Applications of Aqueous Equilibria Chapter 8 Web-site:

Applications of Aqueous Equilibria– ch Which of the following combinations will result in a buffer solution upon dissolving in 1 L of water? Calculate the pH of the solutions that are buffers? a. 0.1 mol HClO 3 and 0.1 mol NaClO 3 b. 0.5 mol HF and 0.8 mol NaF c mol NH 4 I and 0.02 mol NH 3

Applications of Aqueous Equilibria– ch. 8

Buffers are classic examples of Le Chatlier’s Principle pH<pKapH=pKa pH>pKa

Applications of Aqueous Equilibria– ch A 65.0-mL sample of 0.12 M HNO 2 (K a = 4.0 x 10 –4 ) is titrated with 0.11 M NaOH. What is the pH after 28.4 mL of NaOH has been added? A) B) 7.00 C) 3.57 D) 3.00 E) 3.22

Applications of Aqueous Equilibria– ch. 8 Generic Acid/Base Titration Reactions: 1. Strong Acid and Strong Base H + + OH –  H 2 O 2. Weak Acid and Strong Base HA + OH –  H 2 O + A – 3. Strong Acid and Weak Base H + + A –  HA

Applications of Aqueous Equilibria– ch. 8

3. A 84.0-mL sample of 0.04 M KNO 2 (K a for HNO 2 = 4.0 x 10 –4 ) is titrated with 0.2 M HI. What is the pH after 13.6 mL of HI has been added?

Applications of Aqueous Equilibria– ch A 63.0 mL sample of 0.11 M HCN (K a = 6.2 x 10 –10 ) is titrated with 0.06 M KOH. What is the pH at the equivalence point?

Applications of Aqueous Equilibria– ch A 75.0 mL of 0.22M NH 3 (K b = 1.8 x 10 –5 ) is titrated with 0.50M HCl. What is the pH at the equivalence point?

Applications of Aqueous Equilibria– ch A 15.0 mL of 0.40M HF (K a = 7.2 x 10 –4 ) is titrated with 0.85M LiOH. What is the pH after 8 mL of LiOH has been added?

Applications of Aqueous Equilibria– ch A 120 mL of 0.14M NaCH 3 COO is titrated with 0.36 M HCl. What is the pH after 50 mL of HCl has been added?

Applications of Aqueous Equilibria– ch Draw the titration curves for (1) a weak acid being titrated by a strong base and (2) a weak base being titrated by a strong acid and (3) label the following points: a. the equivalence point (predict if pH = 7, pH > 7 or pH <7) b. the region with maximum buffering c. where pH = pK a d. the buffer region e. where the pH only depends on [HA] f. where the pH only depends on [A - ]

Applications of Aqueous Equilibria– ch. 8 Volume of strong base pH >7 Titration Curves Titration of a weak acid with a strong base Titration of a weak base with a strong acid Volume of strong acid pH <7 Buffer Zone

Applications of Aqueous Equilibria– ch A solution contains 0.36 M HA (K a = 2.0 x ) and 0.24 M NaA. Calculate the pH after 0.04 mol of NaOH is added to 1.00 L of this solution. a b c d e. 6.70

Applications of Aqueous Equilibria– ch How many moles of NaCN must be added to 1.2 L solution that is 0.6 M in HCN in order to get a pH of 10.1?

Applications of Aqueous Equilibria– ch Determine the solubility in mol/L and g/L for the following compounds: a. BaCO 3 (K sp = 1.6 x ) b. Ag 2 S (K sp = 2.8 x )

Applications of Aqueous Equilibria– ch. 8 Solubility ⇒ the maximum amount of solute that can dissolve into a given amount of solvent at any one temperature at this point the solution is said to be saturated and in equilibrium

Applications of Aqueous Equilibria– ch Determine the K sp values for the following compounds: a. Ag 3 PO 4 (solubility = 1.8 x mol/L) b. MgF 2 (solubility = g/L)

Applications of Aqueous Equilibria– ch What is the solubility of Ca 3 (PO 4 ) 2 (K sp = 1 x ) in 0.02M solution of Na 3 PO 4 ?

Applications of Aqueous Equilibria– ch What is the solubility of Zn(OH) 2 (K sp = 4.5 x ) in a solution with pH of 11? Is Zn(OH) 2 more soluble in acidic or basic solutions?

Applications of Aqueous Equilibria– ch Will BaCrO 4 (K sp = 8.5 x ) precipitate when 200 mL of 1 x M Ba(NO 3 ) 2 is mixed with 350 mL of 3 x M KCrO 4 ?

