/[HCHO 2 ] –1.8 x = (0.417)[H + ]/0.217 –[H + ] = 9.4 x 10 -5, and pH = 4.03

25 1/18/ M ethylamine and M ethyl ammonium chloride –K b = [C 2 H 5 NH 3 + ][OH - ]/[C 2 H 5 NH 2 ] –4.3 x = (0.0965)[OH - ]/ –[OH - ] = 1.5 x 10 -4, pOH = 3.82 and pH = M hydrazine and M hydrazine hydrochloride –K b = [N 2 H 5 + ][OH - ]/[N 2 H 4 ] –9.6 x = (0.321)[OH - ]/0.125 –[OH - ] = 3.7 x 10 -7, pOH = 6.43 and pH = 7.57

26 1/18/ Shortcuts in buffer calculations Equilibrium law used for buffers involves a ratio of the concentrations of the conjugate acid and conjugate base We may use the moles of conjugate acids and moles of conjugate base instead of the acid and base molarities If the concentration of the conjugate acid and conjugate base are equal, their ratio is exactly 1.00 –With equal molarities or equal numbers of moles of conjugate acid and conjugate base K a = [H + ] and pK a = pH for a buffer made from a weak acid and its conjugate base K b = [OH - ] and pK b = pOH for a buffer made from a weak base and its conjugate acid

27 1/18/ pH changes in buffers Addition of a strong acid to a buffer will decrease the pH slightly Addition of a strong base to a buffer will increase the pH slightly To determine the amount that the pH changes when a strong acid or base is added to a buffer, we must calculate the change in concentration of the conjugate acid or base in the buffer

28 1/18/ Steps –Determine either the molarity or the number of moles of the conjugate acid and conjugate base in the original buffer –Determine the amount of strong acid or base added in the same units as the conjugate acid and base in step 1 –If a strong acid is added to the buffer, add the value from step 2 to the conjugate acid and subtract it from the conjugate base –If a strong base is added to the buffer, add the value from step 2 to the conjugate base and subtract it fro the conjugate acid –Substitute the new values for the conjugate acid and conjugate base into the equilibrium law and calculate the pH as shown before

29 1/18/ In an acetate buffer, acetic acid, HC 2 H 3 O 2, is the conjugate acid, and the acetate ion, C 2 H 3 O 2 -, is the conjugate base –Addition of a strong acid increases the concentration of acetic acid and decreases the concentration of acetate ions. –Addition of a strong base to this buffer increases the acetate ion concentration and decreases the acetic acid concentration HC 2 H 3 O 2 + OH - C 2 H 3 O 2 - An acetate buffer is prepared with M acetic acid and M NaC 2 H 3 O 2. If mol of solid NaOH is added to 100 mL of this buffer, calculate the change in pH of the buffer due to the addition of the NaOH. –To calculate the change in pH, the pH of the original buffer is needed, along with the final pH after the base is added –Dissociation reaction for acetic acid HC 2 H 3 O 2 H + + C 2 H 5 O 2 - –Equilibrium law K a = [H + ][ C 2 H 5 O 2 - ]/[ HC 2 H 3 O 2 ] –Value for K a and the acetate and acetic acid concentrations are substituted into the equilibrium law 1.8 x = [H + ])(0.100)/0.250

30 1/18/ –Hydrogen ion concentration is calculated as [H + ] = 4.5 x M, and the pH is 4.35 –To calculate the pH of the buffer after the NaOH is added, we must first convert either the molarities of the conjugate acids and bases to moles or the moles of NaOH to molarity so that all the units are the same Molarity NaOH = mol NaOH/0.100 L = M –Adding this value to the acetate concentration gives us M C 2 H 3 O 2 - Subtracting it from the acetic acid concentration yields M HC 2 H 3 O 2 Substituting these new values into the equilibrium law –1.8 x = [H + ](0.120)/0.230 –[H + ] = 3.45 x M –pH = 4.46 –There is an increase of 0.11 pH units when the NaOH is added The answer is reasonable since the pH is expected to rise slightly when a base is added

