 # Chapter 15 Applications of Aqueous Equilibria Addition of base: Normal human blood pH is 7.4 and has a narrow range of about +/- 0.2.

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Chapter 15 Applications of Aqueous Equilibria Addition of base: Normal human blood pH is 7.4 and has a narrow range of about +/- 0.2

When an acid or base is added to water, the pH changes drastically. A buffer solution resists a change in pH when an acid or base is added. Buffer Solutions Buffers: Mixture of a weak acid and its salt Mixture of a weak base and its salt pH = 4 pH = 10.5 pH = 6.9 pH = 7.1

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion Understanding Buffer Solutions

Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid] pH of a Buffer:

Using Henderson-Hasselbalch equation pH = pK a + log [HCOO - ] [HCOOH] HCOOH : K a = 1.67 X 10 -4, pK a = 3.78 pH = 3.77 + log [0.52] [0.30] = 4.02 What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (K a = 1.67 X 10 -4 ).

A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution

Blood Plasma pH = 7.4

Buffer Solutions02

= 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? (K a = 5.62 X 10 -10 ). NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) Initial (moles) End (moles) 0.36 X0.0 800 =0.0290.050 x 0.0200 = 0.001 0.30 X.0800 = 0.024 0.029-0.001 = 0.0280.001-0.001 =0.0 0.024 + 0.001 = 0.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL, 0.1 L [NH 3 ] = 0.025 0.10 -0.001 +0.001 Change

Titrations Slowly add base to unknown acid UNTIL The indicator changes color (pink)

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

Acid–Base Titration Curves01

Strong Acid-Strong Base Titrations Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

Strong Acid-Strong Base Titrations 1.Before Addition of Any NaOH 0.100 M HCl

2.Before the Equivalence Point Strong Acid-Strong Base Titrations 2H 2 O(l)H 3 O 1+ (aq) + OH 1- (aq) 100% Excess H 3 O 1+

3.At the Equivalence Point Strong Acid-Strong Base Titrations The quantity of H 3 O 1+ is equal to the quantity of added OH 1-. pH = 7

4.Beyond the Equivalence Point Strong Acid-Strong Base Titrations Excess OH 1-

Strong Acid-Strong Base Titrations

Weak Acid-Strong Base Titrations

1.Before Addition of Any NaOH 0.100 M CH 3 CO 2 H H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l)

Weak Acid-Strong Base Titrations 2.Before the Equivalence Point H 2 O(l) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + OH 1- (aq) 100% Excess CH 3 CO 2 H Buffer region

Weak Acid-Strong Base Titrations 3.At the Equivalence Point OH 1- (aq) + CH 3 CO 2 H(aq)CH 3 CO 2 1- (aq) + H 2 O(l) Notice how the pH at the equivalence point is shifted up versus the strong acid titration. pH > 7

Notice that the strong and weak acid titration curves correspond after the equivalence point. Weak Acid-Strong Base Titrations 4.Beyond the Equivalence Point Excess OH 1-

Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): H + (aq) + NH 3 (aq) NH 4 + (aq)

Exactly 100 mL of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

Polyprotic Acids: Titration of Diprotic Acid + Strong Base

Titration of a Triprotic Acid: H 3 PO 4 + Strong Base

Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Fractional Precipitation01

Solubility Equilibria AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2 - (aq) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 4 3 - ] 2 Dissolution of an ionic solid in aqueous solution: Q (IP) = K sp Saturated solution Q (IP) < K sp Unsaturated solution No precipitate Q (IP) > K sp Supersaturated solution Precipitate will form K sp = 1.8 x 10 -10 K sp : Solubility Product (IP) : Ion product

Measuring K sp and Calculating Solubility from K sp

Molar solubility (mol/L): Number of moles of a solute that dissolve to produce a litre of saturated solution Solubility (g/L) is the number of grams of solute that dissolve to produce a litre of saturated solution

Measuring K sp If the concentrations of Ca 2+ (aq) and F 1- (aq) in a saturated solution of calcium fluoride are known, K sp may be calculated. Ca 2+ (aq) + 2F 1- (aq)CaF 2 (s) K sp = [Ca 2+ ][F 1- ] 2 = (3.3 x 10 -4 )(6.7 x 10 -4 ) 2 = 1.5 x 10 -10 [F 1- ] = 6.7 x 10 -4 M[Ca 2+ ] = 3.3 x 10 -4 M (at 25 °C)

What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.8 x 10 -10 Calculating Solubility from K sp

Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Fractional Precipitation01

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 4.7 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.100 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form If this solution also contained 0.100 M AlCl3, Aluminum could be precipitated, with Calcium remaining in the solution.

