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Chapter 19 – Ionic Equilibria in Aqueous Solutions 19.1 – Equilibria of Acid-Base Buffer Systems 19.2 – Acid-base Titration Curves 19.3 – Equilibria of.

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Presentation on theme: "Chapter 19 – Ionic Equilibria in Aqueous Solutions 19.1 – Equilibria of Acid-Base Buffer Systems 19.2 – Acid-base Titration Curves 19.3 – Equilibria of."— Presentation transcript:

1 Chapter 19 – Ionic Equilibria in Aqueous Solutions 19.1 – Equilibria of Acid-Base Buffer Systems 19.2 – Acid-base Titration Curves 19.3 – Equilibria of Slightly Soluble Ionic Compounds 19.4 – Equilibria Involving Complex Ions 19.5 – Application of Ionic Equilibria to Chemical Analysis 1

2 19.1 - Equilibria of Acid-Base Buffer Systems 2 The effect of addition of acid or base to … an unbuffered solution or a buffered solution acid addedbase added acid addedbase added

3 19.1 - Equilibria of Acid-Base Buffer Systems 3 Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH 3 COOH] initial [CH 3 COO - ] added % Dissociation*pH * % Dissociation = [CH 3 COOH] dissoc [CH 3 COOH] initial x 100 0.100.00 0.10 0.050 0.10 0.15 1.3 0.036 0.018 0.012 2.89 4.44 4.74 4.92

4 19.1 - Equilibria of Acid-Base Buffer Systems 4 Figure 19.3 How a buffer works. Buffer with equal concentrations of conjugate base and acid OH - H3O+H3O+ Buffer after addition of H 3 O + H 2 O + CH 3 COOH H 3 O + + CH 3 COO - Buffer after addition of OH - CH 3 COOH + OH - H 2 O + CH 3 COO -

5 19.1 - Equilibria of Acid-Base Buffer Systems 5 Sample Problem 19.1Calculating the Effect of Added H 3 O + or OH - on Buffer pH PROBLEM:Calculate the pH: (a) of a buffer solution consisting of 0.50M CH 3 COOH and 0.50M CH 3 COONa (b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a) (c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a) K a of CH 3 COOH = 1.8x10 -5. (Assume the additions cause negligible volume changes. PLAN:We know K a and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system. Initial Change Equilibrium 0.50 + x 0.50-x - - - 0.500 + x 0.50 +xx - x SOLUTION: CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) (a)

6 19.1 - Equilibria of Acid-Base Buffer Systems 6 Sample Problem 19.1Calculating the Effect of Added H 3 O + and OH - on Buffer pH continued (2 of 4) [CH 3 COOH] equil ≈ 0.50M[CH 3 COO - ] initial ≈ 0.50M[H 3 O + ] = x K a = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [H 3 O + ] = x = K a [CH 3 COO - ] [CH 3 COOH] = 1.8x10 -5 M Check the assumption:1.8x10 -5 /0.50 X 100 = 3.6x10 -3 % CH 3 COOH( aq ) + OH - ( aq ) CH 3 COO - ( aq ) + H 2 O ( l )Concentration (M) Before addition Addition After addition (b) [OH - ] added = 0.020 mol 1.0L soln = 0.020M NaOH 0.50- - - - -0.020- 0.4800.52

7 19.1 - Equilibria of Acid-Base Buffer Systems 7 Sample Problem 19.1Calculating the Effect of Added H 3 O + and OH - on Buffer pH continued (3 of 4) Set up a reaction table with the new values. CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) Initial Change Equilibrium 0.48- - x 0.48 -x - - 0.520 x + x 0.52 +x [H 3 O + ] = 1.8x10 -5 0.48 0.52 = 1.7x10 -5 pH = 4.77 Before addition Addition After addition (c)[H 3 O + ] added = 0.020 mol 1.0L soln = 0.020M H 3 O + 0.50- - - - -0.020- 0.4800.52

8 19.1 - Equilibria of Acid-Base Buffer Systems 8 Sample Problem 19.1Calculating the Effect of Added H 3 O + and OH - on Buffer pH continued (4 of 4) Set up a reaction table with the new values. CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) Initial Change Equilibrium 0.52- - x 0.52 -x - - 0.480 x + x 0.48 +x [H 3 O + ] = 1.8x10 -5 0.48 0.52 = 2.0x10 -5 pH = 4.70