31 1/18/ Calculations Involving Buffered Solutions Containing Weak Acids "Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14. This is not a new type of problem." When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be presented as follows:

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33 1/18/ OH - + HA A - + H 2 O weak acid conjugate base OH - ions are not allowed to accumulate but are replaced by A - ions

34 1/18/ K a = [H + ][A - ] / [HA] [H + ] = K a x [HA] / [A - ] If the amounts of HA and A - originally present are very large compared with the amount of OH - added, the change in [HA]/[A - ] will be small. Therefore the pH change will be small

35 1/18/ Buffering also works for addition of protons instead of hydroxide ions –H + + A - HA Conjugate base Weak acid Buffering with a Weak Base and Its Conjugate Acid –Weak base B reacts with any H + added –B + H + BH + Base Conjugate acid Conjugate acid BH+ reacts with any added OH BH + + OH - B + H 2 O Conjugate acid Base

36 1/18/ Summary Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base and the conjugate acid BH+ When H + is added to a buffered solution, it reacts essentially to completion with the weak base present When OH - is added to a buffered solution, it reacts essentially to completion with the weak acid present The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and the weak base. The pH remains relatively unchanged as long as the concentrations of buffering materials are large compared with the amounts of H + or OH - added

37 1/18/ Buffer Capacity

38 1/18/ Buffering Capacity The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. –The pH of a buffered solution is determined by the ratio [A - ]/[HA] –The capacity of a buffered solution is determined by the magnitudes of [HA] and [A - ]

39 1/18/ Preparing a Buffer Preparation of a buffer starts with the selection of the desired pH Then a table of K a and K b values for weak acids and bases is consulted to find an appropriate conjugate acid-base pair to use These 2 steps determine the ratio of the salt to the weak acid or base Next, the moles of acid or base that need to be buffered are estimated The total number of moles of the conjugate acid-base pair should be at least 20 times the amount of the acid or base that needs to be buffered The volume of buffer solution needed is determined next

40 1/18/ Method for preparing the buffer is decided on –Conjugate acid and conjugate base are measured and dissolved to the desired volume –Measuring the desired amount of conjugate acid and then adding the appropriate amount of a strong base to convert some of the conjugate acid into its conjugate base –Conjugate base is measured and the necessary amount of strong acid is added to convert some of the conjugate base into its conjugate acid Optimal buffering occurs when [HA] is equal to [A - ] ( [A - ]/[HA] = 1 ) The pK a of the weak acid to be used should be as close as possible to the desired pH

41 1/18/ Titration Titration technique used for chemical analysis utilizing reactions of 2 solutions

42 1/18/ One reactant solution is placed in a beaker and the other in a buret (long, graduated tube with stopcock) –The stopcock (valve that allows chemist to add controlled amounts of solution from buret to beaker) An indicator is added to solution in beaker (substance which undergoes a color change in the pH interval of the equivalence point)

43 1/18/ Chemist reads volume of solution in buret at start of experiment and again at point where indicator changes color –Difference in these volumes represents the volume of reactant delivered from the buret –Controlled addition of a solution of known concentration (titrant) in order to determine concentration of solution of unknown concentration The crucial point about the titration experiment is that the indicator is designed to change color when the amount of reactant delivered from the buret is exactly the amount needed to react with the solution in the beaker (Equivalence Point (Stoichiometric Point)-point in a titration at which the reaction between titrant and unknown has just been completed.) Classic chemical reaction-Fe 2+ and permanganate ion MnO 4 - –5Fe 2+ + MnO H + Mn Fe H 2 O –The purple permanganate ion is the indicator of the point where the correct amount has been added to completely react all of the Fe 2+ ions in the sample

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45 1/18/ Titration Curve (pH Curve)-plotting of the pH of the solution as a function of the volume of titrant added Monitors the pH of the solution as an acid- base titration proceeds from beginning to well beyond the equivalence point –pH is plotted on the y-axis and volume of base (or acid) delivered is plotted on the x-axis –Shape of titration curve makes it possible to identify the equivalence point and is an aid in selecting an appropriate indicator –It can also be used to calculate the Ka or Kb of a weak acid or base

