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More Acid/Base Equilibria!!! 17.1 – 17.3. 17.1 The Common Ion Effect Consider the ionization of a weak acid, acetic acid: HC 2 H 3 O 2 (aq)  H + (aq)

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Presentation on theme: "More Acid/Base Equilibria!!! 17.1 – 17.3. 17.1 The Common Ion Effect Consider the ionization of a weak acid, acetic acid: HC 2 H 3 O 2 (aq)  H + (aq)"— Presentation transcript:

1 More Acid/Base Equilibria!!! 17.1 – 17.3

2 17.1 The Common Ion Effect Consider the ionization of a weak acid, acetic acid: HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 – (aq) If we increase the [C 2 H 3 O 2 – ] ions by adding NaC 2 H 3 O 2, the equilibrium will shift to the left. (Le Chatelier) This reduces the [H + ] and raises the pH (less acidic) This phenomenon is called the common-ion effect. Common ion equilibrium problems are solved following the same pattern as other equilibrium problems (ICE charts) EXCEPT the initial concentration of the common ion must be considered (it is NOT zero).

3 Example 1:Does the pH increase, decrease, or stay the same on addition of each of the following? (a) NaNO 2 to a solution of HNO 2 (b) (CH 3 NH 3 )Cl to a solution of CH 3 NH 2 (c) sodium formate to a solution of formic acid (d) potassium bromide to a solution of hydrobromic acid (e) HCl to a solution of NaC 2 H 3 O 2 (a)HNO 2  H + + NO 2 - increases (b)CH 3 NH 2 + H 2 O  CH 3 NH OH - decreases (c)HCHO 2  H + + CHO 2 - increases (d) HBr  H + + Br - no change (e) C 2 H 3 O H 2 O  HC 2 H 3 O 2 + OH -1 decreases

4 Example 2: Using equilibrium constants from Appendix D, calculate the pH of the solution containing M KC 3 H 5 O 2 and M HC 3 H 5 O 2 change:-x+x+x Equilibrium: xx.060+ x x x-(1.105*10 -6 )=0 x=1.84*10 -5 pH = -log(1.84*10 -5 ) = 4.74

5 17. 2 Buffered Solutions A buffered solution or buffer is a solution that resists a change in pH after addition of small amounts of strong acid or strong base. A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X – ) or weak base (B) and its conjugate acid (HB + ) Thus a buffer contains both: an acidic species to neutralize added OH – When a small amount of OH – is added to the buffer solution, the OH – reacts with the acid in the buffer solution. a basic species to neutralize added H + When a small amount of H + is added to the buffer solution, the H + reacts with the base in the buffer solution.

6 Composition of a Buffer - 4 ways to make a buffer solution: 1.)Weak acid + salt of the acid HCN and NaCN weak acid: HCN weak base: CN -1 2.)Weak base + salt of the base NH 3 and NH 4 Cl weak acid: NH 4 +1 weak base: NH 3 3.) EXCESS Weak acid + strong base 2 mol HCN + 1 mol NaOH  1 mol HCN + 1 mol NaCN + H 2 O weak acid: HCN weak base: CN -1 2 mol NH 4 Cl + 1 mol NaOH  1 mol NH 4 Cl + 1 mol NH 3 + NaCl weak acid: NH 4 +1 weak base: NH 3 4.)EXCESS Weak base + strong acid 2 mol NH mol HCl  1 mol NH 3 and 1 mol NH 4 Cl weak acid: NH 4 +1 weak base: NH 3 2 mol NaF + 1 mol HCl  1 mol NaF + 1 mole HF + NaCl weak acid: HF weak base: F -1

7 Example 3: Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture of HC 2 H 3 O 2 and NaC 2 H 3 O 2 does. HCl is a strong acid - Cl -1 is a negligible base and will NOT react with added H + - added H + will significantly change the pH of the solution HC 2 H 3 O 2 and C 2 H 3 O 2 -1 are a weak conjugate acid/base pair which act as a buffer HC 2 H 3 O 2 reacts with added base C 2 H 3 O 2 -1 reacts with added acid leaving the [H +1 ] and pH relatively unchanged

8 Buffer Capacity and pH Buffer capacity is the amount of acid or base that can be neutralized by the buffer before there is a significant change in pH. Buffer capacity depends on the concentrations of the components of the buffer - the greater the concentrations of the conjugate acid-base pair, the greater the buffer capacity. The pH of the buffer is related to K a and to the relative concentrations of the acid and base.