9 19.1 - Equilibria of Acid-Base Buffer Systems 9 The Henderson-Hasselbalch Equation HA + H 2 O H 3 O + + A - K a = [H 3 O + ] [A - ] [HA] [H 3 O + ] = K a [HA] [A - ] - log[H 3 O + ] = - log K a + log [A - ] [HA] pH = pK a + log [base] [acid]

10 19.1 - Equilibria of Acid-Base Buffer Systems 10 Buffer capacity is the ability to resist pH change. Buffer range is the pH range over which the buffer acts effectively. The more concentrated the components of a buffer, the greater the buffer capacity. The pH of a buffer is distinct from its buffer capacity. A buffer has the highest capacity when the component concentrations are equal. Buffers have a usable range within ± 1 pH unit of the pK a of its acid component.

11 19.1 - Equilibria of Acid-Base Buffer Systems 11 Figure 19.4 The relation between buffer capacity and pH change.

12 19.1 - Equilibria of Acid-Base Buffer Systems 12 Sample Problem 19.2Preparing a Buffer SOLUTION: PROBLEM:An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limsetone-rich soils. How many grams of Na 2 CO 3 must she add to 1.5L of freshly prepared 0.20M NaHCO 3 to make the buffer? K a of HCO 3 - is 4.7x10 -11. PLAN:We know the K a and the conjugate acid-base pair. Convert pH to [H 3 O + ], find the number of moles of carbonate and convert to mass. HCO 3 - ( aq ) + H 2 O( l ) CO 3 2- ( aq ) + H 3 O + ( aq ) K a = [CO 3 2- ][H 3 O + ] [HCO 3 - ] pH = 10.00; [H 3 O + ] = 1.0x10 -10 4.7x10 -11 = [CO 3 2- ](0.20) 1.0x10 -10 [CO 3 2- ] = 0.094M moles of Na 2 CO 3 = (1.5L)(0.094mols/L)= 0.14 = 15 g Na 2 CO 3 0.14 moles 105.99g mol

13 19.2 – Acid-base Titration Curves 13 pH Figure 19.5 Colors and approximate pH range of some common acid- base indicators.

14 14 Figure 19.7 Curve for a strong acid-strong base titration

15 19.2 – Acid-base Titration Curves 15 Figure 19.8 Curve for a weak acid-strong base titration Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH [HPr] = [Pr - ] pH = 8.80 at equivalence point pK a of HPr = 4.89 methyl red

16 19.2 – Acid-base Titration Curves 16 Sample Problem 19.3Calculating the pH During a Weak Acid- Strong Base Titration PROBLEM:Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; K a = 1.3x10 -5 ) after adding the following volumes of 0.1000M NaOH: (a) 0.00mL(b) 30.00mL(c) 40.00mL(d) 50.00mL PLAN:The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION:(a) Find the starting pH using the methods of Chapter 18. K a = [Pr - ][H 3 O + ]/[HPr][Pr - ] = x = [H 3 O + ] x = 1.1x10 -3 ; pH = 2.96 (b) Before addition Addition After addition 0.04000 0.03000 0.01000 0- - -0 - -- HPr( aq ) + OH - ( aq ) Pr - ( aq ) + H 2 O ( l )Amount (mol)

17 19.2 – Acid-base Titration Curves 17 Sample Problem 19.3Calculating the pH During a Weak Acid- Strong Base Titration continued [H 3 O + ] = 1.3x10 -5 0.001000 mol 0.003000 mol = 4.3x10 -6 MpH = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be (0.004000 mol) (0.004000L) + (0.004000L) = 0.05000M K a x K b = K w K b = K w /K a = 1.0x10 -14 /1.3x10 -5 = 7.7x10 -10 [H 3 O + ] = K w / = 1.6x10 -9 MpH = 8.80 (d) 50.00mL of NaOH will produce an excess of OH -. mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol M = 0.01111[H 3 O + ] = 1.0x10 -14 /0.01111 = 9.0x10 -11 M pH = 12.05

18 18 Figure 19.9 Curve for a weak base- strong acid titration Titration of 40.00mL of 0.1000M NH 3 with 0.1000M HCl pH = 5.27 at equivalence point pK a of NH 4 + = 9.25

