This page has all of the required homework for the material covered in the fourth exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
Use the equation ΔG° = -RTlnK to determine K for the following reaction at 25 °C. R is the gas constant: R = 8.31 J/mole-K.
N2(g) + 3H2(g) ⇔ 2NH3(g)
ΔG° = 2ΔG°NH3(g) - ΔG°N(g) - 3ΔG°H2(g)
ΔG° = 2(-16.66 kJ) - 0 - 3(0) = -33.32 kJ for two moles of NH3
ΔG° = -16.66 kJ/mole = -16,660 J/mole of NH3
lnK = -ΔG°/(RT) = (16,660 J/mole)/[(8.31 J/mole-K)(298 K)] = 6.73
⇒ K = e6.73 = 835
The gaseous reaction
2NO(g) + O2(g) ⇔ 2NO2(g)
has Kc = 1.20 at 1000 K. At equilibrium it was found that there were 7.50 g NO and 26.67 g O2 in a 250 mL container. What is the concentration of NO2 at this equilibrium?Find moles of each and then concentration of each:
(7.50 g NO)(1 mole/30 g) = 0.25 mole NO; (0.25 mole NO)/(0.25 L) = 1.00 M NO
(26.67 g O2)(1 mole/32 g) = 0.83 mole; (0.83 mole O2)/(0.25 L) = 3.33 M O2
For this reaction Kc = [NO2]²/[NO]²[O2]
⇒ 1.2 = [NO2]²/(1²)(3.33) ⇒ [NO2] = 2 M
PCl3(g) + Cl2(g) ⇔ PCl5(g)
At equilibrium there are 0.19 moles of PCl3 present in the container. (a) Calculate the equilibrium concentrations of PCl3, Cl2, and PCl5. (b) Calculate Kc under these conditions.PCl3(g) + Cl2(g) ⇔ PCl5(g) | |||
Init. Moles | 0.03 mole | 0.02 mole | 0 mole |
Init. Conc. | 0.3 M | 0.2 M | 0 M |
Equilib | 0.3 - x | 0.2 - x | x |
Equilib | 0.19 | ? | ? |
(a) Find x: 0.3 - x = 0.19; x = 0.3 - 0.19 = 0.11
⇒ [PCl5] = x = 0.11 M
⇒ [Cl2] = 0.2 - x = 0.2 - 0.11 = 0.09 M
[PCl3] = 0.19 M (given in the problem)
(b) Kc = [PCl5]/{[PCl3][Cl2]} = (0.11 M)/{(0.19 M)(0.09 M)} = 6.43
N2(g) + 2H2O(g) ⇔ 2NO(g) + 2H2(g)
When 0.2 moles of N2 are combined with 0.3 moles of H2O in a 1.0 L container and allowed to go to equilibrium, it is found that there are 0.169 moles of N2 in the equilibrium mixture. What are the equilibrium concentrations of each of the materials and what is the value of the equilibrium constant, Kc?N2(g) + H2O(g) ⇔ 2NO(g) + 2H2(g) | ||||
Init. Moles | 0.2 mole | 0.3 mole | 0 mole | 0 mole |
Init. Conc. | 0.2 M | 0.3 M | 0 M | 0 M |
Equilib | 0.2 - x | 0.3 - 2x | 2x | 2x |
Equilib | 0.169 | ? | ? | ? |
(a) Find x: 0.2 - x = 0.169; x = 0.2 - 0.169 = 0.031
⇒ [H2O] = 0.3 - 2x = 0.238 M
⇒ [H2] = [NO] = 2x = (2)(0.031) = 0.062 M
[N2] = 0.169 M (given in the problem)
(b) Kc = [NO]²[H2]²/{[N2][H2O]²}
Kc = (0.062 M)²(0.062 M)²/{(0.169 M)(0.238 M)²} = 1.54 x 10-3
2NO(g) + Cl2(g) ⇔ 2NOCl(g) Kc = 1.42 x 107
Calculate the concentration of NO at equilibrium when 3.0 moles of NO are reacted with 2.0 moles of Cl2 in a 500 mL container.A large Kc means that the reaction will go to right in the complete reaction line below.