9 Henderson-Hasselbalch equation – used for buffer solutions (on AP equation sheet!!) These equations technically use the equilibrium concentrations of the acid (base) and the conjugate base (acid). However, since the acid/base in the buffer is WEAK – the amount of the conjugate produced by dissociation is generally small compared to the amount of the conjugate added as a salt. IF this is true (it is for all AP buffer problems!) we do not need to do an equilibrium problem – just use the INITIAL concentrations.

10 Example 4 (Example 2 again!): Using equilibrium constants from Appendix D, calculate the pH of the solution containing M KC 3 H 5 O 2 and M HC 3 H 5 O 2

11 Example 5: Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate

12 Example 6: A buffer is prepared by adding 20.0 g of acetic acid, HC 2 H 3 O 2 and 20.0 g of sodium acetate to enough water to form 2.00 L of solution. (a) Determine the pH of the buffer (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide are added to the buffer (b) C 2 H 3 O 2 -1 (aq) + H +1 (aq) + Cl -1 (aq)  HC 2 H 3 O 2 (aq) + Cl -1 (aq) (c) HC 2 H 3 O 2 (aq) + Na +1 (aq) + OH -1 (aq)  C 2 H 3 O 2 -1 (aq) + H 2 O (l) + Na +1 (aq)

13 17.3 Acid-Base Titrations – Titration Curves In an acid-base titration: A solution of base (or acid) of known concentration (called standard) is added to an acid (or base). Acid-base indicators or a pH meter are used to signal the equivalence point (when moles acid = moles base). The plot of pH versus volume during a titration is called a pH titration curve.

14 Strong acid added to strong base equal moles of acid and base present Starts high Ends low Equivalence point = 7 start end

15 Strong base added to strong acid Starts low Ends high Equivalence point = 7 start end

16 Strong acid added to weak base Buffer area – in this area there is weak base and some salt of the weak base Actual pH depends on the salt formed but it will be < 7 Starts med-high Ends low Equivalence point < 7

17 Weak base added to strong acid Starts low Ends med-high Equivalence point < 7 start end

18 Weak acid added to strong base Actual pH depends on the salt formed but it will be > 7 Starts high Ends med-low Equivalence point > 7 start end

19 Weak acid added weak base Starts med-high Ends med-low Equivalence point = 7 start end

20 Strong base added to strong diprotic acid (H 2 SO 4 )

21 Example 7: Predict whether the equivalence point of each of the following titrations is below, above or at pH 7: a) NaHCO 3 titrated with NaOH b) NH 3 titrated with HCl c) KOH titrated with HBr At the equivalence point, only products are present in solution, so determine the products of the reaction and then determine if the solution is acidic, basic or neutral a) NaHCO 3 + NaOH  Na 2 CO 3 + H 2 O weak acid strong base pH > 7 CO 3 -2 is basic, Na + is neutral, H 2 O is neutral b) NH 3 + HCl  NH 4 Cl weak base strong acidpH < 7 NH 4 +1 is acidic, Cl - is neutral c) KOH + HBr  KBr + H 2 O strong base strong acidpH = 7 K + and Br - are both neutral

22 Example 8: How many mL of M NaOH solution is required to titrate 40.0 mL of M HNO 3 ? ? mL 40.0 mL M M NaOH + HNO 3  H 2 O + NaNO 3 1 mole

23 Example 9: A 20.0 mL sample of M HBr solution is titrated with M NaOH solution. Calculate the pH of the solution after the following volumes of base solution have been added: (a) 15.0 mL (b) 19.9 mL (c) 20.0 mL (d) 20.1 mL (e) 35.0 mL mLmLmLmol H +1 mol OH -1 M ofpH HBrNaOHTotal(M) (V)(M) (V)excess ion (mol / tot vol) (a) M H (b) M H (c) x M H +1* 7.0 (d) M OH (e) M OH When molarity of H + (or OH - ) is less than we must consider the autoionization of water! (H + = 1.0*10 -7 )

24 Example 10: Calculate the pH at the equivalence point for titrating M solutions of each of the following bases with M HBr: (a) NaOH (b) NH 2 OH (a) strong acid/strong base titration so pH = 7 (b) HBr + NH 2 OH  Br - + NH 2 OH 2 + strong acid weak base.200M all product at equivalence point – no excess & Br - is neutral and will have no affect on pH Volume doubles (equal molarity and 1:1 stoich ratio) so molarity halves [NH 2 OH 2 + ] = 0.200mol / 2 = M NH 2 OH 2 +  H +1 + NH 2 OH K b = 1.1 x (appendix) I C -x +x +x E0.100 – x x x K a = K w / K b = 1  / 1.1  = 9.1  (x 2 ) / (0.100-x) = 9.1 x x = 3.0 x M = [H +1 ] pH = - log(3.0 x ) = 3.52

25 Acid – Base Indicators The equivalence point of an acid-base titration can be determined by measuring pH, but it can also be determined by using an acid-base indicator which marks the end point of a titration by changing color. Although the equivalence point (defined by the stoichiometry) is not necessarily the same as the end point (where the indicator changes color), careful selection of the indicator can ensure that the difference between them is negligible. Acid-base indicators are complex molecules that are themselves, weak acids (represented by HIn). They exhibit one color when the proton is attached and a different color when the proton is absent.