19 19 pK a = 7.19 pK a = 1.85 Figure 19.10 Curve for the titration of a weak polyprotic acid. Titration of 40.00mL of 0.1000M H 2 SO 3 with 0.1000M NaOH

20 19.3 – Equilibria of Slightly Soluble Ionic Compounds 20 Sample Problem 19.4Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds SOLUTION: PROBLEM:Write the ion-product expression for each of the following: (a) Magnesium carbonate(b) Iron ( II ) hydroxide (c) Calcium phosphate(d) Silver sulfide PLAN:Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). K sp = [Mg 2+ ][CO 3 2- ](a) MgCO 3 ( s ) Mg 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Fe 2+ ][OH - ] 2 (b) Fe(OH) 2 ( s ) Fe 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 (c) Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3- ( aq ) (d) Ag 2 S( s ) 2Ag + ( aq ) + S 2- ( aq ) S 2- ( aq ) + H 2 O( l ) HS - ( aq ) + OH - ( aq ) Ag 2 S( s ) + H 2 O( l ) 2Ag + ( aq ) + HS - ( aq ) + OH - ( aq ) K sp = [Ag + ] 2 [HS - ][OH - ]

21 19.3 – Equilibria of Slightly Soluble Ionic Compounds 21 Sample Problem 19.5Determining K sp from Solubility PROBLEM:(a) Lead ( II ) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10 -3 g/100mL solution. What is the K sp of PbSO 4 ? (b) When lead ( II ) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64g/L. Calculate the K sp of PbF 2. PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. K sp = [Pb 2+ ][SO 4 2- ] = 1.40x10 -4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 = SOLUTION:PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq )(a) 1000mL L 4.25x10 -3 g 100mL soln 303.3g PbSO 4 mol PbSO 4 1.96x10 -8

22 19.3 – Equilibria of Slightly Soluble Ionic Compounds 22 Sample Problem 19.5Determining K sp from Solubility continued (b) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 = 2.6x10 -3 M K sp = (2.6x10 -3 )(5.2x10 -3 ) 2 = 0.64g L soln245.2g PbF 2 mol PbF 2 7.0x10 -8

23 19.3 – Equilibria of Slightly Soluble Ionic Compounds 23 Table 19.2 Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, FormulaK sp Aluminum hydroxide, Al(OH) 3 Cobalt ( II ) carbonate, CoCO 3 Iron ( II ) hydroxide, Fe(OH) 2 Lead ( II ) fluoride, PbF 2 Lead ( II ) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn( I O 3 ) 2 3 x 10 -34 1.0 x 10 -10 4.1 x 10 -15 3.6 x 10 -8 1.6 x 10 -8 4.7 x 10 -29 8 x 10 -48 Mercury ( I ) iodide, Hg 2 I 2 3.9 x 10 -6

24 19.3 – Equilibria of Slightly Soluble Ionic Compounds 24 Sample Problem 19.6Determining Solubility from K sp PROBLEM:Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5x10 -6. PLAN:Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ][OH - ] 2 -Initial Change Equilibrium - - 00 +S+ 2S S2S K sp = (S)(2S) 2 S == 1.2x10x -2 M Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration (M)

25 19.3 – Equilibria of Slightly Soluble Ionic Compounds 25 Table 19.3 Relationship Between K sp and Solubility at 25 0 C No. of IonsFormulaCation:AnionK sp Solubility (M) 2MgCO 3 1:13.5 x 10 -8 1.9 x 10 -4 2PbSO 4 1:11.6 x 10 -8 1.3 x 10 -4 2BaCrO 4 1:12.1 x 10 -10 1.4 x 10 -5 3Ca(OH) 2 1:25.5 x 10 -6 1.2 x 10 -2 3BaF 2 1:21.5 x 10 -6 7.2 x 10 -3 3CaF 2 1:23.2 x 10 -11 2.0 x 10 -4 3Ag 2 CrO 4 2:12.6 x 10 -12 8.7 x 10 -5

26 19.3 – Equilibria of Slightly Soluble Ionic Compounds 26 Figure 19.12 The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2+ ( aq ) + CrO 4 2- ( aq ) CrO 4 2- added