2NO(g) + Cl2(g) ⇔ 2NOCl(g) | |||
Init. Moles | 3.0 mole | 2.0 mole | 0 moles |
Init. Conc. | 6 M | 4 M | 0 M |
Complete Rxn | 0 | 1 | 6 |
Equilib Conc | 0 + 2x | 1 + x | 6 - 2x |
Kc = 1.42 x 107 = [NOCl]²/{[NO]²[Cl2]} = (6-2x)²/{(2x)²(1+x)}
Assume x << 1
⇒ x = {36/[(4)(1.42 x 107)]}½ = 8.02 x 10−4
⇒ [NO] = 2x = 16 x 10−4 M
2NOCl(g) ⇔ 2NO(g) + Cl2(g) Kc = 7.04 x 10-8
Calculate the concentration of NO at equilibrium when 3.0 moles of NOCl and 2.0 moles of Cl2 are initially placed in a 500 mL container.A small Kc means that the reaction will go to left in the complete reaction line below.
2NOCl(g) ⇔ 2NO(g) + Cl2(g) | |||
Init. Moles | 3.0 mole | 0 mole | 2.0 moles |
Init. Conc. | 6 M | 0 M | 4 M |
Complete Rxn | 6 | 0 | 4 |
Equilib Conc | 6 - 2x | 0 + 2x | 4 + x |
Kc = 7.04 x 10−8 = {[NO]²[Cl2]}/[NOCl]² = {(2x)²(4+x)}/(6-2x)²
Assume x << 4
⇒ x = {(36)(7.04 x 10−8}/[(4)(4)]}½ = (1.584 x 10−7}½ = 3.98 x 10−4
⇒ [NO] = 2x = 7.96 x 10−4 M
2NOCl(g) ⇔ 2NO(g) + Cl2(g) Kc = 7.04 x 10-8
Calculate the concentration of NO at equilibrium when 3.0 moles of NOCl, 2.0 moles of Cl2, and 2.0 moles of NO are initially placed in a 500 mL container.A small Kc means that the reaction will go to left in the complete reaction line below.
2NOCl(g) ⇔ 2NO(g) + Cl2(g) | |||
Init. Moles | 3.0 mole | 2.0 mole | 2.0 moles |
Init. Conc. | 6 M | 4 M | 4 M |
Complete Rxn | 10 | 0 | 2 |
Equilib Conc | 10 - 2x | 0 + 2x | 2 + x |
Kc = 7.04 x 10−8 = {[NO]²[Cl2]}/[NOCl]² = {(2x)²(2+x)}/(10-2x)²
Assume x << 2 ⇒ 7.04 x 10−8 = (2x)²(2)/10 = (0.8)x²
⇒ x = {(7.04 x 10−8)/(0.8)}½ = 2.97 x 10−4
⇒ [NO] = 2x = 5.9 x 10−4 M
Ka = 1.5 x 10-5 means that the reaction will go toward the left in the complete reaction line below.
HC4H7O2 ⇔ H+ + C4H7O2- | |||
Init. Moles | 1.5 mole | 0 mole | 0 mole |
Init. Conc. | 5 M | 0 M | 0 M |
Complete Rxn | 5 | 0 | 0 |
Equilib Conc | 5 - x | x | x |
Ka = 1.5 x 10-5 = [H+][C4H7O2-]/[HC4H7O2] = x²/(5 - x) Assume x << 5
⇒ x = [(5)(1.5 x 10-5)]½ = 0.00866 = [H+]
⇒ pH = -log(0.00866) = 2.06
Ka = 1.8 x 10-4 means that the reaction will go toward the left in the complete reaction line below. Notice that the Na+ ion is a spectator ion and doesn't show up in the equilibrium equation.
