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Answer Keys to Selected Problems

Chapter 15 Key

15.1.

(a)

AgI(s) Ag+(aq) + I(aq)
change (x) +x +x

(b)

CaCO3(s) Ca2+(aq) + CO32−(aq)
change (x) +x +x


(c)

Mg(OH)2(s) Mg2+(aq) + 2 OH(aq)
change (x) +x +2x

(d)

Mg3(PO4)2(s) 3 Mg2+(aq) + 2 PO43−(aq)
change (x) +x +2/3x

(e)

Ca5(PO4)3OH(s) 5 Ca2+(aq) + 3 PO43−(aq) + OH(aq)
change (x) +5x +3x +x

 

15.3.  There is no change. A solid has an activity of 1 whether there is a little or a lot.

 

15.5.  The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.

 

15.7.  CaF2, MnCO3, and ZnS

 

15.9.

(a)  LaF3(s) ⇌ La3+(aq) + 3 F(aq)  Ksp = [La3+] [F]3

(b)  CaCO3(s) ⇌ Ca2+(aq) + CO32−(aq)  Ksp = [Ca2+] [CO32−]

(c)  Ag2SO4(s) ⇌ 2 Ag+(aq) + SO42−(aq)  Ksp = [Ag+]2 [SO42−]

(d)  Pb(OH)3(s) ⇌ Pb2+(aq) + 2 OH(aq)  Ksp = [Pb2+] [OH]2

 

15.11.  (a)  1.77 × 10–7(b)  1.6 × 10–6(c)  2.2 × 10–9(d)  7.91 × 10–22

 

15.13.  (a)  2 × 10–2 M(b)  1.5 × 10–3 M(c)  2.27 × 10–9 M(d)  2.2 × 10–10 M

 

15.15.  CaSO4∙2H2O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L.

 

15.17.  4.8 × 10–3 M = [SO42−] = [Ca2+]; Since this concentration is higher than 2.60 × 10–3 M, “gyp” water does not meet the standards.

 

15.19.  Mass (CaSO4·2H2O) = 0.72 g/L

 

15.21.

(a)  [Ag+] = [I] = 1.3 × 10–5 M

(b)  [Ag+] = 2.88 × 10–2 M[SO4​2−] = 1.44 × 10–2 M

(c)  [Mn2+] = 3.7 × 10–5 M, [OH] = 7.4 × 10–5 M

(d)  [Sr2+] = 4.3 × 10–2 M, [OH] = 8.6 × 10–2 M

(e)  [Mg2+] = 1.3 × 10–4 M,  [OH] = 2.6 × 10–4 M

 

15.23.  (a)  1.45 × 10–4(b)  8.2 × 10–55(c)  1.35 × 10–4(d)  1.18 × 10–5(e)  1.08 × 10–10

 

15.25.  (a)  CaCO3 does precipitate.(b)  The compound does not precipitate.(c)  The compound does not precipitate.(d)  The compound precipitates.

 

15.27.  3.03 × 10−7 M

 

15.29.  9.2 × 10−13 M

 

15.31.  [Ag+] = 1.8 × 10–3 M

 

15.33.  [F] = 6.3 × 10–4 M

 

15.35.  (a)  2.25 L(b)  7.2 × 10–7 g

 

15.37.  100% of it is dissolved

 

15.39.

(a)  [Ag+] = 6.4 × 10−9 M ,  [Cl] = 0.025 M.  Check: [latex]\frac {6.4 \times 10^{-9}}{0.025} \times 100 \% = 2.6 \times 10^{-5}\%[/latex], an insignificant change

(b)  [Ca2+] = 2.2 × 10−5 M ,  [F] = 0.0013 M.  Check: [latex]\frac {2.26 \times 10^{-5}}{0.00133} \times 100 \% = 1.70 \%[/latex]. This value is less than 5% and can be ignored.

(c)  [SO42−] = 0.2238 M[Ag+] = 7.4 × 10–3 M.  Check: [latex]\frac {3.7 \times 10^{-3}}{0.2238} \times 100 \% = 1.64 \times 10^{-2} \%[/latex].  The condition is satisfied.

(d)  [OH] = 2.8 × 10–3 M,  5.7 × 10−12 M = [Zn2+].  Check: [latex]\frac {5.7 \times 10^{-12}}{2.8 \times 10^{-3}} \times 100 \% = 2.0 \times 10^{-7}\%[/latex]x is less than 5% of [OH] and is, therefore, negligible.

 

15.41.

(a)  [Cl] = 7.6 × 10−3 M.  Check: [latex]\frac {7.6 \times 10^{-3}}{0.025} \times 100 \% = 30 \%[/latex].  This value is too large to drop x. Therefore solve by using the quadratic equation:  [Ti+] = 3.1 × 10–2 M,  [Cl] = 6.1 × 10–3

(b)  [Ba2+] = 7.7 × 10–4 M.  Check: [latex]\frac {7.7 \times 10^{-4}}{0.0313} \times 100 \% = 2.4\%[/latex].  Therefore, the condition is satisfied.  [Ba2+] = 7.7 × 10–4 M,  [F] = 0.0321 M

