Important topics for Chapter 14 & 15  Batamo
Chapter 14
1. Dynamic equilibrium:
For a hypothetical reaction: aA + bB <==> cC + dD (kf, kr)
When it’s at equilibrium, kf[A]a[B]b = kr[C] c[D]d
and thus, the equilibrium constant K = kf /kr = [C] c[D]d/[A]a[B]b
Important information:
(1)* Equilibrium can be approached either from the forward or the reverse direction.
(2)* K (can be Kc or Kp or Ka or Kb) is always unitless and is affected by temperature only. While k
possessing unit varies with the reaction order and is affected by both temperature and catalyst.
(3)** The concentration or pressure used to calculate Kc or Kp must be at equilibrium.
If the concentration or pressure used in calculation is not at equilibrium, the value is called the reaction
quotient, Q.
(4)** Liquids (l) and solids (s) can not appear in the equilibrium constant.
(5) The rate constant, k, can not be obtained from the balanced chemical equation; however, the
equilibrium constant, K, is always obtained from the balanced chemical equation.
(6)** When K > 1, reaction lies to the right, which leads to a product-rich mixture at equilibrium.
(When K >> 1, reaction lies far to the right, which leads to nearly completion.)
When K = 1, reaction is at equilibrium.
When K < 1, reaction lies to the left, which leads to a reactant-rich mixture at equilibrium. (When K
<< 1, reaction lies far to the left, indicating nearly no reaction.)
2. ***Characteristics of K:
For a hypothetical reaction: aA + bB <==> cC + dD
(a) cC + dD <==> aA + bB
K = [C] c[D]d/[A]a[B]b
K(a) = [A]a[B]b/[C] c[D]d = 1/K
(b) xaA + xbB <==> xcC + xdD
K(b) = [C] xc[D]xd/[A]xa[B]xb
= {[C] c[D]d/[A]a[B]b}x = Kx
(c) Given
(1)
(2)
aA <==> cC
bB <==> dD
K(1) = [C] c /[A]a
K(2) = [D] d /[B]b
When add equation (1) and equation (2) together, one obtains the resultant equation as
aA + bB <==> cC + dD (3)
the equilibrium constant, K(3), for the resultant
equation, (3), K(3) = [C] c[D]d/[A]a [B]b = K(1) x K(2) = K
3.
Relationship between Kp and Kc:
For a hypothetical reaction: aA(g) + bB(g) <==> cC(g) + dD(g)
Kc = [C] c[D]d/[A]a[B]b since the reactants and products are all gases, one can write the
equilibrium constant as Kp = (PC) c(PD)d/(PA)a(PB)b and
from PV = nRT, one can write P = (n/V)RT = MRT where M is molarity.
Because all gases are in the same container and thus they have same temperature and volume.
Therefore
PA = (nA/V)RT = [A]RT, PB = (nB/V)RT = [B]RT
PD = (nD/V)RT = [D]RT
PC = (nC/V)RT = [C]RT, and
By substituting all into Kp = ([C]RT) c([D]RT)d/([A]RT)a([B]RT)b
= ([C]c[D]d/[A]a[B]b)(RT)c+d-a-b = Kc(RT)∆n(g)
where ∆n(g) = (c+d) – (a+b)
Note: When ∆n(g) > 0, then Kp > Kc. When ∆n(g) = 0, then Kp = Kc.. When ∆n(g) < 0, then Kp < Kc.
4. ***Le Chấtelier’s principle:
(a) Adding a reactant or removing a product: reaction runs to the right (i.e. the product side) to restore
the equilibrium.
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Removing a reactant or adding a product: reaction runs to the left.
Note: Adding a gas which does not react with reactant(s) or product(s) and is neither a reactant nor
a product will not disturb the equilibrium.
Note: Adding a solid or a liquid does not disturb the equilibrium, as its concentration is always a
constant value in equilibrium. Thus, adding a catalyst does not disturb equilibrium.