HCHO2 ⇔ H+ + CHO2- | |||
Init. Moles | 1.2 mole | 0 mole | 2.4 mole |
Init. Conc. | 0.6 M | 0 M | 1.2 M |
Complete Rxn | 0.6 | 0 | 1.2 |
Equilib Conc | 0.6 - x | x | 1.2 + x |
Ka = 1.8 x 10-4 = [H+][CHO2-]/[HCHO2] = (x)(1.2 + x)/(0.6 - x)
Assume x << 0.6
⇒ x = [(0.6)(1.8 x 10-4)]/1.2 = 9 x 10-5 = [H+]
⇒ pH = -log(9 x 10-5) = 4.05
Ksp = 6.5 x 10−6 means that there will not be very many ions in the solution. Because the solid (concentration set to one) is in equilibrium with the ions, we only need to consider what the condition is at equilibrium as shown in the table.
Ca(OH)2(s) ⇔ Ca2+ + 2OH- | |||
Equilib Conc | - | x | 2x |
Ksp = 6.5 x 10-6 = [Ca2+][OH−]² = (x)(2x)² = 4x³
⇒ x = [(6.5 x 10-6)/4]⅓ = 0.0118
⇒ [Ca2+] = 0.0118 M and [OH-] = 0.0236 M
(0.0085 L NaOH)(1.0 mole/L NaOH) = 0.0085 moles of OH−
NaOH reacts with acetic acid in a one-to-one ratio. The NaOH reacts with both the dissociated and nondissociated hydrogen ions from the acetic acid solution, so the moles of NaOH used is the initial moles of HC2H3O2 that would have been in the solution before any dissociation of the acid.
Ka = 1.8 x 10-5 means that the reaction will go toward the left in the complete reaction line below.
HC2H3O2 ⇔ H+ + C2H3O2- | |||
Init. Moles | 0.0085 mole | 0 mole | 0 mole |
Init. Conc. | 1.417 M | 0 M | 0 M |
Complete Rxn | 1.417 | 0 | 0 |
Equilib Conc | 1.417 - x | x | x |
Ka = 1.8 x 10-5 = [H+][C2H3O2-]/[HC2H3O2] = (x)(x)/(1.417 - x)
Assume x << 1.417
⇒ x = [(1.417)(1.8 x 10-5)]½ = 0.005 = [H+]
⇒ pH = -log(0.005) = 2.3
H+ + OH− ⇔ H2O Kc = 1/Kw = 1 x 1014 = 1/([H+][OH−])
After 4 mL HCl:
(0.004 L HCl)(0.5 mole/L HCl) = 0.002 moles of H+
⇒ Initial [H+] = (0.002 moles H+)/(0.014 L soln) = 0.143 M HCl.
(0.01 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH−
⇒ Initial [OH−] = (0.0015 moles OH−)/(0.014 L soln) = 0.107 M OH−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
H+ + OH− ⇔ H2O | |||
Init. Moles | 0.002 mole | 0.0015 mole | - |
Init. Conc. | 0.143 M | 0.107 M | - |
Complete Rxn | 0.036 | 0 | - |
Equilib Conc | 0.036 + x | x | - |
Kc = 1 x 1014 = 1/([H+][OH−]) = 1/{(0.036 + x)(x)} Assume x << 0.036
⇒ [H+] = 0.036 M ⇒ pH = -log(0.036) = 1.44
H+ + OH− ⇔ H2O Kc = 1/Kw = 1 x 1014 = 1/([H+][OH−])
After 3 mL HCl:
(0.003 L HCl)(0.5 mole/L HCl) = 0.0015 moles of H+
⇒ Initial [H+] = (0.0015 moles H+)/(0.013 L soln) = 0.115 M HCl.