Applications of Aqueous Equilibria– ch. 8 You have completed ch. 8

1. What is the pH at the equivalence point for the following titrations? a. NaOH (strong base) with HBr (strong acid) pH = 7 pH > 7 pH < 7 b. HCl (strong acid) with NaCH 3 COO (weak base) pH = 7 pH > 7 pH < 7 c. KOH (strong base) with HCN (weak acid) pH = 7 pH > 7 pH < 7 Answer Key – ch. 8

2. Which of the following will result in a buffer solution upon mixing? a. 0.1 mol HClO 3 and 0.1 mol NaClO 3 are put into 1L of water Strong acid and it’s conjugate base is not a buffer b. 0.5 mol H 2 S and 0.8 mol NaHS are put into 1L of water weak acid and it’s conjugate base is a buffer c mol KHC 2 O 4 and 0.02 mol K 2 C 2 O 4 are put into 1L of water weak acid and it’s conjugate base is a buffer d. 0.1 mol LiOH and 0.2 mol H 3 PO 4 are put into 1L of water strong base and weak acid will be a buffer as long as the strong base is the limiting reagent e. 0.1 mol HNO 3 and 0.04 mol NH 3 are put into 1 L of water strong acid and weak base could be a buffer if the strong as is limiting – however in this case the weak base is limiting so no buffer Answer Key – ch. 8

3. Which of the following 50 mL solutions can absorb the most acid without changing the pH? a. 0.4 M HF /0.5 M NaF b. 0.8 M HF /0.7 M NaF c. 0.3 M HF/0.7 M NaF Answer Key – ch. 8 In order for a buffer to absorb an acid you want [base] to be high and the [acid] to be low and vice versa if absorbing base

Answer Key – ch. 8 … continue to next slide

4. b. …continued n A - ⇒ (80 mL)(0.1 mol/L) = 8 mmol n HA ⇒ (30 mL)(0.2 mol/L) = 6 mmol pH = - log(1.9 x 10 –5 ) + log(8/6) pH = 4.8 Answer Key – ch. 8

5. Calculate the pH for the following: In all 3 scenarios we’re adding a strong acid (HCl) to a salt with a weak base (NO 2 – ) ⇒ since we have a strong substance (HCl) we can assume the neutralization reaction goes to completion ⇒ work stoichiometrically in moles a. 25 mL of 0.1 M HCl /25 mL of 0.2 M KNO 2 H+H+ NO 2 –  HNO 2 I2.550 ∆ F mmol H + 5 mmol NO 2 – Limiting Reagent

5. b. 50 mL of 0.1 M HCl /25 mL of 0.2 M KNO 2 Answer Key – ch. 8 H+H+ NO 2 –  HNO 2 I550 ∆-5 +5 F005 5 mmol H + 5 mmol NO 2 – Note ⇒ Equivalence Point HNO 2 H2OH2O ⇌ H3O+H3O+ NO 2 – I0.067N/A00 ∆-xN/A+x E0.067N/Axx

Answer Key – ch. 8 c. 60 mL of 0.1 M HCl/25 mL of 0.2 M KNO 2 H+H+ NO 2 –  HNO 2 I650 ∆-5 +5 F105 6 mmol H + 5 mmol NO 2 – Limiting Reagent

Answer Key – ch Calculate the pH of the following: In all 3 scenarios we’re adding a strong base (OH – ) to a weak acid (HClO) ⇒ since we have a strong substance (OH – ) we can assume the neutralization reaction goes to completion ⇒ work stoichiometrically in moles a. 40 mL of 0.2 M KOH/40 mL of 0.5 M HClO 8 mmol OH – 20 mmol HClO OH – HClO  H2OH2OClO – I820N/A0 ∆- 8 N/A+8 F012N/A8 Limiting Reagent

Answer Key – ch b. 80 mL of 0.25 M KOH/40 mL of 0.5 M HClO 20 mmol OH – 20 mmol HClO OH – HClO  H2OH2OClO – I20 N/A0 ∆- 20 N/A20 F00N/A20 Note ⇒ Equivalence Point ClO – H2OH2O ⇌ OH – HClO I0.167N/A00 ∆- xN/A+x F0.167N/Axx

Answer Key – ch c. 100 mL of 0.25 M KOH/40 mL of 0.5 M HClO 25 mmol OH – 20 mmol HClO OH – HClO  H2OH2OClO – I2520N/A0 ∆- 20 N/A20 F50N/A20 Limiting Reagent

Answer Key – ch Calculate the pH of the following: a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M CH 3 CH 2 COOH (pKa = 4.87) and 0.6 M CH 3 CH 2 COONa. A strong acid (HCl) will react with the base in the buffer (CH 3 CH 2 COO – ) CH 3 CH 2 COO – H+H+  CH 3 CH 2 COOH I mmol ∆ F50060

Answer Key – ch. 8 7 b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M CH 3 CH 2 COOH and 0.6 M CH 3 CH 2 COONa. The strong base (OH – ) will react with the acid in the buffer (CH 3 CH 2 COOH) CH 3 CH 2 COOHOH -  CH 3 CH 2 COO – I50 mmol4 mmol60 mmol ∆ F46064

