Ch. 16: Ionic Equilibria Buffer Solution An acid/base equilibrium system that is capable of maintaining a relatively constant pH even if a small amount.

Presentation on theme: "Ch. 16: Ionic Equilibria Buffer Solution An acid/base equilibrium system that is capable of maintaining a relatively constant pH even if a small amount."— Presentation transcript:

Ch. 16: Ionic Equilibria Buffer Solution An acid/base equilibrium system that is capable of maintaining a relatively constant pH even if a small amount of strong acid or base is added. a) Components of a buffer solution: a mixture of a weak acid and its conjugate base e.g. acetic acid & sodium acetate (HA & A – ) HA (aq) H + (aq) + A – (aq) or, ammonium chloride & ammonia (NH 4 + & NH 3 ) NH 4 + (aq) H + (aq) + NH 3(aq)

pH of a Buffer Solution (b) pH of buffer solution: (pH ≈ pK a of HA) HA (aq) H + + A – (aq) Initial “M” 0 “A” Change –x +x +x Equil M – x x A + x K a = [H + ][A – ] [HA] = x(A + x) (M – x) Since K a is usually small (< ~ 10 –4 ): M – x ≈ M and A + x ≈ A thus, K a ≈ x(A)/M or simply, [H + ] ≈ K a [HA] [A – ] (general buffer system) and [H + ] ≈ K a moles HA moles A –

Henderson-Hasselbach Eqn [H + ] ≈ K a [HA] [A – ] (general buffer system) Again assuming that x is small, you can take the log of both sides and rearrange to get the Henderson-Hasselbach equation; pH = pK a + log [acid] [base ] The HH equation is optional; if x is small approximation does not work, you need to use the more general method anyway...

Buffering Action (LeChatelier’s Principle) –The major equilibrium in a buffer system is: HA (aq) H + (aq) + A – (aq) 1. What if some acid [H + ] is added? Since H + is a product of above equilibrium, the reaction will shift in reverse, so that [HA] will increase and [A – ] will decrease But, as long as some A – remains, the system is still a mixture of HA and A –. Therefore, the general equation, [H + ] ≈ K a [HA]/[A – ] still applies. Thus, [H + ] is still ≈ K a.

Buffering Action (cont) 2. What if some base (OH-) is added instead? The added base (OH-) will neutralize some of the acid HA and produce more of the conjugate base A –. HA + OH – --> H 2 O + A – [HA] will decrease and [A-] will increase But, as long as some HA remains, the system is still a mixture of HA and A –, and ∴ still a buffer solution. buffer capacity: is the amount of acid or base you can add without causing a large change in pH. buffer range: the pH values for which a buffer system is the most effective. (usually ±1 pH unit of either side of pK a )

Sample Buffer Solution Problems 1.(a) The pK a of HF is 3.17. Calculate the pH of a 1.00-L solution that is 1.00 M HF and 1.50 M NaF. (b) What is the pH of this solution after addition of 50.0 mL of 10.0 M HCl? 2.What mass of solid NaCN must be added to 500 mL (3 sig fig) of 0.50 M HCN to prepare a buffer solution with a pH of 9.50? pK a for HCN is 9.21.

Sample Buffer Solution Problems 1.(a) The pK a of HF is 3.17. Calculate the pH of a 1.00-L solution that is 1.00 M HF and 1.50 M NaF. (b) What is the pH of this solution after addition of 50.0 mL of 10.0 M HCl? Answer: (a) pH = 3.35 (b) pH = 2.99 2.What mass of solid NaCN must be added to 500 mL (3 sig fig) of 0.50 M HCN to prepare a buffer solution with a pH of 9.50? pK a for HCN is 9.21. Answer: 24 g NaCN

Another Sample Problem Write the balanced chemical equation for the important equilibrium that is occurring in an aqueous solution of NaNO 2 and HNO 2. Write the appropriate equilibrium constant expression for the reaction. Determine the pH of a solution of 0.100 M of each compound. (K a HNO 2 = 4.6 x 10 – 4 )