In summary, adding a substance that is not shown in an equilibrium equation will not disturb
equilibrium presumably this substance is non-reactive to any substance shown in the equilibrium
equation
(b) For gaseous reactions only:
Increasing the pressure (i.e. decreasing the volume) causes reaction run to the side of small moles.
(i.e. compare the sum of coefficients between the reactant side and the product side).
Decreasing the pressure or increasing the volume causes reaction run to the side of more moles.
(c) For an exothermic reaction, increasing temperature will shift the reaction to the left (i.e. reactant)
side. For an endothermic reaction, increasing temperature will shift the reaction to the right.
5. Calculations: Given equilibrium concentration (or pressure), calculate Kc (or Kp).
Practice Questions for Chapter 14
1. Given the equilibrium below: PCl3(g) + Cl2(g)  PCl5(g) Kc = 0.18
(a) What is the equilibrium constant for PCl5(g)  PCl3(g) + Cl2(g)? 5.56
(b) What is the equilibrium constant for ⅓ PCl5(g)  ⅓ PCl3(g) + ⅓ Cl2(g)? 1.77
2. Given the equilibria:
Kc = 2.40 x 10-18
N2(g) + ½ O2(g)  N2O(g)
Kc = 4.1 x 10-31
N2(g) + O2(g)  2NO(g)
What is the equilibrium constant for N2O(g) + ½ O2(g)  2NO(g)? 1.708x10-13
3. Write the equilibrium law according to Kc for the following reaction:
CaCO3(s) + 2HCl(aq)  Ca2+(aq) + 2Cl-(aq) + H2O(l) + CO2(g)
Kc = [Ca2+(aq)][Cl-(aq)]2[CO2(g)]/[HCl(aq)]2
4. A mixture of 5.00 x 10-3 mole of H2 and 1.00 x 10-2 mole of I2 is placed in a 5.00 L container at 448 oC
and allowed to come to equilibrium. At equilibrium, the concentration of HI is 1.87 x 10-3 M. What is the
Kc at 448 oC for the reaction H2(g) + I2(g)  2HI(g)? Kc= (1.87x10-3)2/(6.5x10-5)(1.065x10-3) = 50.5
5. At 25oC, Kc = 4.500 for the reaction SO3(g) + NO(g)  NO2(g) + SO2(g).
(a) If 0.50 mole of SO3 and 0.60 mole of NO are placed in a 2.00 L container and allowed to react,
what will be the equilibrium concentration of NO(g) and SO2(g), respectively?
Need to solve quadric equation. 0 < x < 0.25 M and thus x = 0.1845 M, NOT 0.523 M. At
equilibrium, [NO2]=[SO2] = 0.1845 M, [SO3] = 0.0655 M, [NO] = 0.1155 M.
(b) What is the Kp at the same temperature? Given Kp = Kc(RT)n(g) 4.500
Ho < 0
6. Consider the equilibrium: 2SO2(g) + O2(g)  2SO3(g)
How will each of the following affect an equilibrium mixture of the three gases? Shift to left, right or no
change. (See Examples 14.12, 14.13, p. 636: 14.58, 14.62)
(a) O2(g) is added to the system R (b) Remove O2(g) from the system L (c) The reaction mixture is heated
L (d) The volume of the reaction vessel is doubled L (e) A catalyst is added to the mixture N (f) The total
pressure of the system is increased by adding a noble gas N (g) If SO3(g) is removed from the system R (h)
A Helium gas is added to the mixture. N (i) H2O(g) is added to the mixture. R. Note that SO3(g) does react
with H2O(g) to produce H2SO4(g)
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Chapter 15
1. Definitions:
(a) Arrhenius acids are the ones give H+ (or H3O+) in water while Arrhenius bases are the ones that
give OH- in water. Without water as solvent, there is no Arrhenius acids and bases.