1) Use the difference between Ksp’s to separate one precipitate at a time. 2) Use Common Ion effect to Precipitate one Ion at a time Fractional Precipitation01

The Common-Ion Effect and Solubility01

Factors That Affect Solubility Mg 2+ (aq) + 2F 1- (aq)MgF 2 (s) Solubility and the Common-Ion Effect

1. Use the difference between Ksp’s to separate one precipitate at a time. 2. Use Common Ion effect to Precipitate one Ion at a time 3. Use pH to Increase or decrease solubility and separate one ion at a time Fractional Precipitation01

Factors That Affect Solubility Solubility and the pH of the Solution Ca 2+ (aq) + HCO 3 1- (aq) + H 2 O(l)CaCO 3 (s) + H 3 O 1+ (aq)

1. Use the difference between Ksp’s to separate one precipitate at a time. 2. Use Common Ion effect to Precipitate one Ion at a time 3. Use pH to Increase or decrease solubility and separate one ion at a time. 4. Use complex formation to prevent certain Ions to precipitate Fractional Precipitation01

Factors That Affect Solubility Solubility and the Formation of Complex Ions Ag(NH 3 ) 2 1+ (aq)Ag(NH 3 ) 1+ (aq) + NH 3 (aq) Ag(NH 3 ) 2 1+ (aq)Ag 1+ (aq) + 2NH 3 (aq)K f = 1.7 x 10 7 K 1 = 2.1 x 10 3 Ag(NH 3 ) 1+ (aq)Ag 1+ (aq) + NH 3 (aq) K 2 = 8.1 x 10 3 K f is the formation constant. How is it calculated? (at 25 °C)

Factors That Affect Solubility Solubility and the Formation of Complex Ions = (2.1 x 10 3 )(8.1 x 10 3 ) = 1.7 x 10 7 [Ag(NH 3 ) 2 1+ ] [Ag 1+ ][NH 3 ] 2 K 1 = [Ag(NH 3 ) 1+ ] [Ag 1+ ][NH 3 ] K 2 = [Ag(NH 3 ) 2 1+ ] [Ag(NH 3 ) 1+ ][NH 3 ] K f = K 1 K 2 =

Factors That Affect Solubility Solubility and the Formation of Complex Ions Ag(NH 3 ) 2 1+ (aq)Ag 1+ (aq) + 2NH 3 (aq)

Factors That Affect Solubility Solubility and Amphoterism Al(OH) 4 1- (aq)Al(OH) 3 (s) + OH 1- (aq) Al 3+ (aq) + 6H 2 O(l)Al(OH) 3 (s) + 3H 3 O 1+ (aq) In base: In acid: Aluminum hydroxide is soluble both in strongly acidic and in strongly basic solutions.

Factors That Affect Solubility Solubility and Amphoterism

The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? AgBr (s) Ag + (aq) + Br - (aq) K sp = 5.4 x 10 -13 s 2 = K sp s = 7.3 x 10 -7 NaBr (s) Na + (aq) + Br - (aq) [Br - ] = 0.0010 M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = 0.0010 + s  0.0010 K sp = 0.0010 x s s = 5.4 x 10 -10 AgBr (s) Ag + (aq) + Br - (aq) M M

The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solution. pH and Solubility

Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 5.6 x 10 -12 K sp = (s)(2s) 2 = 4s 3 4s 3 = 5.6 x 10 -12 s = 1.1 x 10 -4 M [OH - ] = 2s = 2.2 x 10 -4 M pOH = 3.65 pH = 10.35 At pH less than 10.35 Lower [OH - ], Increase solubility of Mg(OH) 2 OH - (aq) + H + (aq) H 2 O (l) remove At pH greater than 10.35 Raise [OH - ] add Decrease solubility of Mg(OH) 2 What is the pH of a solution containing Mg(OH) 2 ? what pH Would reduce the solubility of this precipitate?

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- K f = [CoCl 4 ] [Co 2+ ][Cl - ] 4 2- The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. Co(H 2 O) 6 2+ CoCl 4 2- KfKf stability of complex

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Qualitative Analysis

Qualitative Analysis of Cations

Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Fractional Precipitation01

Flame Test for Cations lithium sodium potassiumcopper Nessler's is K 2 [HgI 4 ] in a basic solution, [HgI 4 ] 2- reacts with the NH4 + to form a yellow brown precipitate.

Fractional Precipitation01

Separation of Ions by Selective Precipitation MS(s) + 2 H 3 O + (aq) ae M 2+ (aq) + H 2 S(aq) + 2 H 2 O(l) (see page 631) : K spa = [Cd +2 ] [H 2 S ] / [H 3 O + ] 2 K spa = For ZnS, K spa = 3 x 10 ­-2 ; for CdS, K spa = 8 x 10 ­-7 [Cd +2 ] = [Zn +2 ] = 0.005 M Because the two cation concentrations are equal, Q spa is :[Cd +2 ] [H 2 S ] / [H 3 O + ] 2 =(0.005).(0.1)/ (0.3) 2 Q spa = 6 x 10 ­-3 For CdS, Q spa > K spa ; CdS will precipitate. For ZnS, Q spa < K spa ; Zn +2 will remain in solution. Determine whether Cd +2 can be separated from Zn +2 by bubbling H 2 S through ([H 2 S ] = 0.1 ) a 0.3 M HCl solutions that contains 0.005 M Cd +2 and 0.005 M Zn +2 ?

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (K a = 1.67 X 10 -4 ). HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 Mixture of weak acid and conjugate base! X (0.52 + X) 0.30 - X = 1.67 X 10 -4 X = 9.77 X 10 -5 = [ H 3 O + ] pH = - log ( 9.77 X 10 -5 ) = 4.01

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