1AO3
TUTORIAL CHEMISTRY #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ The following problems (1-13) are based on material from Chapters 5, 16 and class notes. 1. Complete and balance the following reaction in basic solution: MnO2 (s) + ClO3- (aq) MnO4- (aq) + Cl- (aq) Start with determining the changes in the oxidation numbers for all elements in molecules, as well as identifying oxidizing/reducing agents: Reactants: MnO2 (Mn: +4; O:-2) and ClO3- (Cl: +5; O: -2) Products: MnO4- (Mn: +7) and Cl- (Cl: -1). Therefore the MnO2 is being oxidized (Mn +4 +7; reducing agent) and ClO3- is being reduced (Cl +5 -1; oxidizing agent). Then construct balanced half-reaction under basic conditions (balance redox atoms, add electrons, balance charge with OH-. balance H/O with H2O, multiple by a factor to eliminate electrons): (NOTE: this is only one possible approach another would be to balance H with H+, then add H2O to balance O, then add OH- to both sides to convert H+ to H2O) Oxidation Half-reaction: Reduction Half-reaction: ( MnO2 (s) + 4OH- (aq) MnO4- (aq) + 3e- + 2H2O (l) ) X 2 ClO3- (aq) + 6e- + 3H2O (l) Cl- (aq) + 6OH- (aq)
Net Redox Reaction: 2MnO2 (s) + ClO3- (aq) + 2OH- (aq) 2MnO4- (aq) + Cl- (aq) + H2O (l)
2. Complete and balance the following reaction in acidic solution: CH3CH2OH(l) + Cr2O72(aq) CH3COOH(aq) + Cr3+(aq) Hint: In order to determine the oxidation number of alcohol and carboxylic acid, consider the carbon species -CH2OH and -COOH, respectively, since the CH3 end group does not change. This is a redox reaction. We are using a strong oxidizing agent with a primary alcohol, so the alcohol (ethanol) will be converted to a carboxylic acid (acetic acid). The orange Cr2O72-(aq) ion is reduced to green Cr3+(aq) ion. Start with determining the changes in the oxidation numbers for all elements in molecules, as well as identifying oxidizing/reducing agents: Reactants: CH3CH2OH (C of the CH2OH group: -1; O:-2; H: +1) and Cr2O72- (Cr: +6; O: -2) Products: CH3COOH (C of the COOH group: +3), Cr3+ (Cr: +3, O: -2). Therefore the Cr2O72- is being reduced (Cr +6 +3; oxidizing agent) and CH2OH is being oxidized (C -1 +3; reducing agent). Then construct balanced half-reaction under basic conditions (balance redox atoms, add electrons, balance H/O with H2O, then balance charge with H+, multiple by a factor to eliminate electrons): Page 1 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Oxidation Half-reaction: ( CH3CH2OH (aq) + H2O (l) CH3COOH (aq) + 4H+ (aq) + 4e- ) X3 Reduction Half-reaction: ( Cr2O72- (aq) + 6e- + 14H+ (aq) 2Cr3+ (aq) + 7H2O (l) ) X2
Net Redox Reaction: 3CH3CH2OH (aq) + 2Cr2O72- (aq) + 16H+ (aq) 3CH3COOH (aq) + 4Cr3+ (aq) + 11H2O (l)