(a) Brønsted-Lowry acids donate H+ while Brønsted-Lowry bases accept H+. Brønsted-Lowry acids
and bases apply to any physical phases (s, l, g, or aq).
(b) Lewis acids accept an electron pair (i.e. 2 electrons exact) while Lewis bases donate an electron
pair (i.e. 2 electrons exact). Note: Lewis acids and bases apply to any physical phases (s, l, g, or
aq).
***Note: Lewis acids can be any of the following: cations, central atom with empty valence orbitals,
central atom is less than an octet. While Lewis bases can be any of the following: anions, central atom
with nonbonding electrons (i.e. lone pairs), central atom is more than an octet. All reactions in organic
chemistry can be considered as Lewis acid-base reactions.
However, Lewis acid-base reactions in inorganic chemistry, such CHEM 1412, will produce a
coordinate covalent bond. Why it’s called coordinate? Because the two electrons used in bond
generation are from same central atom. The two electrons used in bond generation in CHEM 1411 are
from two central atoms belonging to two different reactants: each central atom contributes one electron
only.
Note: **Metal oxides and hydroxides are basic. Nonmetal oxides are acidic. All metal ions are acidic
except IA metal ions and heavy IIA metal ions (Ca2+, Sr2+ and Ba2+) as these metal ions are neutral. Be
aware of the controversial one, Mg2+: some argue it’s acidic while others argue it’s neutral.
2. **Conjugate acids and bases: These are derived from Brønsted-Lowry acids and bases.
Adding an H+ to obtain a conjugate acid; subtracting and H+ to obtain a conjugate base. What is the
conjugate acid and conjugate base for OH-?
Note: A substance that has both conjugate acid and conjugate base concurrently is called amphoteric
substance.This is because this substance can functions as either an acid or a base.
Examples: H2O, H2PO4-, HPO42-, HCO3-, HS-, HSO4-, etc.
Reaction direction for acid-base: Always proceeds from stronger acid/base to weaker acid/base.
3. Calculations:
Definition of Kw, pH, pOH, pKw and their relations. Note: Only temperature can change the value of
Kw..
4. More important calculations:
(a) pH of diluted monoprotic strong acid:
(b) pH of diluted di- strong acid: considering the ionization of the first proton only.
(c) pH of diluted monobasic strong base:
(d) pH of diluted dibasic strong base: considering the ionization of both hydroxide ions.
****Note: In any aqueous acid solution, there are two sources of H+: one is from acid and the other is
from water. By the same token, in any basic solution, there are two sources of OH-: one is from base and
the other is from water. Therefore, you cannot dilute an acid into a base; you cannot dilute a base into
an acid.
5. Weak Acids and Bases (pp. 656-670)
1. Ka and Kb definitions:
*Only weak acids have Ka values and weak bases have Kb values.
*Both are affected by temperature only.
Weak Acids (sect. 15.5): (Also referring to the bottom section of this page)
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(a) For mono-protic acid (HA): HA(aq)  H+(aq)+ A-(aq); Ka = [H+][A-]/[HA]
Sect 15.8, p. 666: Polyprotic acids:
(b) For di-protic acid (H2A): H2A(aq)  H+(aq)+ HA-(aq); Ka1 = [H+][HA-]/[H2A]
HA-(aq)  H+(aq)+ A2-(aq); Ka2 = [H+][A2-]/[HA-]
See: Example 15.11 and p. 689: 15.64
(c) For tri-protic acid (H3A): H3A(aq)  H+(aq)+ H2A-(aq); Ka1 = [H+][H2A-]/[H3A]
H2A-(aq)  H+(aq)+ HA2-(aq); Ka2 = [H+][HA2-]/[H2A-];
HA2-(aq)  H+(aq)+ A3-(aq); Ka3 = [H+][A3-]/[HA2-]
See: p. 670 phosphoric acid
**Note: (1) pKa = -logKa and pKb = -logKb
(2) The stronger the weak acid, smaller the pKa value; while the stronger the weak base, smaller
the pKb value.