46 1/18/ There are four major points of interest during a titration experiment –Start of the titration, where the solution contains only one acid or base pH is calculated as previously described –Region where titrant is added up to the end point, and the solution now contains a mixture of unreacted sample and products Mixture of a conjugate acid and its conjugate base If weak acid or base is involved, this is a buffer solution If only strong acids and bases are used, this region is unbuffered –End point, where all the reactant has been converted into product Solution is the salt of the acid or base pH is calculated as previously described titration curve used to determined pH during titration –Region after end point, where solution contains product and excess titrant pH depends on excess titrant used

47 1/18/ Points on titration curve –Midpoint of most vertical part of graph corresponds to exact endpoint. This will also correspond to the equivalence point, or the point at which the equivalents of acid equals the equivalents of base. In addition, the midpoint will also determine the pH of the salt that was formed during the titration.

48 1/18/ –Half-equivalence point is at the center of the buffer region pH increases more quickly at first, then levels out into buffer region At the half-equivalence point, enough base has been added to convert exactly half of the acid into conjugate base The concentration of the acid is equal to the concentration of the conjugate base The curve remains fairly flat until just before the equivalence point, when the pH increases sharply

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51 1/18/ Strong Acid-Strong Base Titration-titration curve can be divided into three segments –Prior to the equivalence point (0-x-amount at equivalence point) The pH rises slowly at first because the solution contains a sizeable excess of H 3 O + ions (buffering effect created by relatively high concentration of H+ in solution) [H + ] (thus the pH) can be calculated We begin to see significant rise in pH close to the equivalence point –At the equivalence point (x) All excess H 3 O + ions have been removed [H 3 O + ] is now x M, which corresponds to pH of –Beyond the equivalence point (x – total amount used) Once the endpoint has been passed, the rate of pH change diminishes again-resembles first part of graph except at higher pH The solution now contains an excess of OH - ions and as pH rises, it approaches that of the titrant [OH-] can be calculated from excess OH- and volume of solution [H+] and pH can be calculated from [OH-] and Kw

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53 1/18/ Strong Base with Strong Acid

54 1/18/ Weak Acids-Strong Bases Titration-segments of titration curve –Before titrant added pH of solution from given –Prior to the equivalence point (0-x) As NaOH is added below the equivalence point –HC 2 H 3 O 2 (aq) + OH - (aq) C 2 H 3 O 2 - (aq) + H 2 O(l) Solution is now a buffer system that contains both the weak acid (HC 2 H 3 O 2 ) and its conjugate base (C 2 H 3 O 2 - ) At the point of half-neutralization, [HC 2 H 3 O 2 ] = [C 2 H 3 O 2 - ] and [H 3 O + ] = K a –At the equivalence point (x) All of the HC 2 H 3 O 2 has been exhausted and the total volume is now double the original volume of the base The equivalence point corresponds to the pH-always > 7 –The stronger the basic anion, the higher the pH of the equivalence point –Beyond the equivalence point (x-total) Solution now contains excess of OH - ions and as the pH rises, it approaches that of the titrant If the acid being titrated is weak, then the graph will not be nearly as vertical at the endpoint. –Weaker acid is, the more the graph deviates from being vertical.

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56 1/18/ Weak Bases-Strong Acids Titration –Before HCl has been added-pH of solution calculated from given –Prior to the equivalence point (0-x) As HCl is added below the equivalence point –NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l) The solution is now a buffer system that contains both the weak base (NH 3 ) and its conjugate acid (NH 4 + ) At the point of half-neutralization, [NH 3 ] = [NH 4 + ] and [OH - ] = K b –At the equivalence point (x) All of the NH 3 has been exhausted and the total volume is now double the original volume of the acid The equivalence point corresponds to pH (always < 7.00) –The conjugate acid of the weak base lowers the pH –Beyond the equivalence point (x- total) The solution now contains an excess of H 3 O + ions and as the pH falls, it approaches that of the titrant

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58 1/18/ Polyprotic Acid-Strong Base Titration –When titrating a polyprotic acid, the graph will show a steep rise, or endpoint, for each of the protons in the acid. An acid with 2 protons will have two endponts, one for each hydrogen. An acid with 3 protons will have three endpoints, one for each proton. –With each successive ionization, the characteristic shape of the titration curve becomes less distinct First endpoint is fairly obvious, the second endpoint is not as well defined, the third endpoint is worse still, and so on.