3. The following redox reaction is an important one used in mining and metallurgy applications for the transformation/detoxification of toxic cyanide ion into innocuous nitrogen gas and carbonate ion. Complete and balance the following reaction in basic solution: CN- (aq) + ClO- (aq) N2 (g) + CO32- (aq) + Cl- (aq) Start with determining the changes in the oxidation numbers for all elements in molecules, as well as identifying oxidizing/reducing agents: Reactants: CN- (C: +2; N:-3) and ClO- (Cl: +1; O: -2) Products: N2 (N: 0), CO32- (C: +4, O: -2) and Cl- (Cl: -1). Therefore the ClO- is being reduced (Cl +1 -1; oxidizing agent) and CN- is being oxidized (C +2 +4; reducing agent) and (N -3 0). Then construct balanced half-reaction under basic conditions (balance redox atoms, add electrons, balance charge with OH-, balance H/O with H2O, multiple by a factor to eliminate electrons): Oxidation Half-reaction: 2CN- (aq) + 12OH- (aq) N2 (g) + 2CO32- + 10e- + 6H2O Note we needed 2 CN ions because the product N2 has 2 N atoms. Reduction Half-reaction: ( ClO- (aq) + 2e- + H2O (l) Cl- (aq) + 2OH- (aq) ) X 5
Net Redox Reaction: 2CN- (aq) + 5ClO- (aq) + 2OH- (aq) N2 (g) + 2CO32- (aq) + 5Cl- (aq) + H2O (l) 4. Complete and balance the following disproportionation reaction in basic solution: Cl2 (g) Cl- (aq) + ClO3- (aq) A disproportionation reaction is a type of redox reaction where the same reactant is being oxidized and reduced to produce different products. In this case, chlorine gas (Cl2: O) is oxidized into ClO3- (Cl: +5) and reduced into Cl- (Cl: -1) when in contact with an alkaline solution. Next construct balanced half-reaction under basic conditions (balance redox atoms, add electrons, balance charge with OH-, balance H/O with H2O, multiple by a factor to eliminate electrons): Page 2 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Oxidation Half-reaction: Cl2 (g) + 12OH- (aq) 2ClO3- (aq) + 6H2O (l) + 10eReduction Half-reaction: ( Cl2 (g) + 2e- 2Cl- (aq) )X 5 Net Redox Reaction: 6Cl2 (g) + 12OH- (aq) 2ClO3- (aq) + 10Cl- (aq) + 6H2O (l) or dividing by a factor of 2 3Cl2 (g) + 6OH- (aq) ClO3- (aq) + 5Cl- (aq) + 3H2O (l) Acids & Bases: Assume all calculations are at 25C, so that Kw = 1.0 10-14 (and pKw = 14.00). 5. What is the conjugate acid of each of the following species? Base (a) (b) (c) (d) S2 CH3COO ClO2 SO32 Conjugate acid HS CH3COOH HClO2 HSO3
6.
What is the conjugate base of each of the following species? Acid (a) (b) (c) (d) H2CO3 HNO3 H2PO4 [(CH3)2NH2]+ (dimethylammonium ion) Conjugate base HCO3 NO3 HPO42 (CH3)2NH (dimethylamine)
7. Write balanced chemical reactions and Ka or Kb expressions for the first ionization of the following acids and bases (i.e. assume in each case that only one H+ is ionized or added). (a) HF (b) H3PO4 (c) CH3NH2 (d) CH3COOH (e) HClO4
(a)
HF + H2O
F + H3O+ H2PO4 + H3O+
Ka = [F][ H3O+] [HF] Ka = [H2PO4][ H3O+] [H3PO4] Kb = [CH3NH3+]
(b)
H3PO4 + H2O
[OH ] Page 3 of 12
(c)
CH3NH2 + H2O
CH3NH3+ + OH
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ [CH3NH2] (d) CH3COOH + H2O H3O+ + CH3COO Ka = [H3O+] [CH3COO] [CH3COOH]
(e)
HClO4 is a strong acid and is fully dissociated at all concentrations in water. The equilibrium constant is much, much greater than 1, so we normally do not write it. What is the pH of a 0.035 M solution of Ba(OH)2 (aq)? What is the pH of a 0.0042 M solution of HClO4 (aq)? What is the pOH of the solution in part (b)? A 30.0 mL sample of HClO4 was diluted to 500 mL, and the pH of the final solution was 4.56. What was the concentration of HClO4 in the original sample?
8.