Weak Bases: Limits to mono-basic base (B): p. 663: sect. 15.6.
B(aq) + H2O(l)  BH+(aq) + OH-(aq); Kb = [BH+][OH-]/[B]
**Note: Corrections of the middle section of p. 683: arrow signs should be drawn the opposite ways: tail
from electrons on O and arrowheads toward H and add one arrow with tail from H-O bond to O:+ atom in
carbonic acid.
6. Relationship between conjugate acid-base pairs (Sect. 15.7): Ka  Kb = Kw or pKa + pKb = pKw = 14
***Note: The stronger the (conjugate) acid, the weaker its conjugate base; while the stronger the
(conjugate) base, the weaker its conjugate acid. How weak is weak? How strong is strong? See text (or p.
674: 15.10 and Example 15.14; p. 689: 15.74, 15.76) and Lab: Hydrolysis of Salts, and its pre-lab and
post-lab questions.
7. Calculations involving Ka and Kb utilizing equilibrium table: Be aware of the differences between
examples 15., 15.9 and 15.10.
For mono-protic strong acid, HX, with concentration at least 100 times more than 1x10-7 M, the [H+]
= [HX]; however, for mono-protic weak acid, HA, the [H+] ≠ [HX]
Weak Acids****: More challenging topics (polyprotic acids: p.p. 666-670)
(a) For mono-protic acid: Use equilibrium table once to calculate [H+].
Short-cut: [H+] = {Ka  [HA]}1/2 only valid if [HA]initial ≥ Ka  400 (see p.p. 658-9; example 15.2)
(b) For di-protic acid: Use equilibrium table twice to calculate [H+]total = [H+]first step + [H+]second step ; [HA-] =
Ka1; [A2-] = Ka2. See important Example 15.11 in p. 667 and and p. 689: 15.64
Note: the assumption given in p. 809 text must be double-checked, i.e. [H+]total ≈ [H+]first step is valid
only when Ka1 >> Ka2 (i.e. must be at least 1000 times greater).
Short-cut for a diprotic acid, H2A:
a. [H+] = {Ka1C}1/2 for Ka1/Ka2 > 1000 only; if the ratio is smaller than 1000, [H+] is
obtained by solving quadratic equation.
b. [A2-] = Ka2
(c) For tri-protic acid: Use equilibrium table three times to calculate [H+]total = [H+]first step + [H+]second step +
[H+]third step. See phosphoric acid example.
Short-cut for a triprotic acid, H3A:
a. [H+] = { Ka1C}1/2 for Ka1/Ka2 > 100000 and Ka2/Ka3 > 100000 only; if the ratio is smaller
than 100000, [H+] is obtained by solving quadratic equations twice. Maybe able to simplify
depending on Ka1 and Ka2 values.
Roughly but not absolutely [H+]total = [Ka1C]1/2 + Ka2 + {Ka2 Ka3/[Ka1C]1/2}
b. [H2A-] and [HA2-] are calculated from equilibrium tables. No proper formulas.
c. [A3-] = Ka2 Ka3/{Ka1C}1/2
Weak Bases****: More challenging topics
(a) For mono-basic base: Use equilibrium table once to calculate [OH-]. Or [OH-] = {Kb x [B]}1/2
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(b) For di-basic base as the salt of di-protic acid: Use equilibrium table twice to calculate [OH-]total =
[OH-]first step + [OH-]second step; Kb1 = Kw/Ka2 and Kb2 = Kw/Ka1.
Note: [OH-]total ≈ [OH-]first step is valid only when Kb1 >> Kb2 (i.e. must be at least 1000 times
greater).