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60 1/18/ When a monoprotic acid is completely neutralized by a monohydroxide base, equal numbers of moles of H 3 O + and OH - ions react –H 3 O + (aq) + OH - (aq) 2H 2 O(l) –At neutralization, n H30+ = n OH- M = n/V N = MV (MV) acid = (MV) base –When 25.0 mL of HCl is neutralized by M NaOH, 35.0 mL of the base are consumed. Calculate the molarity of the acid. MV acid = MV base M(25.0 mL) = (0.100 M)(35.0 M) M acid = M

61 1/18/ If the acid contains more than one ionizable proton, and/or the base contains more than one dissociable hydroxide ion, then the equation given above must be modified in order to include these numbers –(M x V x # H3O+ ) acid = (M x V x # OH- ) base where # indicates the number of acidic hydrogens or dissociable hydroxide ions in chemical formula –What volume of M H 2 SO 4 is needed to be completely neutralized by 300. mL of M KOH? MV# acid = MV# base (0.200 M)V(2) = (0.450)(300 mL)(1) V acid = 338 mL

62 1/18/ Acid-Base Indicators

63 1/18/ Determination of Equivalence Point Use a pH meter, find midpoint of vertical line in the titration curve Use of indicators –Indicators used in titrations must have two important properties The pH at the end point of the titration curve must change by at least 2 pH units very rapidly The pK of the indicator must be close to the end-point pH of the titration –Titration curve shows us required large change in pH often occurs –Proper selection of an indicator mandates that the pH at the end point and the pK a of the indicator be close to each other

64 1/18/ How Indicators Work Indicators are weak acids and weak bases whose respective conjugate bases and conjugate acids have different colors –The reason is that the loss or gain of a proton changes the energy of the electrons within their structures –This, in turn, changes the energy of light absorbed, which is observed as a change in color

65 1/18/ –They change colors within a specified pH range If conjugate acid has 1 color, its conjugate base has another The eye will see these colors clearly only if there is 10x more of one color than of the other If acid is yellow/base is blue, we see yellow if conjugate acid is 10x more concentrated than conjugate base, and blue if conjugate base is 10x more conc. than conjugate acid –[conjugate base]/[conjugate acid] < 0.1 (yellow observed) –[conjugate base]/[conjugate acid] > 10 (blue observed Between these 2 ratios various shades of green observed

66 1/18/ Since they are weak acids/bases, they undergo acid-base equilibrium –HIn(aq) + H 2 O(l) H 3 O + + In - (aq) K In = [H 3 O + ][In - ]/[In] –The species HIn is known as the acid form of the indicator and In - is known as the basic form of the indicator –Each form has its own distinct color –We can rearrange the equilibrium constant expression to yield [H 3 O + ] = K In x [HIn]/[In - ] When [HIn] > [In - ], indicator has color associated with acid form When [In - ] > [HIn], indicator has color associated with basic form When [HIn] = [In - ], color of indicator changes –At this point, [H 3 O + ] = K In –Ideally, this condition should represent end point of titration

67 1/18/ Since indicators are weak acids/bases, significance of this color phenomenon best shown w/Henderson- Hasselbalch equation –pH = pK a + log([A - ]/[HA]) = pK a + log([base]/[acid]) [HA] = molar conc. of undissociated weak acid (M) [A - ] = molar conc. of conjugate base (M) –pOH = pK b + log x [HB + ]/[B] [B] = molar conc. of weak base [HB + ] = molar conc. of conjugate acid –Substituting 0.1 into the log term yields pH = pK a + log(0.1) = pK a – 1.0 pH = pK a + log(10) = pK a For a particular buffering system (acid-conjugate base pair), all solutions that have same ratio [A - ]/[HA] will have same pH In order to select a suitable indicator for a titration, it is best if the pK In lies within one pH unit of the equivalence point –pK In = pH equivalence point 1

