CHEMISTRY 31

Fall, 2017 - Dixon

Homework Set 2.1 Solutions

Ch. 6: 8, 15-17, 19, 23, 25, 33, 36, 44, 49

8. For the reaction HCO₃^- ↔ H⁺ + CO₃^2-, ΔG° = +59kJ/mol at 298.15K. Find the value of K for the reaction.

K = exp(-ΔG°/RT) = exp(-59000 J mol^-1/(8.314 J mol^-1 K^-1*298.15 K)) = 5 x 10^-11.

15.  What concentration of Fe(CN)₆^4- (ferrocyanide) is in equilibrium with 1.0 μM Ag⁺ and Ag₄Fe(CN)₆(s).  Express your answer with a prefix from table 1-3.

Since it states “in equilibrium with”, no ICE table is needed.

K_sp = [Ag⁺]⁴[Fe(CN)₆^4-]  =  8.5 x 10^-45  = (1.0 x 10^-6)⁴*x

x = 8.5 x 10^-45/1.0 x 10^-24  = 8.5 x 10^-21 M = 8.5 zM = [Fe(CN)₆^4-]

16.  Find [Cu²⁺] in a solution saturated with Cu₄(OH)₆(SO₄) if [OH^-] is fixed at 1.0 x 10^-6 M.  Note that Cu₄(OH)₆(SO₄) gives 1 mole of SO₄^2- for 4 mol Cu²⁺.

Cu₄(OH)₆(SO₄)(s) ↔ 4Cu²⁺ + 6OH^- + SO₄^2- K_sp = 2.3 x  10^-69

Initial                                                    0      1.0 x 10^-6     0

Change                                                 +4x      +6x          +x

Equilibrium                                           4x    1.0 x 10^-6     +x    (1.0 x 10^-6 is fixed)

Note: since we know [OH^-] at equilibrium is 1.0 x 10^-6 M, we really did not need the ICE table, but the table is useful in remembering that the ratio of Cu²⁺ to SO₄^2- should be 4:1 from the stoichiometry.

K_sp = 2.3 x  10^-69 = [Cu²⁺]⁴[OH^-]⁶[SO₄^2-] = (4x)⁴(1.0 x 10^-6)⁶x = (2.56 x 10^-34)x⁵

x = (2.3 x  10^-69/2.56 x 10^-34)^1/5 = 9.8 x 10^-8 M

[Cu²⁺] = 4x = 3.9 x 10^-7 M

17.  (a)  From the solubility product of zinc ferrocyanide, Zn₂Fe(CN)₆, calculate the concentration of Fe(CN)₆^4- in 0.10 mM ZnSO₄ saturated with Zn₂Fe(CN)₆.  Assume that Zn₂Fe(CN)₆ is a negligible source of Zn²⁺.

Zn₂Fe(CN)₆(s) ↔ 2Zn²⁺ + Fe(CN)₆^4- K_sp = 2.1 x 10^-16 = [Zn²⁺]²[Fe(CN)₆^4-]

At equilibrium, [Zn²⁺] = 0.10 mM (based on the assumption listed above)

[Fe(CN)₆^4-] = 2.1 x 10^-16/[Zn²⁺]² = 2.1 x 10^-16/1.0 x 10^-8 = 2.1 x 10^-8 M

(b)  What concentration of K₄Fe(CN)₆ should be in a suspension of solid Zn₂Fe(CN)₆ in water to give [Zn²⁺] = 5.0 x 10^-7 M?

K_sp = 2.1 x 10^-16 = [Zn²⁺]²[Fe(CN)₆^4-] = (5.0 x 10^-7 M)²[Fe(CN)₆^4-]

[Fe(CN)₆^4-] = 2.1 x 10^-16/2.5 x 10^-13 = 8.4 x 10^-4 M

19.  A solution contains 0.0500 M Ca²⁺ and 0.0300 M Ag⁺.  Can 99% of Ca²⁺ be precipitated by sulfate without precipitating Ag⁺?  What will be the concentration of Ca²⁺ when Ag₂SO₄ begins to precipitate?

