Question Problem 6.1
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), calculate Kc from [N₂]=1.5×10⁻² M, [H₂]=3.0×10⁻² M and [NH₃]=1.2×10⁻² M.
Steps only
Kc ≈ 3.55×10².
Answer
Kc ≈ 3.55×10².
CLASS 11 · CHEMISTRY · CHAPTER 6
Equilibrium
Clear, point-wise answers to all 28 numbered in-text problems and all 73 end-of-chapter exercises.
101 worked answersComplete chapter answers
101 worked answers · short steps first, explanations when you need them
In-text Problems
Textbook pages 176–208
Question Problem 6.2
For N₂(g) + O₂(g) ⇌ 2NO(g), calculate Kc from [N₂]=3.0×10⁻³ M, [O₂]=4.2×10⁻³ M and [NO]=2.8×10⁻³ M.
Steps only
Kc ≈ 0.622.
Answer
Kc ≈ 0.622.
Question Problem 6.3
Find Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), when the equilibrium concentrations are 1.41 M, 1.59 M and 1.59 M respectively.
Steps only
Kc ≈ 1.79.
Answer
Kc ≈ 1.79.
Question Problem 6.4
For CO(g)+H₂O(g) ⇌ CO₂(g)+H₂(g), Kc=4.24. Initially [CO]=[H₂O]=0.10 M and no products are present. Find all equilibrium concentrations.
Steps only
[CO₂]=[H₂]≈0.067 M; [CO]=[H₂O]≈0.033 M.
Answer
[CO₂]=[H₂]≈0.067 M; [CO]=[H₂O]≈0.033 M.
Question Problem 6.5
For 2NOCl(g) ⇌ 2NO(g)+Cl₂(g), Kc=3.75×10⁻⁶ at 1069 K. Find Kp.
Steps only
Kp≈0.033 kPa (equivalently 3.3×10⁻⁴ bar).
Answer
Kp≈0.033 kPa (equivalently 3.3×10⁻⁴ bar).
Question Problem 6.6
For CO₂(g)+C(s) ⇌ 2CO(g), Kp=3.0 at 1000 K. Initially pCO₂=0.48 bar and pCO=0. Find the equilibrium partial pressures.
Steps only
pCO≈0.665 bar; pCO₂≈0.147 bar.
Answer
pCO≈0.665 bar; pCO₂≈0.147 bar.
Question Problem 6.7
For 2A ⇌ B+C, Kc=2×10⁻³. At one moment, [A]=[B]=[C]=3×10⁻⁴ M. Which direction will the reaction move?
Steps only
The reaction moves in reverse, towards A.
Answer
The reaction moves in reverse, towards A.
Question Problem 6.8
13.8 g N₂O₄ is placed in a 1 L vessel at 400 K. For N₂O₄(g) ⇌ 2NO₂(g), the total equilibrium pressure is 9.15 bar. Find Kp, Kc and the equilibrium partial pressures.
Steps only
pN₂O₄≈0.81 bar; pNO₂≈8.34 bar; Kp≈85.9; Kc≈2.6.
Answer
pN₂O₄≈0.81 bar; pNO₂≈8.34 bar; Kp≈85.9; Kc≈2.6.
Question Problem 6.9
3.00 mol PCl₅ is placed in a 1 L vessel at 380 K. For PCl₅(g) ⇌ PCl₃(g)+Cl₂(g), Kc=1.80. Find the equilibrium concentrations.
Steps only
[PCl₅]≈1.41 M; [PCl₃]=[Cl₂]≈1.59 M.
Answer
[PCl₅]≈1.41 M; [PCl₃]=[Cl₂]≈1.59 M.
Question Problem 6.10
For glucose phosphorylation, ΔG°=+13.8 kJ mol⁻¹. Find Kc at 298 K.
Steps only
Kc≈3.81×10⁻³.
Answer
Kc≈3.81×10⁻³.
Question Problem 6.11
Sucrose hydrolysis has Kc=2×10¹³ at 300 K. Find ΔG°.
Steps only
ΔG°≈−76.4 kJ mol⁻¹.
Answer
ΔG°≈−76.4 kJ mol⁻¹.
Question Problem 6.12
Give the conjugate bases of HF, H₂SO₄ and HCO₃⁻.
Steps only
The conjugate bases are F⁻, HSO₄⁻ and CO₃²⁻.
Answer
The conjugate bases are F⁻, HSO₄⁻ and CO₃²⁻.
Question Problem 6.13
Give the conjugate acids of NH₂⁻, NH₃ and HCOO⁻.
Steps only
The conjugate acids are NH₃, NH₄⁺ and HCOOH.
Answer
The conjugate acids are NH₃, NH₄⁺ and HCOOH.
Question Problem 6.14
For H₂O, HCO₃⁻, HSO₄⁻ and NH₃, give each conjugate acid and conjugate base.
Steps only
H₂O: H₃O⁺ / OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻.
Answer
H₂O: H₃O⁺ / OH⁻; HCO₃⁻: H₂CO₃ / CO₃²⁻; HSO₄⁻: H₂SO₄ / SO₄²⁻; NH₃: NH₄⁺ / NH₂⁻.
Question Problem 6.15
Classify OH⁻, F⁻, H⁺ and BCl₃ as Lewis acids or bases.
Steps only
Lewis bases: OH⁻, F⁻. Lewis acids: H⁺, BCl₃.
Answer
Lewis bases: OH⁻, F⁻. Lewis acids: H⁺, BCl₃.
Question Problem 6.16
A soft drink has [H⁺]=3.8×10⁻³ M. Find its pH.
Steps only
pH≈2.42, so the drink is acidic.
Answer
pH≈2.42, so the drink is acidic.
Question Problem 6.17
Find the pH of 1.0×10⁻⁸ M HCl at 25°C.
Steps only
pH≈6.98. The solution is slightly acidic.
Answer
pH≈6.98. The solution is slightly acidic.
Question Problem 6.18
For 0.020 M HF with Ka=3.2×10⁻⁴, find the degree of dissociation, concentrations of HF, H₃O⁺ and F⁻, and pH.
Steps only
α≈0.12 (12%); [H₃O⁺]=[F⁻]≈2.4×10⁻³ M; [HF]≈1.76×10⁻² M; pH≈2.62.
