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Chapter 6  Acids and Bases

6.1  Arrhenius Definition

Arrhenius acid = substance that produces H+ ions in aqueous solution

  1. HA (aq) ⇋ H+ (aq) + A (aq) e.g. HCl (aq) ⇋ H+ (aq) + Cl (aq)
  2. HA is a polar covalent compound behaving like an ionic compound in that, upon dissolution, it dissociates into ions

Arrhenius base = substance that produces OH ions in aqueous solution

  1. MOH (aq) ⇋ M+ (aq) + OH (aq) e.g. NaOH (s) ⇋ Na+ (aq) + OH (aq)
  2. MOH is an ionic compound behaving like an ionic compound where M = metal

6.2  Brønsted-Lowry Definition

Brønsted-Lowry acid = substance that donates protons (H+)

Brønsted-Lowry base = substance that accepts protons (What allows it to accept protons? The presence of lone pairs which are repurposed to form a bond with the H+.)

HA (aq)+H2O (l)A (aq)+H3O+ (aq)
B-L acid B-L base B-L base B-L acid


Conjugate acid-base pair = reactant + conjugate product that differ by only 1 H+

  1. HA (acid) and A (conjugate base) are a conjugate acid-base pair
  2. H2O (base) and H3O+ (conjugate acid) are a conjugate acid-base pair

The production of the proton, H+, in the Arrhenius acid definition is seen as equivalent to the production of the hydronium ion, H3O+, in the Brønsted-Lowry acid definition. H+ and H3O+ are often used interchangeably as Arrhenius and Brønsted-Lowry definitions of acids are used interchangeably. H3O+ is the standard Brønsted-Lowry acid by which we will judge all other acids.

The largest difference between the two definitions is the Brønsted-Lowry base focuses on accepting the H+ donated by the acid whereas the Arrhenius base produces OH (which is a really good H+ acceptor). OH is the standard Brønsted-Lowry base by which we will judge all other bases.

6.3  Equilibrium

Acid ionization = reaction of an acid, HA, with water to produce ions, A and H3O+ (hydronium ion)

HA (aq)+H2O (l)A (aq)+H3O+ (aq)
B-L acid B-L base conjugate base conjugate acid


Acid ionization constant, Ka

Ka = 
[A][H3O+]
[HA]
    (1)

Percent ionization is a measure of how much H3O+ is actually produced relative to the theoretical amount that could be produced from the initial concentration of the acid:

 ionization = 
[H3O+]eq
[H3O+]theoretical
 × 100 = 
[H3O+]eq
n[HA]0
 × 100
    (2)

where n = number of H+ that can be ionized.

Base ionization = reaction of a base, β, with water to produce ions, βH+ and OH (hydroxide ion)

β (aq)+H2O (l)βH+ (aq)+OH (aq)
B-L base B-L acid conjugate acid conjugate base


Base ionization constant, Kb

Kb = 
 
[β H+][OH]
[β]
    (3)

Percent ionization is a measure of how much OH is actually produced relative to the theoretical amount that could be produced from the initial concentration of the base:

 ionization = 
[OH]eq
[OH]theoretical
 × 100 = 
[OH]eq
n[β]0
 × 100
    (4)

where n = number of OH that can be ionized/produced.
Water’s Autoionization = H2O is seen as a B-L acid and base

H2O (l)+H2O (l)H3O+ (aq)+OH (aq)
B-L base B-L acid conjugate acid conjugate base


Ion product constant of water, Kw

Kw =  [H3O+][OH]      (5)

Kw = 1.00 × 10−14 at 25C

RH2O (l)+H2O (l)H3O+ (aq)+OH (aq)
I    0.0 M 0.0 M
C    +x +x
E    x x
1.00 × 10−14 = (x)(x) = x2 
x = 1.00 × 10−7 = [H3O+] = [OH

The concentration of H3O+ and OH in pure water are equal and at 25C are both 1.00 × 10−7 M. Anytime [H3O+] = [OH], the solution is said to be neutral.

6.4  p-Scales

A p-scale conveys the order of magnitude (power of ten) of a quantity, X, by a reciprocal relationship

X = 10pX = 
1
10pX
    (6)
pX = −log(X)     (7)

Significant figures in X are decimal places in pX.