27 19.3 – Equilibria of Slightly Soluble Ionic Compounds 27 Sample Problem 19.7Calculating the Effect of a Common Ion on Solubility PROBLEM:In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5x10 -6. PLAN:Set up a reaction equation and table for the dissolution of Ca(OH) 2. The Ca(NO 3 ) 2 will supply extra [Ca 2+ ] and will relate to the molar solubility of the ions involved. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration(M) Initial Change Equilibrium - - - 0.100 +S+2S 0.10 + S2S K sp = 6.5x10 -6 = (0.10 + S)(2S) 2 = (0.10)(2S) 2 S << 0.10 S = = 4.0x10 -3 Check the assumption: 4.0% 0.10M 4.0x10 -3 x 100 =

28 19.3 – Equilibria of Slightly Soluble Ionic Compounds 28 Sample Problem 19.8Predicting the Effect on Solubility of Adding Strong Acid PROBLEM:Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of these ionic compounds: (a) Lead ( II ) bromide(b) Copper ( II ) hydroxide(c) Iron ( II ) sulfide PLAN:Write dissolution equations and consider how strong acid would affect the anion component. No effect. SOLUTION:(a) PbBr 2 ( s ) Pb 2+ ( aq ) + 2Br - ( aq ) (b) Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) OH - is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS( s ) Fe 2+ ( aq ) + S 2- ( aq )S 2- is the anion of a weak acid and will react with water to produce OH -. Both weak acids serve to increase the solubility of FeS. FeS( s ) + H 2 O( l ) Fe 2+ ( aq ) + HS - ( aq ) + OH - ( aq )

29 19.3 – Equilibria of Slightly Soluble Ionic Compounds 29 Sample Problem 19.9Predicting Whether a Precipitate Will Form PROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO 3 ) 2 is mixed with 0.200L of 0.060M NaF? PLAN:Write out a reaction equation to see which salt would be formed. Look up the K sp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION:CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = 3.2x10 -11 mol Ca 2+ = 0.100L(0.30mol/L) = 0.030mol[Ca 2+ ] = 0.030mol/0.300L = 0.10M mol F - = 0.200L(0.060mol/L) = 0.012mol[F - ] = 0.012mol/0.300L = 0.040M Q = [Ca 2+ ][F - ] 2 =(0.10)(0.040) 2 = 1.6x10 -4 Q is >> K sp and the CaF 2 WILL precipitate.

30 19.5 – Ionic Equilibria in Chemical Analysis 30 Sample Problem 19.12Separating Ions by Selective Precipitation SOLUTION: PROBLEM:A solution consists of 0.20M MgCl 2 and 0.10M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3x10 -10 ; K sp of Cu(OH) 2 is 2.2x10 -20. PLAN:Both precipitates are of the same ion ratio, 1:2, so we can compare their K sp values to determine which has the greater solubility. It is obvious that Cu(OH) 2 will precipitate first so we calculate the [OH - ] needed for a saturated solution of Mg(OH) 2. This should ensure that we do not precipitate Mg(OH) 2. Then we can check how much Cu 2+ remains in solution. Mg(OH) 2 ( s ) Mg 2+ ( aq ) + 2OH - ( aq ) K sp = 6.3x10 -10 Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) K sp = 2.2x10 -20 [OH - ] needed for a saturated Mg(OH) 2 solution = = 5.6x10 -5 M

31 19.5 – Ionic Equilibria in Chemical Analysis 31 Sample Problem 19.12Separating Ions by Selective Precipitation continued Use the K sp for Cu(OH) 2 to find the amount of Cu remaining. [Cu 2+ ] = K sp /[OH - ] 2 = 2.2x10 -20 /(5.6x10 -5 ) 2 =7.0x10 -12 M Since the solution was 0.10M CuCl 2, virtually none of the Cu 2+ remains in solution.

32 19.5 – Ionic Equilibria in Chemical Analysis 32 Figure 19.17 The general procedure for separating ions in qualitative analysis. Add precipitating ion Centrifuge Add precipitating ion Centrifuge

33 19.5 – Ionic Equilibria in Chemical Analysis 33 Figure 19.18 A qualitative analysis scheme for separating cations into five ion groups. Add 6M HCl Centrifuge Acidify to pH 0.5; add H 2 S Centrifuge Add NH 3 /NH 4 + buffer(pH 8) Centrifuge Add (NH 4 ) 2 HPO 4 Centrifuge


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