(c)  Mg(NO3)2 = 0.02444 M,  [C2O42−] = 2.9 × 10−5 MCheck: [latex]\frac {2.9 \times 10^{-5}}{0.02444} \times 100 \% = 0.12 \%[/latex].  The condition is satisfied; the above value is less than 5%.  [C2O42−] = 2.9 × 10−5M[Mg2+] = 0.0244 M

(d)  [OH] = 0.0501 M,  [Ca2+] = 3.15 × 10–3.  Check: [latex]\frac {3.15 \times 10^{-3}}{0.050} \times 100 \% = 6.28 \%[/latex]This value is greater than 5%, so a more exact method, such as successive approximations, must be used.  [Ca2+] = 2.8 × 10–3 M,  [OH] = 0.053 × 10–2 M

 

15.43.  The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.

 

15.45.  (a) Hg2​2+ and Cu2+:  add SO42−(b)  SO42− and Cl:  add Ba2+(c)  Hg2+ and Co2+:  add S2–(d)  Zn2+ and Sr2+:  add OH until [OH] = 0.050 M(e) Ba2+ and Mg2+:  add SO42−(f)  CO32− and OH:  add Ba2+

 

15.47.  AgI will precipitate first.

 

15.49.  1.5 × 10−12 M

 

15.51.  3.99 kg

 

15.53.  (a)  3.1 × 10–11(b)  [Cu2+] = 2.6 × 10–3[IO3] = 5.3 × 10–3

 

15.55.  1.8 × 10–5 g Pb(OH)2

 

15.57Mg(OH)2(s) ⇌ Mg2+ + 2 OH  Ksp = [Mg2+] [OH]2

1.23 × 10−3 g Mg(OH)2

 

15.59.  MnCO3 will form first since it has the smallest Ksp value among these homologous compounds and is therefore the least soluble. MgCO3∙3H2O will be the last to precipitate since it has the largest Ksp value and is the most soluble. Ksp value.

 

15.62.  when the amount of solid is so small that a saturated solution is not produced

 

15.64.  1.8 × 10–5 M

 

15.66.  5 × 1023

 

15.68.

Cd(CN)42– Cd2+ + 4 CN 
I (M) 0.250 0 0
C (M) x +x +4x
E (M)
0.250 – x 4x

[Cd2+] = 9.5 × 10–5 M,  [CN] = 3.8 × 10–4 M

 

15.70.  [Co3+] = 3.0 × 10–6 M[NH3] = 1.8 × 10–5 M

 

15.72.  1.3 g

 

15.74.  0.79 g

 

15.76.

(a)

image

(b)

image

(c)

image

(d)

image

(e)

image

 

15.78.

(a)

image

(b)  H3O+ + CH3 → CH4 + H2O

image

(c)  CaO + SO3 → CaSO4

image

(d)  NH4+ + C2H5O → C2H5OH + NH3

image

 

15.80.  0.0281 g

 

15.82HNO3(l) + HF(l) → H2NO3+ + FHF(l) + BF3(g) → H+ + BF4

 

15.84.  (a)  H3BO3 + H2O → H4BO4 + H+(b)  The electronic and molecular shapes are the same—both tetrahedral.(c)  The tetrahedral structure is consistent with sp3 hybridization.

 

15.86.  0.014 M

 

15.88.  7.2 × 10–15 M

 

15.90.  4.4 × 10−22 M

 

15.93.  [OH] = 4.5 × 10−6[Al3+] = 2 × 10–16 (molar solubility)

 

15.95[SO42−]=0.049 M[Ba2+] = 4.7 × 10–7 (molar solubility)

 

15.97.  [OH] = 7.6 × 10−3 M[Pb2+] = 2.1 × 10–11 (molar solubility)

 

15.99.  7.66

 

15.101.

(a)  Ksp = [Mg2+][F]2 = (1.21 × 10–3)(2 × 1.21 × 10–3)2 = 7.09 × 10–9

(b)  7.09 × 10–7 M

(c)  Determine the concentration of Mg2+ and F that will be present in the final volume using M1V1 = M2V2.

Mg(NO3)2:
[3.00 × 10–3 M Mg(NO3)2](0.1000 L) = M2(0.3000 L)
 M2 = 1.00 × 10–3 M
NaF:
(2.00 × 10–3 M)(0.2000 L) = M2(0.3000 L)
M2 = 1.33 × 10–3 M

Compare the value of the ion product  Qsp with Ksp. If this value is larger than Ksp, precipitation will occur.

Qsp = [Mg2+][F]2 = (1.00 × 10–3)(1.33 × 10–3)2 = 1.77 × 10–9.  This value is smaller than Ksp, so no precipitation will occur.

(d)  MgF2 is less soluble at 27°C than at 18°C. Because added heat acts like an added reagent, when it appears on the product side, the Le Châtelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.

 

15.103.  BaF2, Ca3(PO4)2, ZnS; each is a salt of a weak acid, and the [H3O+] from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations

 

15.105.  Effect on amount of solid CaHPO4, [Ca2+], [OH]:(a)  increase, increase, decrease(b)  decrease, increase, decrease(c)  no effect, no effect, no effect(d)  decrease, increase, decrease(e)  increase, no effect, no effect

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Chapter 15 Key Copyright © by Nicole Bouvier-Brown; Saori Shiraki; J. Ryan Hunt; and Emily Jarvis. All Rights Reserved.