Answer
α≈0.12 (12%); [H₃O⁺]=[F⁻]≈2.4×10⁻³ M; [HF]≈1.76×10⁻² M; pH≈2.62.
Question Problem 6.19
A 0.10 M monobasic acid has pH 4.50. Find [H⁺], [A⁻], [HA] and Ka.
Steps only
[H⁺]=[A⁻]≈3.16×10⁻⁵ M; [HA]≈0.09997 M; Ka≈1.0×10⁻⁸.
Answer
[H⁺]=[A⁻]≈3.16×10⁻⁵ M; [HA]≈0.09997 M; Ka≈1.0×10⁻⁸.
Question Problem 6.20
Find pH and percent dissociation for 0.080 M HOCl when Ka=2.5×10⁻⁵.
Steps only
pH≈2.85; percent dissociation≈1.76%.
Answer
pH≈2.85; percent dissociation≈1.76%.
Question Problem 6.21
A 0.004 M hydrazine solution has pH 9.7. Find Kb and pKb.
Steps only
Using pH 9.7: [OH⁻]≈5.0×10⁻⁵ M, Kb≈6.3×10⁻⁷ and pKb≈6.20. NCERT prints a different result from an inconsistent [H⁺] value.
Answer
Using the given pH exactly: Kb≈6.3×10⁻⁷ and pKb≈6.20. The printed NCERT calculation gives 8.96×10⁻⁷ and 6.04 because its [H⁺] value does not match pH 9.7.
Question Problem 6.22
Find the pH of a solution with 0.20 M NH₄Cl and 0.10 M NH₃. For ammonia, pKb=4.75.
Steps only
pH≈8.95.
Answer
pH≈8.95.
Question Problem 6.23
For 0.050 M NH₃, find its degree of ionisation and pH, then find Ka for NH₄⁺. Use Kb(NH₃)=1.8×10⁻⁵.
Steps only
Degree of ionisation≈1.9%; pH≈10.98; Ka(NH₄⁺)≈5.6×10⁻¹⁰.
Answer
Degree of ionisation≈1.9%; pH≈10.98; Ka(NH₄⁺)≈5.6×10⁻¹⁰.
Question Problem 6.24
Find the pH of 0.10 M NH₃, then after 50.0 mL is mixed with 25.0 mL of 0.10 M HCl. Use Kb=1.77×10⁻⁵.
Steps only
Initially pH≈11.12; after mixing pH≈9.25.
Answer
Initially pH≈11.12; after mixing pH≈9.25.
Question Problem 6.25
For ammonium acetate, pKa(acetic acid)=4.76 and pKb(ammonia)=4.75. Find pH.
Steps only
pH≈7.005 (about 7.01).
Answer
pH≈7.005 (about 7.01).
Question Problem 6.26
A₂X₃ has Ksp=1.1×10⁻²³ in pure water and neither ion reacts with water. Find its molar solubility S.
Steps only
Molar solubility≈1.0×10⁻⁵ mol L⁻¹.
Answer
Molar solubility≈1.0×10⁻⁵ mol L⁻¹.
Question Problem 6.27
Ksp(Ni(OH)₂)=2.0×10⁻¹⁵ and Ksp(AgCN)=6.0×10⁻¹⁷. Which salt is more soluble?
Steps only
Ni(OH)₂ is more soluble.
Answer
Ni(OH)₂ is more soluble.
Question Problem 6.28
Find the molar solubility of Ni(OH)₂ in 0.10 M NaOH if Ksp=2.0×10⁻¹⁵.
Steps only
Molar solubility≈2.0×10⁻¹³ M.
Answer
Molar solubility≈2.0×10⁻¹³ M.
Chapter 6 Exercises — Part 1
Textbook pages 208–210
Question 6.1
A liquid is in equilibrium with its vapour in a sealed container at fixed temperature. The container volume is suddenly increased. What happens first to vapour pressure, evaporation and condensation? What is the final vapour pressure?
Steps only
Pressure first falls; evaporation > condensation; final pressure is the same at fixed temperature.
Answer
Pressure first falls; evaporation > condensation; final pressure is the same at fixed temperature.
Question 6.2
Find Kc for 2SO₂(g)+O₂(g)⇌2SO₃(g), when [SO₂]=0.60 M, [O₂]=0.82 M and [SO₃]=1.90 M at equilibrium.
Steps only
Kc=(1.90)²/[(0.60)²(0.82)]≈12.2 M⁻¹.
Answer
Kc=(1.90)²/[(0.60)²(0.82)]≈12.2 M⁻¹.
Question 6.3
At a total pressure of 10⁵ Pa, iodine vapour contains 40% I atoms by volume. For I₂(g)⇌2I(g), calculate Kp.
Steps only
Kp=(4.0×10⁴)²/(6.0×10⁴)=2.67×10⁴ Pa.
Answer
Kp=(4.0×10⁴)²/(6.0×10⁴)=2.67×10⁴ Pa.
Question 6.4
Write Kc for: (i) 2NOCl(g)⇌2NO(g)+Cl₂(g); (ii) 2Cu(NO₃)₂(s)⇌2CuO(s)+4NO₂(g)+O₂(g); (iii) CH₃COOC₂H₅(aq)+H₂O(l)⇌CH₃COOH(aq)+C₂H₅OH(aq); (iv) Fe³⁺(aq)+3OH⁻(aq)⇌Fe(OH)₃(s); (v) I₂(s)+5F₂(g)⇌2IF₅(g).
Steps only
Use products over reactants; omit pure solids and liquids.
Answer
Use products over reactants; omit pure solids and liquids.
Question 6.5
Find Kc from Kp for: (i) 2NOCl(g)⇌2NO(g)+Cl₂(g), Kp=1.8×10⁻² at 500 K; (ii) CaCO₃(s)⇌CaO(s)+CO₂(g), Kp=167 at 1073 K.
Steps only
Both have Δn=1, so divide Kp by RT.
Answer
Both have Δn=1, so divide Kp by RT.
Question 6.6
For NO(g)+O₃(g)⇌NO₂(g)+O₂(g), Kc=6.3×10¹⁴ at 1000 K. What is Kc for the reverse reaction?
Steps only
Kreverse=1/Kforward=1.59×10⁻¹⁵.