68 1/18/ We have a buffer solution with concentrations of 0.20 M HC 2 H 3 O 2 and 0.50 M C 2 H 3 O 2 -. The acid dissociation constant for HC 2 H 3 O 2 is 1.8 x Find the pH of the solution. –pH = pK a + log x [C 2 H 3 O 2 - ]/[ HC 2 H 3 O 2 ] –pH = -log(1.8 x ) + log x (0.50 M)/(0.20) –pH = -log(1.8 x ) + log(2.5) –pH = (4.7) + (0.40) = 5.1 If both concentrations are 0.20M –pH = pK a + log x [C 2 H 3 O 2 - ]/[ HC 2 H 3 O 2 ] –pH = -log(1.8 x ) + log x (0.20 M)/(0.20) –pH = -log(1.8 x ) + log(1) –pH = (4.7) + (0) = 4.7 Notice that when the concentrations of acid and conjugative base in solution are the same, pH = pK a (and pOH = pK b ) –When you choose an acid for a buffer solution, it is best to pick an acid with a pK a that is close to the desired pH –That way you can have almost equal amounts of acid and conjugate base in the solution, which will make the buffer as flexible as possible in neutralizing both added H + and OH -.

69 1/18/ Indicators Indicator color changes will be sharp, occurring with the addition of a single drop of titrant Suitable indicators must be selected based on the equivalence point –Strong acid-strong base titrations may use indicators with end points as far apart as pH 5 and pH 9 –Titration of weak acids or weak bases requires more careful selection of an indicator with appropriate transition interval

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71 1/18/ Solubility Equilibria and the Solubility Product

72 1/18/ Why Do Some Solids Dissolve in Water? Sugar is a molecular solid, in which the individual molecules are held together by relatively weak intermolecular forces. When sugar dissolves in water, the weak bonds between the individual sucrose molecules are broken, and these C 12 H 22 O 11 molecules are released into solution.

73 1/18/ –It takes energy to break bonds between C 12 H 22 O 11 molecules It also takes energy to break H bonds in water that must be disrupted to insert one of these sucrose molecules into solution Sugar dissolves in water because energy is given off when slightly polar sucrose molecules form intermolecular bonds with the polar water molecules. The weak bonds that form between the solute and the solvent compensate for the energy needed to disrupt the structure of both the pure solute and the solvent. In the case of sugar and water, this process works so well that up to 1800 grams of sucrose can dissolve in a liter of water.

74 1/18/ Ionic solids (or salts) contain (+)and (-) ions, which are held together by the strong force of attraction between particles w/opposite charges. –When one dissolves in water, ions that form solid are released into solution, where they become associated w/polar solvent molecules. H 2 O NaCl(s) Na+(aq) + Cl-(aq)

75 1/18/ –We can generally assume that salts dissociate into their ions when they dissolve in water. Ionic compounds dissolve in water if the energy given off when the ions interact with water molecules compensates for the energy needed to break the ionic bonds in the solid and the energy required to separate the water molecules so that the ions can be inserted into solution.

76 1/18/ Solubility Equilibria-based on following assumption: When solids dissolve in water, they dissociate to give the elementary particles from which they are formed. –Molecular solids dissociate to give individual molecules H 2 O C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) –Ionic solids dissociate into solutions of (+)/(-) ions they contain H 2 O NaCl(s) Na + (aq)+ Cl - (aq)

77 1/18/ When the salt is first added, it dissolves and dissociates rapidly. –The conductivity of solution therefore increases rapidly at first. Dissolve NaCl(s) Na+(aq) + Cl-(aq) Dissociate

78 1/18/ Concentrations of these ions soon become large enough that the reverse reaction starts to compete with forward reaction, which leads to a decrease in rate at which Na + and Cl - ions enter solution. Associate Na + (aq)+Cl - (aq) NaCl(s) Precipitate Eventually, the Na + and Cl - ion concentrations become large enough that the rate at which precipitation occurs exactly balances the rate at which NaCl dissolves. –Once that happens, there is no change in the concentration of these ions with time and the reaction is at equilibrium. –AT equilibrium, solution is saturated, because it contains the maximum conc. of ions that can exist in equilibrium w/solid salt. –The amount of salt that must be added to a given volume of solvent to form a saturated solution is called solubility of salt.