4.**** The anion from a strong acid or the cation from a strong base is neutral. The anion from a
weak acid is a weak base, while the cation from a weak base is a weak acid. (See sect 15.10 especially
example 15.14 and p. 689: 15.75, 15.77; preferably lab manual: more clear )
8.**** (a) To calculate the conjugate acid’s [H+] from a known weak base: (sect. 15.10)
Solve it from equilibrium table (See p. 689: 15.80) or by short-cut: [H+] = {(Kw  Kb )C}1/2 as the Ka of
the conjugate acid is unknown while the Kb of the weak base is known, i.e. available from Tables 15.4 and
15.5.
(b) To calculate the conjugate base’s [OH-] from a known weak acid: Solve it from equilibrium table (See
Example 15.13 or p. 689: 15.79) or by short-cut: [OH-] = {(Kw  Ka )C}1/2 as the Kb of the conjugate base
is unknown while the Ka of the weak acid is known, i.e. available from Tables 15.4 and 15.5.
9. Binary acids: (p. 670)
(a) Periodic trend: Acidity increases when going to the right and going down the periodic table.
(b) ***Factors that affect the acidity of binary acids (H-X): When determine the acidity, one must
consider ALL these factors concurrently. Hint: To be an acid, the H+ must be released. The easier
for a substance to release its H+, the stronger the acid.
(i) Electronegativity: The greater the electronegativity of X, the stronger the acid.
(ii) Bond strength: The weaker the H-X bond, the stronger the acid.
(iii) Charge density of anion (or say stability of anion, X-): The smaller the charge density of the
anion (i.e. more stable the anion), the stronger the acid. Note: If the anion possesses resonance
structures, the more resonance structures the anion has, the stronger the acid.
Example: Arrange the acidity strength for HF, HCl, HBr, and HI.
<sol> According to electronegativity of X, the acid strength is HF > HCl > HBr > HI
According to bond strength of H-X, the acid strength is HI > HBr > HCl > HF
According to charge density, the acid strength is HI > HBr > HCl > HF
Thus, by considering all factors concurrently, the acid strength is HI > HBr > HCl > HF, which
obeys the experimental data.
10. Oxoacids (i.e. oxyacids): These acids must contain oxygen atoms. (p. 671)
(a) Same central atom, same number of hydrogen atoms but different number of oxygen atoms: More
oxygen, stronger the acid.
Examples:(i)H3PO3<H3PO4; (ii)HClO<HClO2<HClO3<HClO4
(b) Different central atom but from same group, same numbers of hydrogen and oxygen atoms:
Greater the electronegativity of the central atom, stronger the acid.
Example: HIO4<HBrO4<HClO4
(c) Periodic trend: H3PO4 < H2SO4 < HClO4
(d) Inductive effect for organic acids derived from acetic acid, CH3COOH: (p. 673)
(i) More halogens, same kind, attached to C of CH3, the stronger the acid.
(ii) More electronegative halogens attached to C of CH3, the stronger the acid.
11. Acid-base properties of salts (Also see lab: hydrolysis of salts)****
Rules: When consider the acidity or basicity of a salt, one must consider the acidity and basicity of its
component cation and anion, or say, hydrolysis of both cation and anion:
(a) If one is neutral (i.e. one does not hydrolyze), then it’s easy to determine directly without cation’s
Ka and anion’s Kb. See p. 678.
(b) If cation is acidic and anion is basic (i.e. both hydrolyze), then one has to compare the value of Ka
and Kb. See p. 678, Table 15.7. If Ka > Kb, it’s acidic. If Ka < Kb, it’s basic; if Ka = Kb, it’s neutral.
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Baisc Concepts: (a) The ions that do not react with water do not hydrolyze and thus they are neutral: IA
metal ions, heavy IIA metal ions, nitrate, sulfate, perchlorate, chlorate, chloride, bromide, iodide, etc.
(b) The ions that do react with water do hydrolyze: acidic ion: ammonium.
Basic ions: sulfide, floride, phosphate, acetate, carbonate, etc.