Another Sample Problem Write the balanced chemical equation for the important equilibrium that is occurring in an aqueous solution of NaNO 2 and HNO 2. HNO 2(aq) H + (aq) + NO 2 – (aq) Write the appropriate equilibrium constant expression for the reaction. K a = [H + ][NO 2 – ]/[HNO 2 ] Determine the pH of a solution of 0.100 M of each compound. (K a HNO 2 = 4.6 x 10 – 4 ) 3.34

Titrations TitrationAn unknown amount of one reactant is combined exactly with a precisely measured volume of a standard solution of the other. End-pointWhen exactly stoichiometric amounts of two reactants have been combined. IndicatorSubstance added to aid in detection of the endpoint (usually via a color change)

Example Problem Vinegar is an aqueous solution of acetic acid, HC 2 H 3 O 2. A 12.5 mL sample of vinegar was titrated with a 0.504 M solution of NaOH. The titration required 15.40 mL of the base solution in order to reach the endpoint. What is the molar concentration of HC 2 H 3 O 2 in vinegar? NaOH (aq) + HC 2 H 3 O 2(aq) --> NaC 2 H 3 O 2(aq) + H 2 O (l) (15.40 mL NaOH) x x = 0.007762 mol HC 2 H 3 O 2 M = = 0.621 M acetic acid 0.504 mol NaOH 1000 mL NaOH soln 1 mole HC 2 H 3 O 2 1 mole NaOH 0.007762 mol HC 2 H 3 O 2 0.0125 L vinegar

Titration of a Weak Acid by a Strong Base e.g. 25.0 mL of 0.200 M HA (acetic acid) is titrated with 0.200 M NaOH (K a = 1.8 x 10 –5 )

Example e.g. 25.0 mL of 0.200 M HA (acetic acid) is titrated with 0.200 M NaOH. Determine pH in each of four regions: (a) at the start -- before any NaOH is added Just a simple weak acid: pH = 2.72 (b) before the equivalence point: The acid is partially neutralized by the added OH – HA + OH – --> H 2 O + A – Since both HA and A – are now present, the mixture is a buffer solution ! and [H + ] ≅ K a (mole HA left) / (mole A – produced) e.g. after addition of 10.0 mL of NaOH, pH = 4.57

Example, cont. (c) at the equivalence point: The acid (HA) is completely neutralized, and only the anion A – is present ∴ the solution is now the salt of a weak acid! Total volume is now 50.0 mL, so [A – ] = 0.10 M and pH = 8.88 (d) after the equivalence point: All acid (HA) is gone and excess OH – is present ∴ the pH is determined by how much OH – is left after all the acid has been neutralized e.g. after 40.0 mL of NaOH added: pH = 12.66

Titration Curve

16 Indicators: Phenolphthalein

Choosing an Indicator

SUMMARY of Acid-Base Equilibrium Concepts (Chapters 15 and 16) 1.pH scale and the autoionization of water (K w ) 2.Strong acids and bases 3.Weak acids and bases (K a or K b ) 4.Salts of weak acids and salts of weak bases (conjugate pairs: K w ≅ K a K b ) 5.Polyprotic acids (K a1 and K a2 etc) 6.Buffer solutions (K a ) (mixture of weak acid and its conjugate base) 7.Acid-Base Titrations (especially weak acid by strong base, or weak base by strong acid)

Sample Problems 1. A 0.625 g sample of an unknown weak acid (call it HA for short) is dissolved in enough water to make 25.0 mL of solution. This weak acid solution is then titrated with 0.100 M NaOH, and 45.0 mL of the NaOH solution is required to reach the equivalence point. Using a pH meter, the pH of the solution at the equivalence point is found to be 8.25. (a) Determine the molecular mass of the unknown acid. (b) Determine the pK a value of the unknown acid. 2. The pK b value for ammonia (NH 3 ) is 4.74. Calculate the mass (in grams) of solid ammonium chloride (NH 4 Cl, 53.5 g/mole) that must be added to 300 mL of 0.25 M NH 3 solution to make a buffer solution with a pH equal to 10.00.