(0.01 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH−
⇒ Initial [OH−] = (0.0015 moles OH−)/(0.013 L soln) = .115 M OH−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
H+ + OH− ⇔ H2O | |||
Init. Moles | 0.0015 mole | 0.0015 mole | - |
Init. Conc. | 0.115 M | 0.115 M | - |
Complete Rxn | 0 | 0 | - |
Equilib Conc | x | x | - |
Kc = 1 x 1014 = [H+][OH−] = 1/x² ⇒ x = 1 x 10−7 = [H+]
⇒ pH = -log(10−7) = 7
H+ + OH− ⇔ H2O Kc = 1/Kw = 1 x 1014 = 1/([H+][OH−])
After 2 mL HCl:
(0.002 L HCl)(0.5 mole/L HCl) = 0.001 moles of H+
⇒ Initial [H+] = (0.001 moles H+)/(0.012 L soln) = 0.0833 M HCl.
(0.01 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH−
⇒ Initial [OH−] = (0.0015 moles OH−)/(0.012 L soln) = 0.125 M OH−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
H+ + OH− ⇔ H2O | |||
Init. Moles | 0.001 mole | 0.0015 mole | - |
Init. Conc. | 0.0833 M | 0.125 M | - |
Complete Rxn | 0 | 0.0417 | - |
Equilib Conc | x | 0.0417 + x | - |
Kc = 1 x 1014 = 1/([H+][OH−]) = 1/{(x)(0.0417 + x)} Assume x << 0.0417
⇒ [OH−] = 0.0417 M ⇒ pOH = -log(0.0417) = 1.38
⇒ pH = 14- 0.067 = 12.62
HC6H5O + OH− ⇔ H2O + C6H5O− Kc = Ka/Kw = 1.3 x 104 = [C6H5O−]/([HC6H5O][OH−])
After 16 mL NaOH:
(0.016 L NaOH)(0.6 mole/L NaOH) = 0.0096 moles of OH−
⇒ Initial [OH−] = (0.0096 moles OH−)/(0.037 L soln) = 0.259 M OH−.
(0.021 L HC6H5O)(0.4 mole/L NaOH) = 0.0084 moles of HC6H5O
⇒ Initial [HC6H5O] = (0.0084 moles HC6H5O)/(0.037 L soln) = 0.227 M OH−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HC6H5O + OH− ⇔ H2O + C6H5O− | ||||
Init. Moles | 0.0084 mole | .0096 mole | - | 0 moles |
Init. Conc. | 0.227 M | 0.259 M | - | 0 |
Complete Rxn | 0 | 0.03246 | - | 0.227 |
Equilib Conc | x | 0.03246 + x | - | 0.227 - x |
Kc = 1.3 x 104 = (0.227 - x)/{(x)(0.03246 + x)} Assume x << 0.03246
⇒ [OH−] = 0.03246 M ⇒ pOH = -log(0.03246) = 1.49
⇒ pH = 14 - 1.49 = 12.5
HC6H5O + OH− ⇔ H2O + C6H5O− Kc = Ka/Kw = 1.3 x 104 = [C6H5O−]/([HC6H5O][OH−])
After 14 mL NaOH:
(0.014 L NaOH)(0.6 mole/L NaOH) = 0.0084 moles of OH−
⇒ Initial [OH−] = (0.0084 moles OH−)/(0.035 L soln) = 0.24 M OH−.