(c) Cations from strong bases: neutral; anions from strong acids: neutral. (See Lab)
(d) Cations from weak bases are weak acids; anions from weal acids are weak bases. (See Lab)
Practice Questions for Chapter 15 (Continued): Keys are highlighted.
1. Give the equilibrium as PO43- (aq) + H2O(l)  HPO42- (aq) + OH- (aq)
(a) Identify the conjugate acid-base pairs. PO43- and HPO42- ; H2O and OH- (Example 15.1)
(b) Specify the position of the given equilibrium lie to the left or to the right. Left (Example 15.7)
(c) What are the conjugate acid and conjugate base respectively for OH-? Conjugate acid: H2O; conjugate
base: O2-. (p. 646)
2. Calculate the pH for the following solutions: (a) 2.5 x 10–2 M HNO3(aq) 1.60
(b) 2.5 x 10–2 M NaOH (aq) 12.40 (c) 4 x 10–10 M HNO3(aq) 7.00 (d) 4 x 10–10 M NaOH (aq) 7.00
Hint: There are two sources of H+ and OH-. When concentration is very small, the contribution from
water can not be ignored. See p. 3 of this handout.
3. A solution of HNO3 has a pH of 3.5. (a) What is its OH- concentration in molarity? 3.16x10-11 M
(b) How many grams of HNO3 are there in 300 mL of this solution? 0.006 grams
4. A solution was made by dissolving 2.5 grams Ca(OH)2 in 200 mL final volume. (Example 15.6)
(a) What is the molar concentration of OH- in the solution? 0.33 M (b) What is its pH? 13.52
5. The Kb for dimethylamine, (CH3)2NH, at 25oC is 9.6×10-4. What’s the Ka of its conjugate acid?
(B) 9.6×10-4
(C) 1.04×10-10 (D) 1.04×10-11 (E) 1.0×10-14
(A) 9.6×10-3
6. The pKb for dimethylamine, (CH3)2NH, at 25oC is 3.018. What’s the pH of a 0.1 M aqueous solution of
this compound?
(A) 2.01
(B) 3.46
(C) 6.72
(D) 7.00
(E) 11.99
7. A solution of acetic acid has a pH of 3.45. What is the concentration of acetic acid in this solution? Ka
for CH3COOH is 1.8×10-5.
(B) 1.80×10-5
(C) 6.99×10-3
(D) 3.45
(E) 7.00
(A) 3.55×10-8
8. What is the pH for 0.5 M NaCN? The Ka for HCN is 6.2×10-10.
(A) 0.30
(B) 0.50
(C) 2.55
(D) 11.45
(E) 2.84×10-3
9. Arrange the acid strength for the followings according to an increasing order. (p.p. 670-674)
(a) CI3COOH, CBr3COOH, CCl3COOH, CF3COOH, CH3COOH (b) CI3COOH, CI2HCOOH, CIH2COOH,
CH3COOH, CICl2COOH (c) H2SeO3, H2SeO4, H2SO4, HClO4, H2O
(a) CH3COOH < CI3COOH < CBr3COOH < CCl3COOH < CF3COOH
(b) CH3COOH < CIH2COOH < CI2HCOOH < CI3COOH < CICl2COOH
(c) H2O < H2SeO3 < H2SeO4 < H2SO4 < HClO4
10. Which of the following 0.1M aqueous solutions is neutral? (See Example 15.4, p. 679)
(C) NH4Cl
(D) KI
(E) KCN
(A) NaF
(B) NaC2H3O
11. Which of the following 0.1 M aqueous solutions is more soluble in a base? (Application of Example
15.4, p. 679)
(C) BaSO4
(D) Na2S
(E) NH4Cl
(A) NaCl
(B) CaCl2
12. Which of the following is a Lewis acid? (See Example 15.5, p. 683)
(B) CO32(C) Al3+
(D) BF4(E) CN(A) NH3
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