(a) (b) (c) (d)
SOLUTION: (a) Ba(OH)2 (aq) is a soluble hydroxide, and a strong base. It dissociates fully in water, so Ba(OH)2 (aq) produces Ba2+ (aq) and 2 OH (aq). The original 0.035 M sample of Ba(OH)2(aq) therefore produces 0.035 M Ba2+ ions, and 2 (0.035M) OH ions (according to the formula of our hydroxide). [OH] = 2(0.035M) = 0.070 M
= -log[OH] = -log[0.070] = 1.15 From pH + pOH = 14.00, rearrange to get (2 sig figs. becomes 2 dec. places) From this we can find pOH: pOH pH = 14.00 pOH (b) = 14.00 1.15 = 12.85
HClO4 is a strong acid, and dissociates fully in water to give: HClO4(aq) + H2O(l) ClO4(aq) + H3O+(aq)
The original 0.0042 M sample of HClO4(aq) therefore produces 0.0042 M ClO4 ions, and 0.0042 M H3O+ ions pH = -log[H3O+] (c) = -log(0.0042) = 2.38
pOH = 14.00 pOH = 14.00 2.38 = 11.62 Page 4 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ (d) Since HClO4 is a strong acid, it dissociates fully in solution. In the final solution, pH = 4.56, and we can find [H3O+] directly from pH: Since pH = log[H3O+], rearrange to give pH = log[H3O+] and take the antilog to give [H3O+] =10pH Thus, [H3O+] = 104.56 = 2.75 105 mol/L (2 decimal places in the log becomes 2 significant figures) To find the initial concentration of HClO4, we need the number of moles of HClO4 present in solution: mol HClO4 = = (2.75 105 mol/L)(0.500 L) 1.38 105 mol
Original concentration of HClO4 = mol HClO4/original volume = 1.38 105 mol 0.0300 L 9. Predict and rationalize the pH of a solution of the salt ammonium cyanide (NH4CN) if the Ka of NH4+ is 5.55 1010 and the Ka of HCN is 6.2 1010 = 4.6 104 M
In this case, the salt is highly soluble and will dissociate completely in solution (dissolution):
NH4CN (s) NH4+ (aq) + CN- (aq)
Then, the ammonium and cyanide ions will each undergo a hydrolysis reaction with water as a weak acid and weak base, respectively: pKa NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq)
CN (aq) + H2O (l)
-
pKb
HCN (aq) + OH- (aq)
The final pH of the solution will depend on the relative magnitudes of the two acid/base reactions above (pKa of ammonium VS pKb of cyanide, or Ka or Kb compare either set of values), since it involves 1:1 stoichiometry. pKa (NH4+) = -log(Ka) = 9.26 VS pKb (CN-) = pKw - pKa = 14.0 (-log(6.2 x 10-10) = 4.79 Snce pKb << pKa, or Ka << Kb , a solution of this salt will be basic due to the stronger basic properties (electron donating) of CN- compared to the acidic properties (electron accepting) of NH4+. Page 5 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ 10. Hydrofluoric acid (HF) is a weak acid with Ka = 7.2 104 at 25 C. For a 0.45 M solution of HF at 25 C, determine: (a) (b) (c) (d) SOLUTION: This is a weak acid ionization problem. Lets start with the chemistry: HF (aq) Initial conc. Change conc. Equilibrium Ka = 0.45 M -x 0.45-x [F][ H3O+] [HF] = = 7.2 104 + H2O (l) 0 +x x F (aq) + H3O+ (aq) [H3O+] pOH percent dissociation of HF Kb for F
(107 M) +x x
(x)(x) (0.45-x)
7.2 104
Assume x << 0.45, and thus (0.45-x) 0.45 x2 x Check: (7.2 104)(0.45) 0.018 M 4.0% <5%, assumption is okay 2 sig. figs is appropriate.
0.018 100% = 0.045
Now we can answer the parts of the question: (a) [H3O+] = x = 0.018 M
(b) pOH we can determine from pH: pH = -log[H3O+] = -log(0.018) = 1.74
since pH + pOH = 14.00, rearrange to get pOH = 14.00 pH pOH = 14.00 1.76 = 12.26 Page 6 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Alternatively, we can find [OH-] = Kw / [H3O+] = 1.0 1014 / 0.018 = 5.55 1013 = -log(5.55 1013) = 12.26 Then pOH = -log[OH-] (c) percent dissociation of HF This is similar to the calculation we do when we check our small x assumption. In other words, compare the concentration of what you get at equilibrium (H3O+, in this case,) to the concentration of what you started with (HF). [H3O+]equilibrium 100% [HF]initial (d) Kb for F Recall the equation that relates Ka and Kb for a conjugate acid-base pair: Ka Kb = Kw Kb = Kw Ka We can derive this by considering the three relevant reactions. H2O (l) + HA (aq) 2H2O (l) H3O+ (aq) + A- (aq) Ka = [H3O+][A-] / [HA] Kw = [H3O+][OH-] H3O+ (aq) + OH- (aq) which can be rearranged to = 0.018 100% = 0.045 4.0% dissociated
To get the reaction of interest, which is: H2O (l) + A- (aq) OH- (aq) + HA (aq) we reverse the first reaction, K = 1 / Ka and add that to the second (KKw). So Kb = Kw / Ka = {[H3O+][OH-] } / {[H3O+][A-] / [HA]} = [OH-][HA] / [A-] Thus, Kb = 1.0 1014 7.2 104 = 1.4 1011
11.