79 1/18/ Solubility Rules-patterns obtained from measuring solubility of different salts and are based on the following definitions of the terms A salt is soluble if it dissolves in water to give a solution with a concentration of at least 0.1 moles per liter at room temperature. A salt is insoluble if the concentration of an aqueous solution is less than M at room temperature. Slightly soluble salts give solutions that fall between these extremes.

80 1/18/ The Solubility Product Expression Silver chloride is so insoluble in water ( g/L) that a saturated solution contains only about 1.3 x moles of AgCl per liter of water. H 2 O AgCl(s) Ag + (aq) +Cl - (aq) Equilibrium constant expressions for this reaction gives following result. –Water isn't included because it is neither consumed nor produced in this reaction, even though it is a vital component of the system.

81 1/18/ [Ag + ] and [Cl - ] terms represent concentrations of the Ag + and Cl - ions in moles per liter when this solution is at equilibrium. [AgCl] is more ambiguous. –It doesn't represent conc. of AgCl dissolved in water- assume that AgCl dissociates into Ag + /Cl - ions when it dissolves in water –It can't represent amount of solid AgCl in system-equilibrium is not affected by the amount of excess solid added to the system. –The [AgCl] term has to be translated quite literally as the number of moles of AgCl in a liter of solid AgCl.

82 1/18/ Concentration of solid AgCl calculated from density & molar mass of AgCl. –This quantity is a constant, however. –The number of moles per liter in solid AgCl is the same at the start of the reaction as it is when the reaction reaches equilibrium.

83 1/18/ Since the [AgCl] term is a constant, which has no effect on the equilibrium, it is built into the equilibrium constant for the reaction. –[Ag + ][Cl - ] = K c x [AgCl] –This equation suggests that product of the equilibrium concentrations of Ag + and Cl - ions in this solution is equal to a constant. –Since this constant is proportional to solubility of the salt, it is called the solubility product equilibrium constant for reaction, or K sp. K sp = [Ag + ][Cl - ] –K sp expression for a salt is product of the concentrations of the ions, with each concentration raised to power equal to coefficient of that ion in the balanced equation for the solubility equilibrium.

84 1/18/ The Relationship Between K sp And the Solubility of a Salt K sp is called the solubility product because it is literally the product of the solubilities of the ions in moles per liter. Solubility product of salt can be calculated from its solubility, or vs.

85 1/18/ Photographic films are based on the sensitivity of AgBr to light. When light hits a crystal of AgBr, a small fraction of the Ag + ions are reduced to silver metal. The rest of the Ag + ions in these crystals are reduced to silver metal when the film is developed. AgBr crystals that do not absorb light are then removed from the film to "fix" the image. –Example: Let's calculate the solubility of AgBr in water in grams per liter, to see whether AgBr can be removed by simply washing the film. We start with the balanced equation for the equilibrium. H 2 O AgBr(s) Ag + (aq) + Br - (aq) –We then write the solubility product expression for this reaction. K sp = [Ag + ][Br - ] = 5.0 x –One equation can't be solved for two unknowns-Ag + / Br - ion conc. We can generate a second equation, however, by noting that one Ag + ion is released for every Br - ion. Because there is no other source of either ion in this solution, concentrations of these ions at equilibrium must be the same. –[Ag + ] = [Br - ]

86 1/18/ –Substituting this equation into K sp expression gives following result. [Ag + ] 2 = 5.0 x –Taking the square root of both sides of this equation gives the equilibrium concentrations of the Ag + and Br - ions. [Ag + ] = [Br - ] = 7.1 x M –Once we know how many moles of AgBr dissolve in a liter of water, we can calculate the solubility in grams per liter. –The solubility of AgBr in water is only gram per liter. It therefore isn't practical to try to wash the unexposed AgBr off photographic film with water.