K_sp(CaSO₄) = 2.4 x 10^-5 = [Ca²⁺][SO₄^2-]

K_sp(Ag₂SO₄) = 1.5 x 10^-5 = [Ag⁺]²[SO₄^2-]

It may not be clear which cation will precipitate first (since the K_sp values are similar).  This can be calculated by calculating minimum equilibrium sulfate concentrations to start precipitating out the cation (through rearranging the above equations).

For calcium sulfate, [SO₄^2-] = K_sp(CaSO₄)/[Ca²⁺] = 4.8 x 10^-4 M

For silver sulfate, [SO₄^2-] = K_sp(Ag₂SO₄)/[Ag⁺]²= 1.67 x 10^-2 M, so calcium precipitates first (lowest concentration is the first to precipitate as sulfate concentration is increased).

When calcium has been 99% precipitated, 1% of the calcium will be left.  This means that [Ca²⁺] = 0.01*0.0500 M = 5.00 x 10^-4 M; thus [SO₄^2-] = 0.0480 M

The equilibrium [Ag⁺] = {K_sp(Ag₂SO₄)/[SO₄^2-]}^0.5 = 0.0177 M. Since this is less than 0.0300 M (the initial silver concentration), 99% of Ca²⁺ can not be removed without precipitating Ag⁺.

The concentration of Ca²⁺ when [Ag⁺] starts precipitating is solved by:

[Ca²⁺] = K_sp(CaSO₄)/[SO₄^2-] = 2.4 x 10^-5/1.67 x 10^-2 = 1.4 x 10^-3 M

23.  Identify Lewis acids in the following reactions:

a)      BF₃ + :NH₃ ↔ F₃B^--NH₃⁺ BF₃ is the electron pair acceptor (Lewis Acid)

b)      :F^- + AsF₅ ↔ AsF₆^- AsF₅ is the electron pair acceptor (Lewis Acid)

25.  Given the following equilibria, calculate the concentration of each zinc species in a solution saturated with Zn(OH)₂(s) and containing [OH^-] at a fixed concentration of 3.2 x 10^-7 M.

Zn(OH)₂(s)  K_sp = 3 x 10^-16            Zn(OH)⁺ b₁ = 1 x 10⁴             Zn(OH)₂(aq)  b₂ = 2 x 10¹⁰

Zn(OH)₃^- b₃ = 8 x 10¹³ Zn(OH)₄^2- b₄ = 3 x 10¹⁵

Rxn 1: K_sp = 3 x 10^-16 = [Zn²⁺][OH^-]² or [Zn²⁺] = 3 x 10^-16/[OH^-]² = 2.9 x 10^-3 M

Rxn 2: b₁ = 1 x 10⁴ = [Zn(OH)⁺]/[Zn²⁺][OH^-] or [Zn(OH)⁺] = (1 x 10⁴)(2.9 x 10^-3)(3.2 x 10^-7)

[Zn(OH)⁺] = 9 x 10^-6 M

Rxn 3: b₂ = 2 x 10¹⁰ = [Zn(OH)₂(aq)]/[Zn²⁺][OH^-]² or

[Zn(OH)₂(aq)] = (2 x 10¹⁰)(2.9 x 10^-3 M)(3.2 x 10^-7)² =  6 x 10^-6 M

Rxn 4: b₃ = 8 x 10¹³ = [Zn(OH)₃^-]/[Zn²⁺][OH^-]³ or

[Zn(OH)₃^-] = (8 x 10¹³)(2.9 x 10^-3)(3.2 x 10^-7)³ = 8 x 10^-9 M

Rxn 5: b₄ = 3 x 10¹⁵ = [Zn(OH)₄^2-]/[Zn²⁺][OH^-]⁴ or

[Zn(OH)₄^2-] = (3 x 10¹⁵)(2.9 x 10^-3)(3.2 x 10^-7)⁴ = 9 x 10^-14 M

33.  Identify the Bronsted-Lowry acids among the reactants in the following reactions.

(a) KCN + HI ↔ HCN + KI     acid = HI

(b) PO₄^3- + H₂O ↔ HPO₄^2- + OH^- acid = H₂O

36  Calculate [H⁺] and pH for the following solutions:

a) 0.010 M HNO₃ [H⁺] = 0.010 M (HNO₃ is a strong acid), pH = -log(0.010) = 2.00

b) 0.035 M KOH         [OH^-] = 0.035 M [H⁺] = 1.0x10^-14/0.035 = 2.86 x 10^-13 M, pH = 12.54