Answer
Kreverse=1/Kforward=1.59×10⁻¹⁵.
Question 6.7
Why are pure liquids and solids left out of equilibrium constant expressions?
Steps only
Pure solids/liquids have constant activity and are omitted.
Answer
Pure solids/liquids have constant activity and are omitted.
Question 6.8
N₂ and O₂ react by 2N₂(g)+O₂(g)⇌2N₂O(g). Initially 0.482 mol N₂ and 0.933 mol O₂ are placed in a 10 L vessel. Kc=2.0×10⁻³⁷. Find the equilibrium mixture.
Steps only
Kc is tiny, so almost no reaction occurs; N₂O formed≈1.3×10⁻¹⁹ mol.
Answer
Kc is tiny, so almost no reaction occurs; N₂O formed≈1.3×10⁻¹⁹ mol.
Question 6.9
For 2NO(g)+Br₂(g)⇌2NOBr(g), 0.087 mol NO and 0.0437 mol Br₂ are mixed. At equilibrium 0.0518 mol NOBr is present. Find the equilibrium amounts of NO and Br₂.
Steps only
Subtract consumed amounts: NO=0.0352 mol; Br₂=0.0178 mol.
Answer
Subtract consumed amounts: NO=0.0352 mol; Br₂=0.0178 mol.
Question 6.10
For 2SO₂(g)+O₂(g)⇌2SO₃(g), Kp=2.0×10¹⁰ bar⁻¹ at 450 K. Find Kc.
Steps only
Kc=KpRT=(2.0×10¹⁰)(0.08314×450)≈7.5×10¹¹ L mol⁻¹.
Answer
Kc=KpRT=(2.0×10¹⁰)(0.08314×450)≈7.5×10¹¹ L mol⁻¹.
Question 6.11
A sample of HI(g) starts at 0.20 atm. At equilibrium, p(HI)=0.04 atm for 2HI(g)⇌H₂(g)+I₂(g). Find Kp.
Steps only
p(H₂)=p(I₂)=0.08 atm; Kp=(0.08×0.08)/(0.04²)=4.
Answer
p(H₂)=p(I₂)=0.08 atm; Kp=(0.08×0.08)/(0.04²)=4.
Question 6.12
A 20 L vessel contains 1.57 mol N₂, 1.92 mol H₂ and 8.13 mol NH₃ at 500 K. For N₂+3H₂⇌2NH₃, Kc=1.7×10². Is the mixture at equilibrium? If not, which way will it change?
Steps only
Qc≈2.38×10³, greater than Kc; net reaction goes in reverse.
Answer
Qc≈2.38×10³, greater than Kc; net reaction goes in reverse.
Question 6.13
An equilibrium expression is Kc=[NH₃]⁴[O₂]⁵/([NO]⁴[H₂O]⁶). Write the balanced reaction.
Steps only
Denominator = reactants; numerator = products: 4NO+6H₂O⇌4NH₃+5O₂.
Answer
Denominator = reactants; numerator = products: 4NO+6H₂O⇌4NH₃+5O₂.
Question 6.14
One mole each of H₂O(g) and CO(g) are heated in a 10 L vessel. At equilibrium, 40% of the water has reacted: H₂O+CO⇌H₂+CO₂. Find Kc.
Steps only
Kc=(0.040)²/(0.060)²≈0.444.
Answer
Kc=(0.040)²/(0.060)²≈0.444.
Question 6.15
For H₂(g)+I₂(g)⇌2HI(g), Kc=54.8 at 700 K. At equilibrium [HI]=0.50 M. The mixture began with HI only. Find [H₂] and [I₂].
Steps only
x=√(0.50²/54.8)≈0.0675 M for both H₂ and I₂.
Answer
x=√(0.50²/54.8)≈0.0675 M for both H₂ and I₂.
Question 6.16
Initially [ICl]=0.78 M for 2ICl(g)⇌I₂(g)+Cl₂(g), with Kc=0.14. Find all equilibrium concentrations.
Steps only
Set x²/(0.78−2x)²=0.14; x≈0.167 M.
Answer
Set x²/(0.78−2x)²=0.14; x≈0.167 M.
Question 6.17
Ethane starts at 4.0 atm in a flask and reaches equilibrium for C₂H₆(g)⇌C₂H₄(g)+H₂(g). At 899 K, Kp=0.04 atm. Find the equilibrium concentration of C₂H₆.
Steps only
Solve x²/(4−x)=0.04: p(C₂H₆)≈3.62 atm; c=p/(RT)≈0.0491 mol L⁻¹.
Answer
c(C₂H₆)=p/(RT)≈0.0491 mol L⁻¹.
Question 6.18
For CH₃COOH(l)+C₂H₅OH(l)⇌CH₃COOC₂H₅(l)+H₂O(l): (i) write Qc; (ii) 1.00 mol acid and 0.18 mol ethanol give 0.171 mol ester at equilibrium—find Kc; (iii) with 1.00 mol acid and 0.50 mol ethanol, 0.214 mol ester is present. Has equilibrium been reached?
Steps only
Kc≈3.92. In (iii), Qc≈0.204<Kc, so more ester forms.
Answer
Kc≈3.92. In (iii), Qc≈0.204<Kc, so more ester forms.
Question 6.19
Pure PCl₅ is placed in a vessel. At equilibrium [PCl₅]=0.050 M for PCl₅(g)⇌PCl₃(g)+Cl₂(g); Kc=8.3×10⁻³. Find [PCl₃] and [Cl₂].
Steps only
x=√(8.3×10⁻³×0.050)≈0.0204 M.
Answer
x=√(8.3×10⁻³×0.050)≈0.0204 M.
Question 6.20
For FeO(s)+CO(g)⇌Fe(s)+CO₂(g), Kp=0.265 at 1050 K. Initially pCO=1.4 atm and pCO₂=0.80 atm. Find the equilibrium partial pressures.
Steps only
Qc initially exceeds Kp, so shift left; solve (0.80−x)/(1.4+x)=0.265.
Answer
Qc initially exceeds Kp, so shift left; solve (0.80−x)/(1.4+x)=0.265.
Question 6.21
For N₂(g)+3H₂(g)⇌2NH₃(g), Kc=0.061 at 500 K. At one moment [N₂]=3.0 M, [H₂]=2.0 M and [NH₃]=0.50 M. Is the mixture at equilibrium? Which way will it shift?
