Applications of Aqueous Equilibria Chapter 8 Web-site:

Applications of Aqueous Equilibria– ch What is the pH at the equivalence point for the following titrations? a. NaOH with HBr pH = 7 pH > 7 pH < 7 b. HCl with NaCH 3 COO pH = 7 pH > 7 pH < 7 c. KOH with HCN pH = 7 pH > 7 pH < 7

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 H 2 S and 0.8 mol NaHS 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

Applications of Aqueous Equilibria– ch Which of the following buffer solutions can “absorb” the most acid without changing the pH? (All solutions have the same volume) 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

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

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 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 a titration curve for a weak acid being titrated by a strong base and label the following points: a. the equivalence point 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 Draw a titration curve for a weak base being titrated by a strong acid and label the following points: a. the equivalence point 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 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 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