c) 0.030 M HCl           [H⁺] = 0.030 M, pH = 1.52

d) 3.0 M HCl               [H⁺] = 3.0 M, pH = -0.48

e) 0.010 M [(CH₃)₄N⁺]OH^- [H⁺] = 1.0 x 10^-12 M, pH = 12.00

44.  Write the K_a reaction for trichloroacetic acid, Cl₃CCO₂H, for anilinium ion, C₆H₅NH₃⁺, and for lanthanum ion, La³⁺.

Cl₃CCO₂H (aq) ↔ Cl₃CCO₂^- + H⁺;  C₆H₅NH₃⁺ ↔ C₆H₅NH₂ (aq) + H⁺

La³⁺ + H₂O(l)  ↔ LaOH²⁺ + H⁺

49.  Write the K_b reaction of CN^-.  Given that the K_a value for HCN is 6.2 x 10^-10, calculate the K_b.

Reaction = CN^- + H₂O ↔ HCN + OH^-

K_b = K_w/K_a = 1.0 x 10^-14/6.2 x 10^-10 = 1.6 x 10^-5.

Ch. 7: 2, 7, 15

2. Distinguish between the terms end point and equivalence point.

Equivalence point refers to the point in the titration when the reagents have been added to the exact stoichiometric ratio. End point refers to the point when the equivalence is detected.

7. How many milliliters of 0.100 M KI are needed to react with 40.0 mL of 0.0400 M Hg₂(NO₃)₂ if the reaction is Hg₂²⁺ + 2I^- ↔ Hg₂I₂(s)?

n(KI)/n(Hg₂²⁺) = 2/1 or n(KI) = 2[Hg₂(NO₃)₂]V(Hg₂(NO₃)₂) = 2(0.040 mmol/mL)(40.0 mL)

n(KI) = 3.20 mmol = [KI]V(KI) or V(KI) = 3.20 mmol/(0.100 mmol/mL) = 32.0 mL

15. A cyanide solution with a volume of 12.73 mL was treated with 25.00 mL of Ni²⁺ solution (containing excess Ni²⁺) to convert the cyanide into tetracyanonickelate(II):

4CN^- + Ni²⁺ ↔ Ni(CN)₄^2-

The excess Ni is titrated with 10.15 mL of 0.01307 M ethylenediaminetetraacetic acid (EDTA):

Ni²⁺ + EDTA^4- ↔ Ni(EDTA)^2-

Ni(CN)₄^2- does not react with EDTA. If 39.35 mL of EDTA were required to react with 30.10 mL of the original Ni²⁺ solution, calculate the molarity of CN^- in the 12.73 mL cyanide sample.

n(Ni²⁺)_orig = n(Ni²⁺)_reacted + n(Ni²⁺)_excess and n(CN^-)/n(Ni²⁺)_reacted = 4/1

n(Ni²⁺)_reacted = n(Ni²⁺)_orig - n(Ni²⁺)_excess

[Ni²⁺]_origV_orig = [EDTA]V_EDTA

[Ni²⁺]_orig = (0.01307 mmol/mL)(39.35 mL)/(30.10 mL) = 0.017086 M

n(Ni²⁺)_orig = [Ni²⁺]_origV_orig = (0.017086 mmol/mL)(25.00 mL) = 0.42716 mmol

n(Ni²⁺)_excess = (0.01307 mmol/mL)(10.15 mL) = 0.13266 mmol

n(Ni²⁺)_reacted = 0.42716 mmol - 0.13266 mmol = 0.2945 mmol

n(CN^-) = 4n(Ni²⁺)_reacted = 1.178 mmol

[CN^-] = 1.178 mmol/12.73 mL = 0.09254 M