Steps only
Qc=0.0104<Kc; net reaction proceeds right.
Answer
Qc=0.0104<Kc; net reaction proceeds right.
Question 6.22
Initially [BrCl]=3.3×10⁻³ M for 2BrCl(g)⇌Br₂(g)+Cl₂(g), Kc=32 at 500 K. Find [BrCl] at equilibrium.
Steps only
[BrCl]=3.3×10⁻³−2x with x/(3.3×10⁻³−2x)=√32; [BrCl]≈2.7×10⁻⁴ M.
Answer
[BrCl]=3.3×10⁻³−2x with x/(3.3×10⁻³−2x)=√32; [BrCl]≈2.7×10⁻⁴ M.
Question 6.23
At 1127 K and 1 atm, a mixture of CO and CO₂ contains 90.55% CO by mass and is in equilibrium with solid carbon: C(s)+CO₂(g)⇌2CO(g). Calculate Kc.
Steps only
Mass fractions → partial pressures; Kp≈14.1 atm, then Kc=Kp/RT≈0.152 M.
Answer
Mass fractions → partial pressures; Kp≈14.1 atm, then Kc=Kp/RT≈0.152 M.
Question 6.24
Find (a) ΔrG° and (b) K at 298 K for NO(g)+½O₂(g)⇌NO₂(g), given ΔfG°(NO₂)=52.0 kJ mol⁻¹, ΔfG°(NO)=87.0 kJ mol⁻¹ and ΔfG°(O₂)=0.
Steps only
ΔrG°=−35.0 kJ mol⁻¹; K=exp(35000/RT)≈1.4×10⁶.
Answer
ΔrG°=−35.0 kJ mol⁻¹; K=exp(35000/RT)≈1.4×10⁶.
Question 6.25
When pressure is decreased by increasing volume, do product amounts increase, decrease or stay the same for: (a) PCl₅(g)⇌PCl₃(g)+Cl₂(g); (b) CaO(s)+CO₂(g)⇌CaCO₃(s); (c) 3Fe(s)+4H₂O(g)⇌Fe₃O₄(s)+4H₂(g)?
Steps only
Lower pressure favours more gas: (a) right, (b) left, (c) no shift.
Answer
Lower pressure favours more gas: (a) right, (b) left, (c) no shift.
Question 6.26
For each reaction, say whether increasing pressure changes the equilibrium and, if it does, which direction it shifts: (i) COCl₂⇌CO+Cl₂; (ii) CH₄+2S₂⇌CS₂+2H₂S; (iii) CO₂+C(s)⇌2CO; (iv) 2H₂+CO⇌CH₃OH; (v) CaCO₃(s)⇌CaO(s)+CO₂; (vi) 4NH₃+5O₂⇌4NO+6H₂O.
Steps only
Pressure favours fewer gas moles; equal gas mole counts mean no shift.
Answer
Pressure favours fewer gas moles; equal gas mole counts mean no shift.
Question 6.27
For H₂(g)+Br₂(g)⇌2HBr(g), Kp=1.6×10⁵ at 1024 K. Initially only HBr is present at 10.0 bar. Find equilibrium pressures of all gases.
Steps only
Let x be each product pressure: (10−2x)²/x²=1.6×10⁵.
Answer
Let x be each product pressure: (10−2x)²/x²=1.6×10⁵.
Question 6.28
Hydrogen is produced by the endothermic reaction CH₄(g)+H₂O(g)⇌CO(g)+3H₂(g). (a) Write Kp. (b) State effects of increasing pressure, increasing temperature and adding a catalyst on Kp and mixture composition.
Steps only
Pressure: shift left; temperature: Kp rises and shifts right; catalyst: no equilibrium change.
Answer
Pressure: shift left; temperature: Kp rises and shifts right; catalyst: no equilibrium change.
Question 6.29
For 2H₂(g)+CO(g)⇌CH₃OH(g), predict the shift after (a) adding H₂, (b) adding CH₃OH, (c) removing CO, and (d) removing CH₃OH.
Steps only
Added species are consumed; removed species are replaced.
Answer
Added species are consumed; removed species are replaced.
Question 6.30
For PCl₅(g)⇌PCl₃(g)+Cl₂(g), Kc=8.3×10⁻³ at 473 K and ΔrH°=+124.0 kJ mol⁻¹: (a) write Kc; (b) find Kc for the reverse reaction; (c) state how adding PCl₅, increasing pressure and increasing temperature affect Kc.
Steps only
K changes only with temperature; endothermic forward reaction means Kc rises as T rises.
Answer
K changes only with temperature; endothermic forward reaction means Kc rises as T rises.
Question 6.31
For CO(g)+H₂O(g)⇌CO₂(g)+H₂(g), Kp=10.1 at 400°C. Initially pCO=pH₂O=4.0 bar and there are no products. Find pH₂ at equilibrium.
Steps only
x²/(4−x)²=10.1, so x≈3.04 bar.
Answer
x²/(4−x)²=10.1, so x≈3.04 bar.
Question 6.32
For each reaction, say whether equilibrium contains mostly reactants, mostly products, or appreciable amounts of both: (a) Cl₂⇌2Cl, Kc=5×10⁻³⁹; (b) Cl₂+2NO⇌2NOCl, Kc=3.7×10⁸; (c) Cl₂+2NO₂⇌2NO₂Cl, Kc=1.8.
Steps only
Small K → reactants; large K → products; K near 1 → both.
Answer
Small K → reactants; large K → products; K near 1 → both.
Question 6.33
For 3O₂(g)⇌2O₃(g), Kc=2.0×10⁻⁵⁰ at 25°C and [O₂]=1.6×10⁻² M. Find [O₃].
Steps only
[O₃]=√(2.0×10⁻⁵⁰×(1.6×10⁻²)³)≈2.9×10⁻²⁸ M.
Answer
[O₃]=√(2.0×10⁻⁵⁰×(1.6×10⁻²)³)≈2.9×10⁻²⁸ M.
Question 6.34
At 1300 K, CO(g)+3H₂(g)⇌CH₄(g)+H₂O(g) has Kc=3.90. In a 1 L flask, [CO]=0.30 M, [H₂]=0.10 M and [H₂O]=0.020 M. Find [CH₄].