26 Acid – Base Indicators Bromthymol Blue Indicator In Acid In Base

27 Acid – Base Indicators Methyl Orange Indicator In Acid In Base

28 Acid – Base Indicators Phenolphthalein color at different pH values pH values

29 Acid – Base Indicators Consider a hypothetical indicator, HIn, a weak acid with K a =1.0x It has a red color in acid and a blue color in base. HIn(aq)  H +1 (aq) + In -1 (aq) red blue

30 Acid – Base Indicators HIn(aq)  H +1 (aq) + In -1 (aq) red blue This ratio shows that the predominant form of the indicator is HIn, resulting in a red solution. As OH -1 is added (like in a titration) [H +1 ] decreases and the equilibrium shifts to the right, changing HIn to In -. At some point in the titration, enough of the In - form will be present so we start to notice a color change.

31 Acid – Base Indicators It can be shown (using the Henderson-Hasselbalch equation) that for a typical acid-base indicator with dissociation constant, K a, the color transition occurs over a range of pH values given by pK a ± 1. For example, bromthymol blue with K a = 1.0 x (pK a = 7), would have a useful pH range of 7 ± 1 or from 6 to 8. You want to select an indicator whose pKa value is close to the pH you want to detect (usually the pH at the equivalence point)

32 Acid – Base Indicators

33 The pH curve for the titration of mL of 0.10 M HCl with 0.10 M NaOH. Neither of the indicators shown would be useful for a titration. Bromthymol blue (pK a =7) would be useful.

34 The pH curve for the titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH. Here, phenolphthalein is the indicator of choice. It has a pK a value of about 9.

35 Example 11: Use the following table to determine which of the following would be the best indicator to use to indicate the equivalence point of the titrations described in Example 10. a.pH at equivalence point was 7.0 Bromthymol Blue b. pH at equivalence point was 3.52 Methyl Yellow IndicatorKa Methyl Yellow1*10 -4 Methyl Red1*10 -5 Bromthymol Blue1*10 -7 Phenolpthalein1*10 -9

36 Solubility Equilibria & Complex Ions 17.4 – 17.6

37 Know your solubility rules: SOLUBILITY GUIDELINES Soluble CompoundsExceptions NOT precipitatesPRECIPITATES NitratesNone AcetatesNone ChloratesNone ChloridesAg +1, Hg 2 +2, Pb +2 BromidesAg +1, Hg 2 +2, Pb +2 IodidesAg +1, Hg 2 +2, Pb +2 SulfatesCa +2, Sr +2, Ba +2, Hg 2 +2, Pb +2 Insoluble CompoundsExceptions PRECIPITATESNOT Precipitates SulfidesNH 4 +1, Li +1, Na +1, K +1, Ca +2, Sr +2, Ba +2 CarbonatesNH 4 +1, Li +1, Na +1, K +1 PhosphatesNH 4 +1, Li +1, Na +1, K +1 HydroxidesLi +1, Na +1, K +1, Ca +2, Sr +2, Ba +2 ChromatesNH 4 +1, Li +1, Na +1, K +1, Ca +2, Mg +2 We classify these based on the Solubility - maximum amount of solute that dissolves in water.

38 17.4 Solubility Equilibria The Solubility-Product Constant, K sp Consider a saturated solution of BaSO 4 in contact with solid BaSO 4. We can write an equilibrium expression for the dissolving of the solid. BaSO 4 (s)  Ba 2+ (aq) + SO 4 2– (aq) Since BaSO 4 (s) is a pure solid, the equilibrium expression depends only on the concentration of the ions. K sp = [Ba 2+ ][ SO 4 2– ] K sp is the equilibrium constant for the equilibrium between an ionic solid solute and its saturated aqueous solution. K sp is called the solubility-product constant

39 In general: the solubility product is equal to the product of the molar concentration of ions raised to powers corresponding to their stoichiometric coefficients. Al 2 (CO 3 ) 3  2 Al CO 3 -2 K sp = [Al +3 ] 2 [CO 3 -2 ] 3