Sample Problems 1. A 0.625 g sample of an unknown weak acid (call it HA for short) is dissolved in enough water to make 25.0 mL of solution. This weak acid solution is then titrated with 0.100 M NaOH, and 45.0 mL of the NaOH solution is required to reach the equivalence point. Using a pH meter, the pH of the solution at the equivalence point is found to be 8.25. (a) Determine the molecular mass of the unknown acid. (b) Determine the pK a value of the unknown acid. Answer: (a) 139 g/mol HA (b) pK a = 3.70 2. The pK b value for ammonia (NH 3 ) is 4.74. Calculate the mass (in grams) of solid ammonium chloride (NH 4 Cl, 53.5 g/mole) that must be added to 300 mL of 0.25 M NH 3 solution to make a buffer solution with a pH equal to 10.00. Answer: 0.73 g NH 4 Cl

Solubility and Complex Ion Equilibria Solubility Equilibria (a) “Insoluble” salts are actually sparingly soluble equilibrium: solid salt ions in solution e.g. in a saturated solution of Ag 2 CO 3, the following equilibrium is occurring. Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2– (aq) Solubility Product Constant = K sp K sp = [Ag + ] 2 [CO 3 2– ] = 8.1 x 10 –12

Molar Solubility Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2– (aq) K sp = [Ag + ] 2 [CO 3 2– ] = 8.1 x 10 –12 Given the K sp value, determine the “molar solubility” Let x = molar solubility ! = moles Ag 2 CO 3 dissolved/liter [CO 3 2– ] = x and [Ag + ] = 2x K sp = 8.1 x 10 –12 = [Ag + ] 2 [CO 3 2– ] = (2x) 2 (x) K sp = 4x 3 ∴ x = 1.3 x 10 –4 M

Common Ion Effect Common ion effect: Sparingly soluble salts are even less soluble when a “common” ion is present. Le Châtelier’s Principle applies here: salt cation + anion If an excess of one of the ions is present, the equilibrium will shift in reverse and less salt will dissolve!

Example Problem Molar solubility of Ag 2 CO 3 in 0.10 M Na 2 CO 3 ? Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2– (aq) K sp = 8.1 x 10 –12 Let x = molar solubility of Ag 2 CO 3 [Ag + ] = 2x and [CO 3 2– ] = 0.10 + x K sp = 8.1 x 10 –12 = [Ag + ] 2 [CO 3 2– ] = (2x) 2 (0.10 + x) But since Ag 2 CO 3 is not very soluble, assume x << 0.1 K sp ≈ (2x) 2 (0.10) ≈ 8.1 x 10 –12 ∴ x ≈ 4.5 x 10 –6 M

Precipitation When will a precipitate form? The solution must be saturated for a ppt to form. Thus, the initial ion concentrations must be greater than the equilibrium values. e.g. Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 2– (aq) If initial concentrations of the ions are given, calculate the ion product: Q = [Ag + ] 2 [CO 3 2– ] using the initial conc values If Q > K sp then a precipitate will form! In some cases where Q > K sp there is no precipitate unless it is disturbed; these are supersaturated.

precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added Precipitation

Sample Problem A saturated solution of Mg 3 (PO 4 ) 2 has a concentration of 0.939 mg per liter. Calculate the solubility product constant for Mg 3 (PO 4 ) 2.

Sample Problem A saturated solution of Mg 3 (PO 4 ) 2 has a concentration of 0.939 mg per liter. Calculate the solubility product constant for Mg 3 (PO 4 ) 2. Answer: K sp = 6.28 x 10 – 26

Qualitative analysis 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 an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme 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

30 Qualitative Analysis

Multiple Equilibria (d) “Insoluble” salts of weak acids are more soluble in acidic solutions e.g. Ag 2 S (s) 2 Ag + (aq) + S 2– (aq) If the solution is acidic, some S 2– will be converted to HS – and H 2 S, thus causing more Ag 2 S to dissolve The process can be viewed as a combination of multiple equilibria, occurring simultaneously: Ag 2 S (s) 2 Ag + (aq) + S 2– (aq) H + (aq) + S 2– (aq) HS – (aq) H + (aq) + HS – (aq) H 2 S (aq) Net: Ag 2 S (s) + 2 H + (aq) 2 Ag + (aq) + H 2 S (aq)