87 1/18/ Solubility product calculations with 1:1 salts such as AgBr are relatively easy to perform. –In order to extend such calculations to compounds with more complex formulas we need to understand the relationship between the solubility of a salt and concentrations of its ions at equilibrium. –Symbol C s describes the amount of a salt that dissolves in water.

88 1/18/ The Role of the Ion Product (Q sp ) In Solubility Calculations

89 1/18/ Consider a saturated solution of AgCl in water. H 2 O AgCl(s) Ag + (aq)+Cl - (aq) Because AgCl is a 1:1 salt, the concentrations of the Ag + and Cl - ions in this solution are equal. –Saturated solution of AgCl in water: –[Ag + ] = [Cl - ]

90 1/18/ Imagine what happens when a few crystals of solid AgNO 3 are added to this saturated solution of AgCl in water. –According to the solubility rules, silver nitrate is a soluble salt. –It therefore dissolves and dissociates into Ag + and NO 3 - ions. –As a result, there are two sources of the Ag + ion in this solution. AgNO 3 (s) Ag + (aq)+NO 3 - (aq) H 2 O AgCl(s) Ag + (aq) + Cl - (aq) Adding AgNO 3 to a saturated AgCl solution therefore increases the Ag + ion concentrations. –When this happens, the solution is no longer at equilibrium because product of the concentrations of the Ag + and Cl - ions is too large. –In more formal terms, we can argue that the ion product (Q sp ) for the solution is larger than the solubility product (K sp ) for AgCl. –Q sp = (Ag + )(Cl - ) > K sp

91 1/18/ The ion product is literally the product of the concentrations of the ions at any moment in time. When it is equal to the solubility product for the salt, the system is at equilibrium. The reaction eventually comes back to equilibrium after the excess ions precipitate from solution as solid AgCl. When equilibrium is reestablished, however, the concentrations of the Ag + and Cl - ions won't be the same. Because there are two sources of the Ag + ion in this solution, there will be more Ag + ion at equilibrium than Cl - ions: –Saturated solution of AgCl to which AgNO 3 has been added: –[Ag + ] > [Cl - ]

92 1/18/ Now imagine what happens when a few crystals of NaCl are added to a saturated solution of AgCl in water. There are two sources of the chloride ion in this solution. H 2 O NaCl(s) Na + (aq)+ Cl - (aq) H 2 O AgCl(s) Ag + (aq)+Cl - (aq) Once again, the ion product is larger than the solubility product. –Q sp = (Ag + )(Cl - ) > K sp This time, when the reaction comes back to equilibrium, there will be more Cl- ion in the solution than Ag + ion. –Saturated solution of AgCl to which NaCl has been added: –[Ag + ] < [Cl - ]

93 1/18/ The figure below shows a small portion of the possible combinations of the Ag + and Cl - ion concentrations in an aqueous solution. Any point along the curved line in this graph corresponds to a system at equilibrium, because the product of the Ag + and Cl - ion concentrations for these solutions is equal to K sp for AgCl.

94 1/18/ )Point A represents a solution at equilibrium that could be produced by dissolving two sources of the Ag + ion such as AgNO 3 and AgCl in water. 2)Point B represents a saturated solution of AgCl in pure water, in which the [Ag + ] and [Cl - ] terms are equal. 3)Point C describes a solution at equilibrium that was prepared by dissolving two sources of the Cl - ion in water, such as NaCl and AgCl.

95 1/18/ Any point that is not along the solid line in the above figure represents a solution that is not at equilibrium. –Any point below the solid line (such as Point D) represents solution for which the ion product is smaller than the solubility product. Point D: Q sp < K sp If more AgCl were added to solution at Point D, it would dissolve If Q sp < K sp : AgCl(s) Ag + (aq) + Cl - (aq) Points above the solid line (such as Point E) represent solutions for which the ion product is larger than the solubility product. Point E: Q sp > K sp The solution described by Point E will eventually come to equilibrium after enough solid AgCl has precipitated. If Q sp > K sp : Ag + (aq) + Cl - (aq) AgCl(s)

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