−log
5.47 × 10−12 
= −log(5.47) + −log(10−12) = −0.738 + 12.000 = 11.262 
−log
5.47 × 10−11 
= −log(5.47) + −log(10−11) = −0.738 + 11.000 =  10.262 
pX10pXX
7.510−7.53 × 10−8
7.5010−7.503.2 × 10−8
7.50010−7.5003.16 × 10−8

The pH and pOH of a neutral solution at 25C with [H+] ≡ [H3O+] = [OH] = 1.00 × 10−7 M

pH = −log([H3O+]) = −log(1.00 × 10−7) =  7.000 
pOH = −log([OH]) = −log(1.00 × 10−7) = 7.000 

Acidic conditions have [H3O+] > [OH].

For example, [H3O+] = 1.23 × 10−4 M with pH = −log(1.23 × 10−4) = 3.910 has

[OH] = 
Kw
[H3O+]
 = 
1 × 10−14
1.23 × 10−4
 = 8.13 × 10−11 
pOH = −log(8.13 × 10−11) = 10.090 
pH + pOH = 3.910 + 10.090 = 14.000 = pKw = −log(Kw) = −log(1 × 10−14

High [H3O+] results in low [OH] and also low pH < pOH.

Low [H3O+] results in high [OH] and also high pH > pOH.
Acid ionization constants span many orders of magnitude, so you’ll also see the p-scale of the Ka.

pKa = −log(Ka)     (8)
Ka = 10pKa = 
1
10pKa
    (9)

6.5  Substances

6.5.1  Strong Acids

The 7 strong acids have product favored equilibrium with Ka > 1 (pedantic: Ka > KaH3O+ = [H2O][H3O+]/[H3O+] = [H2O] = 55.5) and pKa < 0

  1. monoprotic binary acids = HCl (Ka ≈ 10+7), HBr (Ka ≈ 10+9), HI (Ka ≈ 10+10)
  2. monoprotic oxoacids = HNO3, HClO4, HClO3
  3. diprotic oxoacid = H2SO4

monoprotic = 1 H+ to donate; diprotic = 2 H+ to donate; polyprotic = more than 1 H+ to donate

binary = simple acid

oxoacid = made of oxygen containing polyatomic ion with enough H+ added to make a neutral compound
Because the Ka of strong acids is so large and product favored, ionization is approximated as being 100%. Therefore, the resulting equilibrium concentration of hydronium ion is assumed to be equal to the initial concentration of the strong acid. The strong acid approximation: [H3O+]eq = [strong acid]0
However, only the first ionization, Ka1, of H2SO4 behaves as a strong acid:

H2SO4 (aq) + H2O (l) ⇋ HSO4 (aq) + H3O+ KaH2SO4 ≈ 10+3 = Ka1H2SO4

The second ionization, Ka2, where the second H+ is donated, behaves as a weak acid:

HSO4 (aq) + H2O (l) ⇋ SO42− (aq) + H3O+ KaHSO4 = 1.02 × 10−2 = Ka2H2SO4

The strong acid approximation is not valid for H2SO4 since only 1 of the 2 available protons ionizes 100%.

6.5.2  Weak Acids

Weak acids have reactant favored equilibrium with Ka = 1.00 × 10−14 < Ka < 1 (pedantic: KaH2O = 1.80 × 10−16 < Ka < KaH3O+ = 55.5) and pKa > 0