(0.021 L HC6H5O)(0.4 mole/L NaOH) = 0.0084 moles of HC6H5O
⇒ Initial [HC6H5O] = (0.0084 moles HC6H5O)/(0.035 L soln) = 0.24 M OH−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HC6H5O + OH− ⇔ H2O + C6H5O− | ||||
Init. Moles | 0.0084 mole | .0084 mole | - | 0 moles |
Init. Conc. | 0.24 M | 0.24 M | - | 0 |
Complete Rxn | 0 | 0 | - | 0.24 |
Equilib Conc | x | x | - | 0.24 - x |
Kc = 1.3 x 104 = (0.24 - x)/{(x)(x)} Assume x << 0.24
⇒ x = [(0.24)/(1.3 x 104)]½ = 0.0043 M = [OH−] ⇒ pOH = -log(0.0043) = 2.37
⇒ pH = 14 - 2.37 = 11.63
HC6H5O + OH− ⇔ H2O + C6H5O− Kc = Ka/Kw = 1.3 x 104 = [C6H5O−]/([HC6H5O][OH−])
After 12 mL NaOH:
(0.012 L NaOH)(0.6 mole/L NaOH) = 0.0072 moles of OH−
⇒ Initial [OH−] = (0.0072 moles OH−)/(0.033 L soln) = 0.218 M OH−.
(0.021 L HC6H5O)(0.4 mole/L NaOH) = 0.0084 moles of HC6H5O
⇒ Initial [HC6H5O] = (0.0084 moles HC6H5O)/(0.033 L soln) = 0.255 M OH−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HC6H5O + OH− ⇔ H2O + C6H5O− | ||||
Init. Moles | 0.0084 mole | .0072 mole | - | 0 moles |
Init. Conc. | 0.255 M | 0.218 M | - | 0 |
Complete Rxn | 0.037 | 0 | - | 0.255 |
Equilib Conc | 0.037 + x | x | - | 0.218 - x |
Kc = 1.3 x 104 = (0.218 - x)/{(x)(0.037 + x)} Assume x << 0.03637
x = (0.218)/{(0.037)(1.3 x 104)} = 0.000453 = [OH−]
⇒ pOH = -log(0.000453) = 3.34 ⇒ pH = 14 - 3.34 = 10.66
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 20 mL NaOH:
(0.02 L NaOH)(0.15 mole/L NaOH) = 0.003 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.045 L soln) = 0.0333 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.045 L soln) = 0.0444 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.003 mole | - | 0.002 mole |
Init. Conc. | 0.0333 M | 0.0667 M | - | 0.0444 M |
Complete Rxn | 0 | 0.0334 | - | 0.0777 |
Equilib Conc | x | 0.0334 + x | - | 0.0333 - x |
Kc = 4.9 x 104 = (0.0777 - x)/{(x)(0.0334 + x)} Assume x << 0.0333
⇒ [OH−] = 0.00334 M ⇒ pOH = -log(0.0334) = 1.47
⇒ pH = 14 - 1.47 = 12.5
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 15 mL NaOH:
(0.015 L NaOH)(0.15 mole/L NaOH) = 0.00225 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.04 L soln) = 0.0375 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.04 L soln) = 0.05 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.00225 mole | - | 0.002 mole |
Init. Conc. | 0.0375 M | 0.0563 M | - | 0.05 M |
Complete Rxn | 0 | 0.0188 | - | 0.0875 |
Equilib Conc | x | 0.0188 + x | - | 0.0875 - x |
Kc = 4.9 x 104 = (0.0875 - x)/{(x)(0.0188 + x)} Assume x << 0.0188
⇒ [OH−] = 0.0188 M ⇒ pOH = -log(0.0188) = 1.73
⇒ pH = 14 - 1.73 = 12.3
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 12 mL NaOH:
(0.012 L NaOH)(0.15 mole/L NaOH) = 0.0018 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.037 L soln) = 0.0405 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.037 L soln) = 0.0541 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.0018 mole | - | 0.002 mole |
Init. Conc. | 0.0405 M | 0.0563 M | - | 0.0541 M |
Complete Rxn | 0 | 0.0158 | - | 0.0946 |