Steps only
[CH₄]=3.90×0.30×(0.10)³/0.020=0.0585 M.
Answer
[CH₄]=3.90×0.30×(0.10)³/0.020=0.0585 M.
Question 6.35
What is a conjugate acid–base pair? Give the conjugate acid or base for HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻ and S²⁻.
Steps only
Acid loses H⁺; base gains H⁺.
Answer
Acid loses H⁺; base gains H⁺.
Question 6.36
Which are Lewis acids: H₂O, BF₃, H⁺ and NH₄⁺?
Steps only
Lewis acids here: BF₃ and H⁺.
Answer
Lewis acids here: BF₃ and H⁺.
Question 6.37
Write the conjugate bases of HF, H₂SO₄ and HCO₃⁻.
Steps only
Remove one H⁺ from each acid.
Answer
Remove one H⁺ from each acid.
Question 6.38
Write the conjugate acids of NH₂⁻, NH₃ and HCOO⁻.
Steps only
Add one H⁺ to each base.
Answer
Add one H⁺ to each base.
Question 6.39
H₂O, HCO₃⁻, HSO₄⁻ and NH₃ can each act as a Brønsted acid or base. Give the conjugate acid and conjugate base of each.
Steps only
Conjugate acid/base pairs follow by adding/removing one H⁺.
Answer
Conjugate acid/base pairs follow by adding/removing one H⁺.
Question 6.40
Classify OH⁻, F⁻, H⁺ and BCl₃ as Lewis acids or bases, and show a simple electron-pair reaction for each type.
Steps only
Bases donate pairs (OH⁻, F⁻); acids accept pairs (H⁺, BCl₃).
Answer
Bases donate pairs (OH⁻, F⁻); acids accept pairs (H⁺, BCl₃).
Question 6.41
A soft drink has [H⁺]=3.8×10⁻³ M. Find its pH.
Steps only
Use pH=−log[H⁺]. pH=−log(3.8×10⁻³)≈2.42.
Answer
pH≈2.42.
Question 6.42
Vinegar has pH=3.76. Find [H⁺].
Steps only
Use [H⁺]=10⁻ᵖᴴ. [H⁺]=10⁻³·⁷⁶≈1.74×10⁻⁴ M.
Answer
[H⁺]≈1.74×10⁻⁴ M.
Question 6.43
At 298 K, Ka is 6.8×10⁻⁴ for HF, 1.8×10⁻⁴ for HCOOH and 4.8×10⁻⁹ for HCN. Find Kb for each conjugate base.
Steps only
Use KaKb=Kw=1.0×10⁻¹⁴. Kb(F⁻)=Kw/Ka=1.47×10⁻¹¹; Kb(HCOO⁻)=5.56×10⁻¹¹. Kb(CN⁻)=2.08×10⁻⁶.
Answer
Kb(F⁻)=1.47×10⁻¹¹; Kb(HCOO⁻)=5.56×10⁻¹¹; Kb(CN⁻)=2.08×10⁻⁶.
Question 6.44
Phenol has Ka=1.0×10⁻¹⁰. Find phenolate concentration in 0.050 M phenol and degree of ionisation when 0.010 M sodium phenolate is also present.
Steps only
Without added phenolate, x≈√(KaC)=√(1.0×10⁻¹⁰×0.050)=2.24×10⁻⁶ M. With 0.010 M common ion, added phenolate y≈Ka(0.050)/(0.010)=5.0×10⁻¹⁰ M. Degree of ionisation α=y/C=1.0×10⁻⁸.
Answer
[C₆H₅O⁻]≈2.24×10⁻⁶ M in pure phenol; with common ion, α≈1.0×10⁻⁸.
Question 6.45
For 0.10 M H₂S, Ka₁=9.1×10⁻⁸ and Ka₂=1.2×10⁻¹³. Find [HS⁻] and [S²⁻] (i) in water and (ii) in 0.10 M HCl.
Steps only
In water, [HS⁻]≈√(Ka₁C)=9.5×10⁻⁵ M and [S²⁻]≈Ka₂[HS⁻]/[H⁺]≈1.2×10⁻¹³ M. With 0.10 M HCl, [H⁺]≈0.10 M, so [HS⁻]≈Ka₁C/[H⁺]=9.1×10⁻⁸ M. Then [S²⁻]≈Ka₂[HS⁻]/[H⁺]≈1.1×10⁻¹⁹ M.
Answer
Water: [HS⁻]≈9.5×10⁻⁵ M, [S²⁻]≈1.2×10⁻¹³ M. With HCl: 9.1×10⁻⁸ M and 1.1×10⁻¹⁹ M.
Question 6.46
Ka for acetic acid is 1.74×10⁻⁵. Find degree of dissociation, acetate concentration and pH in 0.050 M solution.
Steps only
x=[H⁺]=[CH₃COO⁻]≈√(KaC)=9.33×10⁻⁴ M. α=x/C≈0.0187 (1.87%). pH=−log x≈3.03.
Answer
α≈1.87%; [CH₃COO⁻]≈9.33×10⁻⁴ M; pH≈3.03.
Question 6.47
A 0.010 M monoprotic organic acid has pH=4.15. Find anion concentration, Ka and pKa.
Steps only
[H⁺]=10⁻⁴·¹⁵≈7.08×10⁻⁵ M, which is also [A⁻]. [HA]≈0.010−0.0000708=0.00993 M. Ka=[H⁺][A⁻]/[HA]≈5.05×10⁻⁷; pKa≈6.30.
Answer
[A⁻]≈7.08×10⁻⁵ M; Ka≈5.05×10⁻⁷; pKa≈6.30.
Question 6.48
Assuming full dissociation, find pH: (a) 0.003 M HCl; (b) 0.005 M NaOH; (c) 0.002 M HBr; (d) 0.002 M KOH.
Steps only
Strong acid: pH=−log C. Strong base: pH=14−(−log[OH⁻]). (a) 2.52; (b) 11.70; (c) 2.70; (d) 11.30.
Answer
(a) 2.52; (b) 11.70; (c) 2.70; (d) 11.30.
Question 6.49
Find pH: (a) 2.0 g TlOH in 2.0 L; (b) 0.30 g Ca(OH)₂ in 0.50 L; (c) 0.30 g NaOH in 0.20 L; (d) 1.0 mL 13.6 M HCl diluted to 1.0 L.