40 Solubility and K sp Solubility is the amount of substance that dissolves to form a saturated solution. This can be expressed as grams of solid that will dissolve per liter of solution. Molar solubility - the number of moles of solute that dissolve to form a liter of saturated solution. Solubility can be used to find K sp and K sp can be used to find solubility (see problems)

41 Example 1: a. If the molar solubility of CaF 2 at 35 o C is 1.24  mol/L, what is K sp at this temperature? CaF 2  Ca F -1 E.00124M actually dissolves K sp = [Ca +2 ] [F -1 ] 2 = ( M)( M) 2 = 7.63 x b. It is found that 1.1  g of SrF 2 dissolves per 100 mL of aqueous solution at 25 o C. Calculate the solubility product of SrF 2. [SrF 2 ] = (.011 g / g/mole) /.100 L = M SrF 2  Sr F -1 E M M 2(.00088) = M K sp = [Sr +2 ] [F -1 ] 2 = ( M)(.00176) 2 = 2.7 x M 2(.00124) = M

42 c. The K sp of Ba(IO 3 ) 2 at 25 o C is 6.0  What is the molar solubility of Ba(IO 3 ) 2 ? Ba(IO 3 ) 2  Ba IO 3 -1 E x (x) (2x) 2 = 4 x 3 = 6.0 x x = 5.3 x M x 2 x

43 17.5 Factors That Affect Solubility Factors that have a significant impact on solubility are: - The presence of a common ion - The pH of the solution Common-Ion Effect Solubility is decreased when a common ion is added. This is an application of Le Châtelier’s principle: Consider the solubility of CaF 2 : CaF 2 (s)  Ca 2+ (aq) + 2F – (aq) If more F – is added (say by the addition of NaF), the equilibrium shifts left to offset the increase. Therefore, more CaF 2 (s) is formed (precipitation occurs).

44 Example 2: Using Appendix D, calculate the molar solubility of AgBr in (a) pure water (b) 3.0  M AgNO 3 solution (c) 0.50 M NaBr solution (a) AgBr  Ag +1 + Br -1 K sp = 5.0  E x x x 5.0  = x 2 x = 7.1  M (b) AgBr  Ag +1 + Br -1 K sp = 5.0  E x x x 5.0  = (.030+x) x x = 1.7  M (c) AgBr  Ag +1 + Br -1 K sp = 5.0  x x.50 + x 5.0  = x (.50+x) x = 1.0  M notice the DECREASED solubility with the common ion in (b) and (c)

45 pH effects Consider: Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH – (aq) If OH – is removed, then the equilibrium shifts right and Mg(OH) 2 dissolves. OH – can be removed by adding a strong acid (lowering the pH): OH – (aq) + H + (aq)  H 2 O(aq) Another example: CaF 2 (s)  Ca 2+ (aq) + 2 F – (aq) If the F – is removed, then the equilibrium shifts right and CaF 2 dissolves. F – can be removed by adding a strong acid (or lowering pH): F – (aq) + H + (aq)  HF(aq)

46 Example 3: Calculate the molar solubility of Mn(OH) 2 at (a) pH 7.0 (b) pH 9.5 (c) pH 11.8 the [OH -1 ] is set by the pH (or pOH) (a) pH = 7.0 so pOH = 7.0 so [OH -1 ] = 1.00  Mn(OH) 2  Mn OH -1 K sp = 1.6  x x 1.00 x K sp = [Mn +2 ][OH -1 ]  = (x) (1.00  ) 2 x = 16 M (b) pH = 9.5 so pOH = 4.5 so [OH -1 ] = 3.16  Mn(OH) 2  Mn OH -1 K sp = 1.6  x x 3.16 x K sp = [Mn +2 ][OH -1 ]  = (x) (3.16  ) 2 x = 1.7  M (c) pH = 11.8 so pOH = 2.2 so [OH -1 ] = 6.31  Mn(OH) 2  Mn OH -1 K sp = 1.6  x x 6.31 x K sp = [Mn +2 ][OH -1 ]  = (x) (6.31  ) 2 x = 4.0  M Common ion effect – increasing [OH - ] decreases solubility

47 Example 4: Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO 3 (b) ZnS (c) BiI 3 (d) AgCN (e) Ba 3 (PO 4 ) 2 If the anion of the salt is the conjugate base of a weak acid, it will combine with H +1, reducing the concentration of the anion and making the salt more soluble ZnCO 3  Zn +2 + CO 3 -2 the CO 3 -2 ion will react with the added H + CO H +  HCO 3 -1 Le Chatelier effect of removing CO 3 -2 more soluble in acid: ZnCO 3, ZnS, AgCN, Ba 3 (PO 4 ) 2