Answer Key – ch. 8

9. How many grams of NaOH need to be added to a 50 mL of 0.3 M HNO 2 (K a = 4.0 x ) to have a solution with a pH = 4? HNO 2 OH -  H2OH2ONO 2 - I15mmolx mmolN/A0 Δ-x N/A+x F15-x0N/Ax

Answer Key – ch Determine the solubility in mol/L and g/L for the following compounds: a. BaCO 3 (K sp = 1.6 x ) ⇒ the solubility is defined as the maximum amount of solute that can be dissolved in a particular amount of solvent at any one temperature or the saturation point ⇒ the solute is at equilibrium for saturated solutions BaCO 3 ⇌ Ba 2+ CO 3 2– IN/A00 ∆ +x EqN/Axx Use K sp to solve for x 1.6 x = (x)(x) x = 4 x 10 –5 since the molar ratio is 1:1 the molar solubility of BaCO 3 = 4 x 10 –5 mol/L or solubility = (4 x 10 –5 mol/L)( g/mol) = g/L

Answer Key – ch b. Ag 2 S (K sp = 2.8 x ) Ag 2 S ⇌ 2 Ag + S 2– IN/A00 ∆ +2x+x EqN/A2xx Use K sp to solve for x 2.8 x = (2x) 2 (x) x = 4.14 x 10 –17 Molar solubility = 4.14 x 10 –17 M or solubility = (4.14 x 10 –17 mol/L)(247.8 g/mol) = 1 x 10 –14 g/L

Answer Key – ch Determine the K sp values for the following compounds: a. Al(OH) 3 (solubility = 5 x mol/L) ⇒ the solubility tells us how much will dissolve in order to get to equilibrium Al(OH) 3 ⇌ Al 3+ 3 OH – IN/A00 ∆ + 5 x (5 x ) EqN/A5 x x K sp = (5 x )(1.5 x ) 3 = 1.7 x 10 –32

Answer Key – ch b. MgF 2 (solubility = g/L) ⇒ first we need the molar solubility ( g/L)/(62.31 g/mol) = mol/L MgF 2 ⇌ Mg 2+ 2 F – IN/A00 ∆ ( ) EqN/A K sp = ( )(0.0024) 2 = 6.6 x 10 –9

Answer Key – ch What is the solubility of Ca 3 (PO 4 ) 2 (K sp = 1 x ) in 0.02M solution of Na 3 PO 4 ? The solute and the solution have the phosphate ion in common ⇒ this will result in lower solubility than if the solute were to be dissolved in pure water aka the common ion effect Ca 3 (PO 4 ) 2 ⇌ 3 Ca 2+ 2 PO 4 3- IN/A00.02 ∆N/A+ 3x+2x EqN/A3x x Insignificantly small Use K sp to solve for x 1 x = (3x) 3 (0.02) 2 x = 4.5 x 10 –18 molar solubility = 4.5 x 10 –18 M

Answer Key – ch What is the solubility of Zn(OH) 2 (K sp = 4.5 x ) in a solution with pH of 11? The solute and the solution have the hydroxide ion in common ⇒ this will result in lower solubility than if the solute were to be dissolved in pure water aka the common ion effect ⇒ since the pH is 11 the pOH is 3 ⇒ [OH - ] = 10 –3 M Zn(OH) 2 ⇌ Zn 2+ 2 OH - IN/A010 –3 M ∆N/A+ x+2x EqN/Ax10 –3 + 2x Insignificantly small Use K sp to solve for x 4.5 x = (x)(10 –3 ) 2 x = 4.5 x molar solubility = 4.5 x M

Answer Key – ch Will BaCrO 4 (K sp = 8.5 x ) precipitate when 200 mL of 1 x M Ba(NO 3 ) 2 is mixed with 350 mL of 3 x M KCrO 4 ? The solution is saturated with BaCrO 4 if [Ba 2+ ][CrO 4 2– ] = 8.5 x however if [Ba 2+ ][CrO 4 2– ] > 8.5 x the solution is supersaturated and will precipitate out some BaCrO 4 or if [Ba 2+ ][CrO 4 2– ] < 8.5 x the solution is unsaturated and there will be no noticeable change We can use M 1 V 1 = M 2 V 2 to get the new concentrations [Ba 2+ ] = (1 x 10 –5 M)(200 mL)/(550 mL) = 3.64 x 10 –6 M [CrO 4 2– ]= (3 x M)(350 mL)/(550 mL) = 1.91 x M [Ba 2+ ][CrO 4 2– ] = (3.64 x 10 –6 M )(1.91 x M )= 6.95 x 10 – 11 the solution is unsaturated and no precipitate will form

Answer Key – ch. 8