A student dissolves 2.15 g of hydrazoic acid, HN3, to form one litre of solution. The pH of the solution is measured and found to be 3.01. What is the ionization constant of hydrazoic acid?
SOLUTION: This is a weak acid ionization problem. Start with the chemistry: HN3(aq) + H2O(l) Ka = [H3O+] [N3] [HN3] Page 7 of 12 H3O+(aq) + N3(aq)
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Ka is unknown, but the pH of the solution is 3.01, and so we can determine the [H3O+]: [H3O+] = antilog(-pH) = 10pH = 103.01 = 9.77 104M
Also, we can determine the initial concentration of the solution: 2.15 g = 0.0500 mol 43.0 g/mol and this is in one litre of solution, so, initially [HN3] = 0.0500 M HN3(aq) + H2O(l) H3O+(aq) + Initial conc. 0.0500 M 0 Change -x +x Equilibrium conc. 0.0500-x x But we calculated [H3O+], and so x = 9.77 104M = [H3O+] = [N3] = 0.00098 and [HN3] = 0.0500 - x = 0.04902 M Ka = (9.77 104)( 9.77 104) = 1.948 10-5 0.04902 12. Para-aminobenzoic acid (NH2C6H4COOH, or HPABA for short) is one of the first major chemical additives in sunscreen lotions, but has since been replaced by other alternatives that are less irritating to the skin. What is the pH of a 0.200 M solution of HPABA? Why would this solution be potentially irritating? (Ka HPABA = 1.2 105). HPABA 0.200 M -x 0.200-x + H2O PABA 0 +x x + H3O+ 0 +x x N3(aq) (107 M) +x x
Initial Change Equilibrium
Ka
=
[PABA][H3O+] [HPABA]
=
(x)(x) (0.200-x)
=
1.2 105
Assuming x << 0.200, solve for x: x pH = = 1.55 103 M = log[H3O+] = [H3O+] 2.81 and the assumption is valid (0.8 %)
Page 8 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Due to the high concentration (required in order to absorb a higher flux of UVA/B solar radiation) and weak acidity properties of PABA, the sunscreen solution has a significantly acidic pH which makes it more irritating to sensitive skin and a potential allergen. Alternative chemicals that effectively absorb broadband UV solar radiation without generating excessively acidic/alkaline pH as lotions on skin would be preferred choices. Challenge and Interest Question: 13. Purified cocaine extracted from the coca leaf is a natural alkaloid (base) that is used as an illegal drug of abuse, as well as a useful topical anesthetic agent in medicinal applications. The acid equilibrium constant for protonated cocaine (symbolized as BH+) in water is Ka = 3.98 1010 at 25C. The base equilibrium reaction for cocaine (symbolized as B) is shown below:
B
O
BH+
O
O
O
N O
+
H2O (l)
NH+ O
+ OH- (aq)
O
O
(a) What is the equilibrium concentration of the conjugate acid form of cocaine (BH+) in a 0.400 M aqueous solution of cocaine (B) as a free base? (b) What is the pH of a 0.400 M aqueous solution of B? (c) What is the equilibrium concentration of the base form of cocaine (B) in a 0.400 M aqueous solution of the conjugate acid form of cocaine(BH+ salt)? (d) What is the pH of a 0.400 M aqueous solution of BH+? (e) Determine the % ionization of BH+ in 0.400 M cocaine (free base, B) solution and % ionization of B in 0.400 M cocaine (BH+ salt) solution. (f) Predict whether the aqueous solubility of cocaine would be higher under acidic conditions or alkaline conditions?