Steps only
(a) [OH⁻]=(2/221.4)/2=0.00452 M; pH≈11.66. (b) [OH⁻]=2(0.30/74.1)/0.50=0.0162 M; pH≈12.21. (c) [OH⁻]=(0.30/40.0)/0.20=0.0375 M; pH≈12.57. (d) [H⁺]=13.6×0.001=0.0136 M; pH≈1.87.
Answer
(a) 11.66; (b) 12.21; (c) 12.57; (d) 1.87.
Question 6.50
A 0.10 M bromoacetic acid solution has α=0.132. Find pH and pKa.
Steps only
[H⁺]=Cα=0.0132 M, so pH≈1.88. [HA]=0.10(1−0.132)=0.0868 M. Ka=(0.0132)²/0.0868≈2.01×10⁻³; pKa≈2.70.
Answer
pH≈1.88 and pKa≈2.70.
Question 6.51
A 0.005 M codeine solution has pH=9.95. Find its ionisation constant Kb and pKb.
Steps only
pOH=14−9.95=4.05, so [OH⁻]=10⁻⁴·⁰⁵≈8.91×10⁻⁵ M. For B+H₂O⇌BH⁺+OH⁻, let x=[OH⁻]=[BH⁺]. Then [B]=0.005−x. Kb=x²/(0.005−x)≈1.62×10⁻⁶; pKb=−log Kb≈5.79.
Answer
Kb≈1.62×10⁻⁶ and pKb≈5.79.
Question 6.52
Find pH and degree of ionisation for 0.001 M aniline, using Kb=4.3×10⁻¹⁰. Also find Ka of its conjugate acid.
Steps only
For weak base aniline, [OH⁻]≈√(KbC)=√(4.3×10⁻¹⁰×0.001)=6.56×10⁻⁷ M. pOH≈6.18, so pH≈7.82. Degree of ionisation α=[OH⁻]/C≈6.56×10⁻⁴, or 0.0656%. For the conjugate acid, Ka=Kw/Kb=1.0×10⁻¹⁴/(4.3×10⁻¹⁰)=2.33×10⁻⁵.
Answer
pH≈7.82; α≈6.56×10⁻⁴ (0.0656%); Ka(anilinium)≈2.33×10⁻⁵.
Question 6.53
Acetic acid has pKa=4.74. Find its degree of ionisation in 0.050 M solution and when the solution also contains (a) 0.010 M HCl or (b) 0.10 M HCl.
Steps only
Ka=10⁻⁴·⁷⁴≈1.82×10⁻⁵. In pure acid, α≈√(Ka/C)=√(1.82×10⁻⁵/0.050)=0.0191, or 1.91%. With 0.010 M strong acid, common H⁺ suppresses ionisation: α≈Ka/[H⁺]=1.82×10⁻³ (0.182%). With 0.10 M HCl, α≈1.82×10⁻⁴ (0.0182%).
Answer
Pure: 1.91%; with 0.010 M HCl: 0.182%; with 0.10 M HCl: 0.0182%.
Question 6.54
Dimethylamine has Kb=5.4×10⁻⁴. Find its degree of ionisation in 0.020 M solution, and the percentage ionised when 0.10 M NaOH is also present.
Steps only
Without added OH⁻, solve x²/(0.020−x)=5.4×10⁻⁴; the positive root is x≈0.00330 M. Degree of ionisation α=x/C=0.00330/0.020≈0.165, or 16.5%. With 0.10 M OH⁻ already present, x≈KbC/[OH⁻]=(5.4×10⁻⁴×0.020)/0.10=1.08×10⁻⁴ M. This is 0.54% of the original base.
Answer
Degree ionised≈16.5% in water and ≈0.54% with 0.10 M NaOH.
Question 6.55
Find [H⁺] for: (a) muscle fluid pH 6.83; (b) stomach fluid pH 1.2; (c) blood pH 7.38; (d) saliva pH 6.4.
Steps only
Use [H⁺]=10⁻ᵖᴴ for each fluid. (a) 10⁻⁶·⁸³=1.48×10⁻⁷ M; (b) 10⁻¹·²=6.31×10⁻² M. (c) 10⁻⁷·³⁸=4.17×10⁻⁸ M; (d) 10⁻⁶·⁴=3.98×10⁻⁷ M.
Answer
(a) 1.48×10⁻⁷ M; (b) 6.31×10⁻² M; (c) 4.17×10⁻⁸ M; (d) 3.98×10⁻⁷ M.
Question 6.56
Find [H⁺] for milk (pH 6.8), black coffee (5.0), tomato juice (4.2), lemon juice (2.2) and egg white (7.8).
Steps only
Use [H⁺]=10⁻ᵖᴴ. Milk: 1.58×10⁻⁷ M; coffee: 1.00×10⁻⁵ M; tomato juice: 6.31×10⁻⁵ M. Lemon juice: 6.31×10⁻³ M; egg white: 1.58×10⁻⁸ M.
Answer
Milk 1.58×10⁻⁷ M; coffee 1.00×10⁻⁵ M; tomato 6.31×10⁻⁵ M; lemon 6.31×10⁻³ M; egg white 1.58×10⁻⁸ M.
Question 6.57
0.561 g KOH is dissolved to make 200 mL solution at 298 K. Find [K⁺], [OH⁻], [H⁺] and pH.
Steps only
M(KOH)≈56.1 g mol⁻¹, so moles=0.561/56.1=0.0100 mol. Concentration=0.0100/0.200=0.0500 M. KOH fully separates: [K⁺]=[OH⁻]=0.0500 M. [H⁺]=Kw/[OH⁻]=1.0×10⁻¹⁴/0.0500=2.0×10⁻¹³ M; pOH=1.301 and pH=12.70.
Answer
[K⁺]=[OH⁻]=0.0500 M; [H⁺]=2.0×10⁻¹³ M; pH=12.70.
Question 6.58
The solubility of Sr(OH)₂ at 298 K is 19.23 g L⁻¹. Find [Sr²⁺], [OH⁻] and pH.