Complex Ion Equilibria 2. Complex Ion Equilibria (increases solubility) Metal ions form complex ions with Lewis bases, such as: Cu 2+ (aq) + 4 NH 3(aq) Cu(NH 3 ) 4 2+ (aq) The reaction is generally very favorable so…K is very large! The equilibrium constant is called a Formation Constant K f = [Cu(NH 3 ) 4 2+ ] [Cu 2+ ][NH 3 ] 4 = 1.1 x 10 13 (large!) Another example: Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K f = [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 = 1.6 x 10 7 If a complex ion can form, then an “insoluble” salt will be much more soluble (Le Châtelier’s Principle again!).

Example Problem Given the above formation constant for Ag(NH 3 ) 2 + and the K sp value for AgBr of 5.0 x 10 –13, calculate the molar solubility of AgBr in (a) water, and (b) 0.10 M NH 3. (a) in water: AgBr (s) Ag + (aq) + Br – (aq) K sp = [Ag + ][Br – ] = x 2 = 5.0 x 10 – 13 ∴ x = molar solubility = 7.1 x 10 – 7 M

Example Problem, cont. (b) in NH 3 solution, two equilibria must be considered: K sp : AgBr (s) Ag + (aq) + Br – (aq) K f : Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K net : AgBr (s) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) + Br – (aq) K net = [Ag(NH 3 ) 2 + ][Br – ]/[NH 3 ] 2 = K sp K f = (5.0 x 10 – 13 )(1.6 x 10 7 ) = 8.0 x 10 – 6 Let x = molar solubility of AgBr [Ag(NH 3 ) 2 + ] = [Br – ] = x [NH 3 ] = 0.10 – 2x K net = x 2 /(0.10 – 2x) 2 = 8.0 x 10 – 6 Since K net is small in this case, assume 2x << 0.10 x 2 / (0.10) 2 ≈ 8.0 x 10 – 6 ∴ x ≈ 2.8 x 10 – 4 Thus, AgBr is about 1,000 times more soluble in NH 3 solution than in water alone.

Amphoteric Metal Hydroxides All metal hydroxides become more soluble in acidic solution (like Ag 2 S), but some are also more soluble in basic solution, e.g. Al(H 2 O) 6 3+ (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) addition of OH − drives the equilibrium to the right and continues to remove H + from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq) Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq) Al(H 2 O) 3 (OH) 3(s) + H 2 O (l) Others include Cr 3+, Zn 2+, Pb 2+, Sb 2+ …

Sample Problem Silver chromate, Ag 2 CrO 4, is an “insoluble” substance with a K sp value of 1.2 x 10 – 12. Silver ion forms a stable complex ion with cyanide ion that has the formula Ag(CN) 2 – and a formation constant (K f ) of 5.3 x 10 18. Calculate the molar solubility of Ag 2 CrO 4 in each of the following solutions. Write balanced chemical equations for any important equilibrium reactions that are occurring. (a) in water (b) in 2.00 M Na 2 CrO 4 (c) in 2.00 M NaCN

Answers (a) Ag 2 CrO 4 2 Ag + (aq) + CrO 4 2– (aq) 6.7 x 10 – 5 M (b) Ag 2 CrO 4 2 Ag + (aq) + CrO 4 2– (aq) 3.9 x 10 – 7 M (c) Ag 2 CrO 4(s) 2 Ag + (aq) + CrO 4 2– (aq) 2 Ag + (aq) + 4 CN – (aq) 2 Ag(CN) 2 – (aq) Ag 2 CrO 4(s) + 4 CN – (aq) 2 Ag(CN) 2 – (aq) + CrO 4 2– (aq) molar solubility ≈ 0.50 M

Download ppt "Ch. 16: Ionic Equilibria Buffer Solution An acid/base equilibrium system that is capable of maintaining a relatively constant pH even if a small amount."

Similar presentations