SOLUTION: (a) This is a weak base ionization problem. Start by writing down the chemistry: B (aq) + H2O (l) BH+ (aq) + OH- (aq) Page 9 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ Set up an ICE table (Initial, Change, Equilibrium concentrations) in order to obtain mathematical expressions that describe the reaction. B (aq) + H2O(l) Initial conc. Change Equilibrium conc. 0.400 M -x 0.400-x BH+ (aq) + OH- (aq) 0 +x x (107 M from water)* +x x
* We ignore the amount of OH- present from the ionization of water, if the value (107 M) is very small, as compared to x. Since B functions as a base in our equilibrium, set up a Kb expression: Kb = [BH+][OH-] [B] = Kw / Ka = 1.0 1014 / 3.98 1010 = 2.51 105
Note that one was provided a Ka value that needed to be converted to Kb since cocaine is a weak base, its conjugate acid is also weak. Substitute the equilibrium concentration values from the ICE table into Kb: (x)(x) (0.400-x) = 2.51 105
Assume x << 0.400, and thus (0.400-x) 0.400 Note that this assumption can be made since: [B]/Kb = 0.40 / 2.51 105 = 1.6 104 >> 20 due to relatively high concentration of cocaine base. x2 x Check: (2.51 105)(0.400) 3.17 103 M = [OH-] or [BH+]
0.00317 100% = 0.8% << 5%, assumption is okay 0.400 Equilibrium [BH+] = x = 3.17 103 M or 3.17 mM
(b)
Recall that pH + pOH = 14.00, which can be rearranged to give: pH = 14.00 pOH We need the value for pOH: pOH = -log[OH] = -log[3.17 103 M] = 2.50
pH = 14.00 2.50 = 11.50 (2 sig figs becomes 2 decimal places in the log) Page 10 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ (c) This is a weak acid ionization problem. Start by writing down the chemistry: BH+ (aq) + H2O (l) B (aq) + H3O+ (aq)
Set up an ICE table (Initial, Change, Equilibrium concentrations) in order to obtain mathematical expressions that describe the reaction. BH+ (aq) Initial conc. Change Equilibrium conc. + H2O (l) B (aq) 0 +x x + H3O+ (aq) (107 M from water)* +x x
0.400 M -x 0.400-x
* We ignore the amount of H3O+ present from the ionization of water, if the value (107 M) is very small, as compared to x. Since B functions as an acid in our equilibrium, set up a Ka expression: Ka = [B][H3O+] [B] = 3.98 1010
Substitute the equilibrium concentration values from the ICE table into Kb: (x)(x) (0.400-x) = 3.98 1010
Assume x << 0.400, and thus (0.400-x) 0.400 Note that this assumption can be made since: [BH+]/Ka = 0.40 / 2.51 1010 >> 20 due to relatively high concentration of cocaine conjugate acid and its very low Ka. x2 x Check: (3.98 1010)(0.400) 1.26 105 M = [H3O+] or [B]
1.26 105 M 100% = 0.0031% << 5%, assumption is okay 0.400 Equilibrium [B] = x = 1.26 105 M or 12.6 M
(d) pH = -log[1.26 105 M] = 4.90 Note carefully that a solution of cocaine as a free base (B) generates a strongly alkaline solution (0.4 M, pH = 11.50), whereas a solution of cocaine-salt (BH+) as its conjugate acid generates a weakly acidic solution (0.4 M, pH 4.90) It is apparent that at equal concentrations, cocaine acts as a stronger base than its conjugate acid form since Kb > Ka Page 11 of 12
CHEMISTRY 1AO3
TUTORIAL #8
November 2-6, 2009
Solutions: Redox Reactions Ch 5 / Acid-Base Chemistry Chapter 16 ______________________________________________________________________________ (e) i) 0.40 M solution of cocaine as free base (B) where pH = 11.50 from part (a): % ionization of BH+ = ([BH+] / Initial conc. of B) x 100 = (3.17 103 M / 0.40 M) x 100 = 0.79% Small fraction exists in ionized state (BH+) at this high pH! ii) 0.40 M solution of cocaine as free base (B) where pH = 4.90 from part (c): % ionization of B = ([B] / Initial conc. of BH+) x 100 = (1.26 105 M / 0.40 M) x 100 = 0.0031% A very small fraction exists in neutral state (B) at this low pH! (f) The aqueous solubility of cocaine is higher under acidic conditions since a larger fraction of cocaine exists in its protonated (charged) conjugate acid form, BH+, whereas under alkaline conditions, especially at pH > pKa, a large fraction of cocaine exists as its less soluble neutral free base form, B. This property is used during the liquid extraction of cocaine from a petroleum fraction containing coca leaves, which is performed under acidic conditions to maximize recovery.
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