Steps only
M(Sr(OH)₂)≈121.6 g mol⁻¹, so molar solubility s=19.23/121.6≈0.158 M. Sr(OH)₂→Sr²⁺+2OH⁻, so [Sr²⁺]=0.158 M and [OH⁻]=0.316 M. pOH=−log(0.316)=0.500, so pH=14−0.500=13.50.
Answer
[Sr²⁺]≈0.158 M; [OH⁻]≈0.316 M; pH≈13.50.
Question 6.59
Propanoic acid has Ka=1.32×10⁻⁵. Find its degree of ionisation and pH in 0.050 M solution, then its degree of ionisation when 0.010 M HCl is added.
Steps only
In pure acid, x≈√(KaC)=√(1.32×10⁻⁵×0.050)=8.12×10⁻⁴ M. Thus α=x/C≈0.0162 (1.62%) and pH≈3.09. With 0.010 M HCl, added H⁺ suppresses ionisation: x≈KaC/[H⁺]=(1.32×10⁻⁵×0.050)/0.010=6.6×10⁻⁵ M. Degree of ionisation is then x/C=0.00132, or 0.132%.
Answer
In water: α≈1.62%, pH≈3.09. With 0.010 M HCl: α≈0.132%.
Question 6.60
A 0.10 M cyanic acid solution (HCNO) has pH=2.34. Find Ka and degree of ionisation.
Steps only
[H⁺]=10⁻²·³⁴≈4.57×10⁻³ M. For the monoprotic acid, [A⁻]≈[H⁺]. [HA]=0.10−0.00457=0.0954 M. Thus Ka=[H⁺][A⁻]/[HA]≈(4.57×10⁻³)²/0.0954=2.19×10⁻⁴. Degree of ionisation α=[A⁻]/C=0.00457/0.10=0.0457, or 4.57%.
Answer
Ka≈2.19×10⁻⁴; degree of ionisation≈4.57%.
Question 6.61
Nitrous acid has Ka=4.5×10⁻⁴. Find pH and degree of hydrolysis for 0.040 M NaNO₂.
Steps only
NO₂⁻ is the conjugate base of weak HNO₂, so Kb=Kw/Ka=1.0×10⁻¹⁴/(4.5×10⁻⁴)=2.22×10⁻¹¹. [OH⁻]≈√(KbC)=√(2.22×10⁻¹¹×0.040)=9.43×10⁻⁷ M. Thus pOH≈6.03 and pH≈7.97. Degree of hydrolysis h≈[OH⁻]/C=2.36×10⁻⁵.
Answer
pH≈7.97; degree of hydrolysis≈2.36×10⁻⁵.
Question 6.62
A 0.020 M pyridinium hydrochloride solution has pH=3.44. Find Kb for pyridine.
Steps only
Pyridinium is a weak acid. [H⁺]=10⁻³·⁴⁴=3.63×10⁻⁴ M. Ka(pyridinium)≈x²/(C−x)=(3.63×10⁻⁴)²/(0.020−0.000363)=6.71×10⁻⁶. For pyridine and pyridinium, KaKb=Kw, so Kb=1.0×10⁻¹⁴/(6.71×10⁻⁶)=1.49×10⁻⁹.
Answer
Kb(pyridine)≈1.5×10⁻⁹.
Question 6.63
Classify aqueous solutions of NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF as acidic, basic or neutral.
Steps only
A salt from strong acid and strong base is neutral. A conjugate ion of a weak acid makes solution basic; a conjugate acid of a weak base makes it acidic. NaCl and KBr are neutral. CN⁻, NO₂⁻ and F⁻ are bases, so NaCN, NaNO₂ and KF are basic. NH₄⁺ is the conjugate acid of weak NH₃, so NH₄NO₃ is acidic.
Answer
NaCl neutral; KBr neutral; NaCN basic; NH₄NO₃ acidic; NaNO₂ basic; KF basic.
Question 6.64
Chloroacetic acid has Ka=1.35×10⁻³. Find the pH of (i) 0.10 M acid and (ii) a solution containing 0.10 M acid and 0.10 M sodium chloroacetate.
Steps only
(i) Let x=[H⁺]. Solve x²/(0.10−x)=1.35×10⁻³. The positive root is x≈0.01096 M, so pH≈1.96. (ii) Equal amounts of weak acid and conjugate base form a buffer. pH=pKa+log(base/acid)=pKa=−log(1.35×10⁻³)≈2.87.
Answer
- (i) pH≈1.96;
- (ii) pH≈2.87.
Question 6.65
At 310 K, Kw=2.7×10⁻¹⁴. Find the pH of neutral water.
Steps only
In neutral water, [H⁺]=[OH⁻]=√Kw. [H⁺]=√(2.7×10⁻¹⁴)=1.64×10⁻⁷ M. pH=−log(1.64×10⁻⁷)≈6.78. Neutral pH need not equal 7 when temperature changes.
Answer
Neutral water at 310 K has pH≈6.78.
Question 6.66
Find the pH after mixing: (a) 10 mL 0.20 M Ca(OH)₂ + 25 mL 0.10 M HCl; (b) 10 mL 0.010 M H₂SO₄ + 10 mL 0.010 M Ca(OH)₂; (c) 10 mL 0.10 M H₂SO₄ + 10 mL 0.10 M KOH.
Steps only
(a) OH⁻ moles=0.010×0.20×2=0.0040; H⁺ moles=0.025×0.10=0.0025. Excess OH⁻=0.0015 mol in 0.035 L, so [OH⁻]=0.0429 M and pH≈12.63. (b) H₂SO₄ supplies 2 H⁺ per mole: 0.010×0.010×2=0.00020 mol H⁺. Ca(OH)₂ supplies 0.00020 mol OH⁻. They neutralise equally; pH≈7 (at 298 K). (c) H₂SO₄ gives 0.0020 mol H⁺ equivalents; KOH gives 0.0010 mol OH⁻. Excess H⁺=0.0010 mol in 0.020 L, so [H⁺]=0.050 M and pH≈1.30.
Answer
(a) pH≈12.63; (b) pH≈7; (c) pH≈1.30.
Question 6.67
Using the chapter-table Ksp values at 298 K (Ag₂CrO₄ 1.1×10⁻¹²; BaCrO₄ 2.1×10⁻¹⁰; Fe(OH)₃ 1.0×10⁻³⁸; PbCl₂ 1.7×10⁻⁵; Hg₂I₂ 4.5×10⁻²⁹), calculate molar solubility and ion concentrations.