48 17.6 Precipitation and Separation of Ions Consider the following: BaSO 4 (s)  Ba 2+ (aq) + SO 4 2– (aq) At any instant in time, Q = [Ba 2+ ][ SO 4 2– ] If Q > K sp, (too many ions) precipitation occurs until Q = K sp. If Q = K sp equilibrium exists (saturated solution) If Q < K sp, (not enough ions) solid dissolves until Q = K sp. Selective Precipitation of Ions Removal of one metal ion from a solution of two or more metal ions is called selective precipitation. Ions can be separated from each other based on the solubilities of their salt compounds. Example: If HCl is added to a solution containing Ag + and Cu 2+, the silver precipitates (as AgCl) while the Cu 2+ remains in solution Generally, the less soluble ion is removed first!

49 Example 5: Will Ca(OH) 2 precipitate if the pH of a M solution of CaCl 2 is adjusted to 8.0? if Q > than K sp then precipitation will occur pH = 8.0 so pOH = 6.0 so [OH -1 ] = 1.0  M Ca(OH) 2  Ca OH -1 K sp = 6.5  x Q = [Ca +2 ] [OH -1 ] 2 Q = (.050)(1.0  ) 2 = 5.0  Q < K so no precipitation occurs

50 Example 6: A solution contains M Ag +1 and M Pb +2. If NaI is added, will AgI or PbI 2 precipitate first? Specify the [I -1 ] needed to begin precipitation for each cation. the cation needing the lower [I -1 ] will precipitate first AgI  Ag +1 + I -1 K sp = 8.3 x x K sp = [Ag +1 ][x] 8.3  = (.00020)[x] 4.2  = x = [I -1 ] PbI 2  Pb I -1 K sp = 1.4 x x K sp = [Pb +2 ][x]  = (.0015)[x]  = x = [I -1 ] AgI will precipitate first at an [I -1 ] = 4.2 

51 Complex Ions Complex ion – a metal ion bonded to one or more Lewis bases. (We saw this with water in chapter 16)

52 It can happen with other Lewis bases (things that have lone pairs of electrons) Rule of thumb: The number of Lewis bases (ligands) that a metal ion attracts is equal to double its charge. (Works about 75% of the time!)

53

54 Extra Stuff Below…

55 Acid – Base Indicators How much In - must be present for the human eye to detect that the color is different? For most indicators, about 1/10 of the initial form must be converted to the other form before a color change is apparent. We can assume that in the titration of an acid with a base, the color change will occur at a pH where

56 Acid – Base Indicators Bromthymol blue, an indicator with a K a = 1.0 x 10 -7, is yellow in its HIn form and blue in its In - form. Suppose we put some strong acid in a flask, add a few drops of bromthymol blue and titrate with NaOH. At what pH will the indicator color change first be visible? HIn(aq)  H +1 (aq) + In -1 (aq) yellow blue

57 Selective Precipitation of Ions (continued) Sulfide ion is often used to separate metal ions. Example: Consider a mixture of Zn 2+ (aq) and Cu 2+ (aq). CuS (K sp = 6 x 10 –37 ) is less soluble than ZnS (K sp = 2 x 10 –25 ). Because CuS is LESS SOLUBLE than ZnS, CuS will be removed from solution before ZnS. As H 2 S is bubbled through the acidified green solution, black CuS forms. When the precipitate is removed, a colorless solution containing Zn 2+ (aq) remains. When more H 2 S is added to the solution, a second precipitate of white ZnS forms.

58 Formula Type of Formula of Hydrolysis equation Hydrolysis equation acid or conjugate of the acid of the base base acid or base HCl HOCl NH 3 Ba(OH) 2 KI NaC 2 H 3 O 2 Strong acid Cl -1 HCl + H 2 O  H 3 O +1 + Cl -1 Cl -1 + H 2 O  X Weak acid OCl -1 HOCl + H 2 O  H 3 O +1 + OCl -1 OCl -1 + H 2 O  HOCl + OH -1 Weak base NH 4 +1 NH H 2 O  H 3 O +1 + NH 3 NH 3 + H 2 O  OH -1 + NH 4 +1 Strong base H2OH2O H 2 O + H 2 O  H 3 O +1 + OH -1 OH -1 + H 2 O  X neutral K +1 or I -1 K +1 + H 2 O  X I -1 + H 2 O  X weak base HC 2 H 3 O 2 HC 2 H 3 O 2 + H 2 O  H 3 O +1 + C 2 H 3 O 2 -1 C 2 H 3 O H 2 O  HC 2 H 3 O 2 + OH -1


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