Equilib Conc | x | 0.0158 + x | - | 0.0946 - x |
Kc = 4.9 x 104 = (0.0946 - x)/{(x)(0.0158 + x)} Assume x << 0.0158
⇒ [OH−] = 0.0158 M ⇒ pOH = -log(0.0158) = 1.8
⇒ pH = 14 - 1.73 = 12.2
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 10 mL NaOH:
(0.010 L NaOH)(0.15 mole/L NaOH) = 0.0015 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.035 L soln) = 0.0429 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.035 L soln) = 0.0571 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.0015 mole | - | 0.002 mole |
Init. Conc. | 0.0429 M | 0.0429 M | - | 0.0571 M |
Complete Rxn | 0 | 0 | - | 0.1 |
Equilib Conc | x | x | - | 0.1 - x |
Kc = 4.9 x 104 = (0.1 - x)/{(x)(x)} Assume x << 0.1
⇒ [OH−] = x = (0.1/49000)½ = 0.00143 M ⇒ pOH = -log(0.00143) = 2.85
⇒ pH = 14 - 2.85 = 11.2
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 8 mL NaOH:
(0.008 L NaOH)(0.15 mole/L NaOH) = 0.00012 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.033 L soln) = 0.0455 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.033 L soln) = 0.0606 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.0012 mole | - | 0.002 mole |
Init. Conc. | 0.0455 M | 0.0364 M | - | 0.0606 M |
Complete Rxn | 0.0091 | 0 | - | 0.097 |
Equilib Conc | 0.0091 + x | x | - | 0.097 - x |
Kc = 4.9 x 104 = (0.097 - x)/{(0.0091 + x)(x)} Assume x << 0.0091
⇒ [OH−] = x = 0.097/{(0.0091)(49000)} = .00022 M; pOH = 3.66
⇒ pH = 14 - 3.66 = 10.34
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 5 mL NaOH:
(0.005 L NaOH)(0.15 mole/L NaOH) = 0.00075 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.030 L soln) = 0.05 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.030 L soln) = 0.0667 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.00075 mole | - | 0.002 mole |
Init. Conc. | 0.05 M | 0.025 M | - | 0.0667 M |
Complete Rxn | 0.025 | 0 | - | 0.0917 |
Equilib Conc | 0.025 + x | x | - | 0.0917 - x |
Kc = 4.9 x 104 = (0.0917 - x)/{(0.025 + x)(x)} Assume x << 0.025
⇒ [OH−] = x = 0.0917/{(0.025)(49000)} = 0.00007 M; pOH = -log[OH−] = 4.13
⇒ pH = 14 - 4.13 = 9.87
HCN + OH− ⇔ H2O + CN− Kc = Ka/Kw = 4.9 x 104 = [CN−]/([HCN][OH−])
After 0 mL NaOH:
(0 L NaOH)(0.15 mole/L NaOH) = 0 moles of OH−
(0.015 L HCN)(0.1 mole/L HCN) = 0.0015 moles of HCN
⇒ Initial [HCN] = (0.0015 moles HCN)/(0.025 L soln) = 0.06 M HCN.
(0.010 L KCN)(0.2 mole/L NaOH) = 0.002 moles of KCN = 0.002 moles of CN−
⇒ Initial [CN−] = (0.002 moles CN−)/(0.025 L soln) = 0.08 M CN−.
A large Kc means that the reaction will go toward the right in the complete reaction line in the following table. The concentration of water is set to one and dashes are placed in the table.
HCN + OH− ⇔ H2O + CN− | ||||
Init. Moles | 0.0015 mole | 0.000 mole | - | 0.002 mole |
Init. Conc. | 0.06 M | 0 M | - | 0.08 M |
Complete Rxn | 0.06 | 0 | - | 0.08 |
Equilib Conc | 0.06 + x | x | - | 0.08 - x |
Kc = 4.9 x 104 = (0.08 - x)/{(0.06 + x)(x)} Assume x << 0.06
⇒ [OH−] = x = 0.08/{(0.06)(49000)} = 0.00003 M; pOH = -log[OH−] = 4.57
⇒ pH = 14 - 4.57 = 9.43