Steps only
For Ag₂CrO₄⇌2Ag⁺+CrO₄²⁻, Ksp=4s³, so s=6.50×10⁻⁵ M; [Ag⁺]=1.30×10⁻⁴ M and [CrO₄²⁻]=6.50×10⁻⁵ M. For BaCrO₄⇌Ba²⁺+CrO₄²⁻, s=√Ksp=1.45×10⁻⁵ M; each ion has this concentration. For Fe(OH)₃⇌Fe³⁺+3OH⁻, Ksp=27s⁴, so s≈1.39×10⁻¹⁰ M; [Fe³⁺]=s and [OH⁻]=3s≈4.17×10⁻¹⁰ M. For PbCl₂, Ksp=4s³, giving s≈1.62×10⁻² M; [Pb²⁺]=s, [Cl⁻]=2s. For Hg₂I₂, Ksp=4s³, giving s≈2.24×10⁻¹⁰ M; [Hg₂²⁺]=s, [I⁻]=2s.
Answer
Ag₂CrO₄: s=6.50×10⁻⁵ M, [Ag⁺]=1.30×10⁻⁴ M, [CrO₄²⁻]=6.50×10⁻⁵ M. BaCrO₄: s=[Ba²⁺]=[CrO₄²⁻]=1.45×10⁻⁵ M. Fe(OH)₃: s=[Fe³⁺]=1.39×10⁻¹⁰ M, [OH⁻]=4.17×10⁻¹⁰ M. PbCl₂: s=[Pb²⁺]=1.62×10⁻² M, [Cl⁻]=3.24×10⁻² M. Hg₂I₂: s=[Hg₂²⁺]=2.24×10⁻¹⁰ M, [I⁻]=4.48×10⁻¹⁰ M.
Question 6.68
Ksp(Ag₂CrO₄)=1.1×10⁻¹² and Ksp(AgBr)=5.0×10⁻¹³. Find the ratio of molarities of their saturated solutions.
Steps only
For Ag₂CrO₄, Ksp=4s³, so s=(1.1×10⁻¹²/4)^(1/3)=6.50×10⁻⁵ M. For AgBr, Ksp=s², so s=√(5.0×10⁻¹³)=7.07×10⁻⁷ M. Ratio s(Ag₂CrO₄)/s(AgBr)≈6.50×10⁻⁵/7.07×10⁻⁷≈92.
Answer
Ag₂CrO₄ saturated solution is about 92 times more concentrated (by formula-unit molarity) than AgBr.
Question 6.69
Equal volumes of 0.002 M sodium iodate and 0.002 M copper(II) chlorate are mixed. Will Cu(IO₃)₂ precipitate? Ksp=7.4×10⁻⁸.
Steps only
Equal-volume mixing halves each concentration: [Cu²⁺]=0.001 M and [IO₃⁻]=0.001 M. For Cu(IO₃)₂, Qsp=[Cu²⁺][IO₃⁻]²=(10⁻³)(10⁻³)²=1.0×10⁻⁹. Since Qsp<Ksp, the solution is not saturated enough to precipitate the salt.
Answer
No. Qsp=1.0×10⁻⁹<Ksp=7.4×10⁻⁸.
Question 6.70
Benzoic acid has Ka=6.46×10⁻⁵ and Ksp(Ag benzoate)=2.5×10⁻¹³. How many times more soluble is silver benzoate in a pH 3.19 buffer than in pure water?
Steps only
In the buffer, only a fraction f=Ka/(Ka+[H⁺]) of dissolved benzoate remains as benzoate ion. At pH 3.19, [H⁺]=6.46×10⁻⁴ M, so f≈0.0909. For the salt, Ksp=s(f s), so sbuffer=√(Ksp/f). In pure water, the weak benzoate hydrolysis leaves approximately f≈1 for this comparison, so spure≈√Ksp. The ratio is sbuffer/spure≈1/√f≈3.32.
Answer
Silver benzoate is about 3.3 times more soluble in the pH 3.19 buffer.
Question 6.71
Equal volumes of equimolar FeSO₄ and Na₂S solutions are mixed. What is the largest initial concentration each solution can have without FeS precipitation? Ksp(FeS)=6.3×10⁻¹⁸.
Steps only
After equal-volume mixing, both ion concentrations are halved: [Fe²⁺]=[S²⁻]=C/2. At the precipitation limit, Qsp=Ksp: (C/2)²=6.3×10⁻¹⁸. C=2√(6.3×10⁻¹⁸)≈5.0×10⁻⁹ M.
Answer
Each starting solution must be at or below about 5.0×10⁻⁹ M.
Question 6.72
How much water is needed to dissolve 1.0 g CaSO₄ at 298 K? Ksp=9.1×10⁻⁶.
Steps only
CaSO₄⇌Ca²⁺+SO₄²⁻, so Ksp=s² and s=√(9.1×10⁻⁶)=3.02×10⁻³ mol L⁻¹. M(CaSO₄)≈136.1 g mol⁻¹, so 1.0 g is 0.00735 mol. Volume needed=amount/solubility=0.00735/(3.02×10⁻³)≈2.44 L.
Answer
About 2.44 L water is needed.
Question 6.73
A 10 mL sample of 0.1 M HCl saturated with H₂S has [S²⁻]=1.0×10⁻¹⁹ M. It is added separately to 5 mL of 0.040 M FeSO₄, MnCl₂, ZnCl₂ and CdCl₂ solutions. In which mixtures will precipitation occur? Use Ksp: FeS 6.3×10⁻¹⁸, MnS 3×10⁻¹⁴, ZnS 2×10⁻²⁵, CdS 8×10⁻²⁷.
Steps only
After mixing, each metal-ion concentration is 0.040×(5/15)=0.0133 M. Sulfide is diluted to 1.0×10⁻¹⁹×(10/15)=6.67×10⁻²⁰ M. For each 1:1 sulfide, Qsp=[M²⁺][S²⁻]=(0.0133)(6.67×10⁻²⁰)=8.9×10⁻²². A precipitate forms when Qsp>Ksp. This exceeds Ksp for ZnS and CdS, but is below Ksp for FeS and MnS.
Answer
ZnS and CdS precipitate; FeS and MnS do not.