  1. monoprotic binary acids = HF (Ka = 6.76 × 10−4), HCN (Ka = 6.17 × 10−10)
  2. diprotic binary acids = H2S (Ka1 = 1.26 × 10−7, Ka2 ≈ 10−19)
  3. monoprotic oxoacids = HClO (Ka = 2.95 × 10−8), HClO2 (Ka = 1.10 × 10−2), HNO2 (Ka = 4.00 × 10−4)
  4. polyprotic oxoacids (both H2CO3 and H2SO3 are the result of CO2 and SO2 dissolving in H2O)
    1. H2CO3 (Ka1 = 5.01 × 10−7, Ka2 = 4.79 × 10−11)
    2. H2SO3 (Ka1 = 1.39 × 10−2, Ka2 = 6.73 × 10−8)
    3. H3PO4 (Ka1 = 7.11 × 10−3, Ka2 = 6.34 × 10−8, Ka3 = 4.80 × 10−13)
  5. carboxylic acids = R-COOH, where R = H or C attached to other stuff; the separated COOH indicates a carboxylic acid with an acidic H
    1. monoprotic acetic acid (R = CH3) = CH3COOH (C2H3O2H) (Ka = 1.74 × 10−5)
    2. monoprotic benzoic acid (R = C6H5) = C6H5COOH (Ka = 6.28 × 10−5)
    3. monoprotic formic acid (R = H) = HCOOH (Ka = 1.70 × 10−4)
    4. triprotic citric acid = C6H8O7 (Ka1 = 7.41 × 10−4, Ka2 = 1.74 × 10−5, Ka3 = 4.07 × 10−7)
    5. amino acids = the building blocks of proteins which are polymers of amino acids
  6. conjugate acids of weak bases = NH4+, CH3NH3+
  7. hydrated, highly charged metal complex ions
    1. [Fe(H2O)6]3+ (aq) + H2O (l) ⇌ [Fe(H2O)5OH]2+ (aq) + H3O+ (aq) (Ka = 6.3 × 10−3)
    2. [Fe(H2O)6]2+ (aq) + H2O (l) ⇌ [Fe(H2O)5OH]+ (aq) + H3O+ (aq) (Ka = 3.2 × 10−10)
    3. [Co(H2O)6]3+ (aq) + H2O (l) ⇌ [Co(H2O)5OH]2+ (aq) + H3O+ (aq) (Ka = 1.3 × 10−9)

Because the Ka of weak acids is reactant favored, ionization is generally much less than 100%. The higher the Ka, the higher the percent ionization.

6.5.3  Not Acids

Not acids have VERY reactant favored equilibrium such that no products are ever present with Ka < Kw = 1.00 × 10−14 (pedantic: Ka < KaH2O = 1.80 × 10−16)

  1. CH4 (Ka1 ≈ 10−56), NH3 (Ka1 ≈ 10−33)
  2. alcohols = R-OH, where R = C attached to other stuff; the separated OH indicates an alcohol which does not have an acidic H
    1. methanol (R = CH3) = CH3OH (Ka = 3.16 × 10−16)
    2. ethanol (R = C2H5 = C2H5OH (Ka = 1.26 × 10−16)
  3. hydrated, low charge metal complex ions
    1. [Na(H2O)n]+ (aq) + H2O (l) ⇌ [Na(H2O)n−1OH (aq)] + H3O+ (aq) (Ka = 7.94 × 10−15)

It is interesting to note that the Ka of alcohols is very similar to that of water. Water being an alcohol if R = H: H-OH. H2O is our standard for deciding whether something is capable of being an acid or a base when dissolved in water. Is it a better acid or base than water? If so, then it is, at least, a weak acid or base.

Because the Ka of not acids is very reactant favored, ionization is assumed to be 0%.

6.5.4  Strong Bases

The strong bases have product favored equilibrium with Kb ≥ 1 (pedantic: KbKbOH = [H2O][OH]/[OH] = [H2O] = 55.5)

  1. monoprotic ionic bases = LiOH, KOH, NaOH (all with Kb = 55.5)
  2. diprotic ionic bases = Ca(OH)2, Ba(OH)2, Sr(OH)2 (all with Kb = 55.5)
  3. diprotic anionic bases = S2− (Kb1 ≈ 10+5, Kb2 = 7.94 × 10−8)

monoprotic = 1 H+ to accept; diprotic = 2 H+ to accept; polyprotic = more than 1 H+ to accept
Because the Kb of strong bases is so large and product favored, ionization is approximated as being 100%.

Therefore, the resulting equilibrium concentration of hydroxide ion is assumed to be equal to the initial concentration of the strong base. The strong base approximation: [OH]eq = n[strong base]0 where n = the number of OH in the chemical formula for ionic bases or the number of OH capable of being produced for anionic bases. Sulfide, S2−, would require the second ionization be treated separately with less than 100% ionization since Kb2 < 1.
Only the first ionization, Kb1, of S2− behaves as a strong base:

S2− (aq) + H2O (l) ⇋ HS (aq) + OH KbS2− ≈ 10+5 = Kb1S2−

The second ionization, Kb2, where the second H+ is accepted, behaves as a weak base:

HS (aq) + H2O (l) ⇋ H2S (aq) + OH KbHS = 7.94 × 10−8 = Kb2S2−

The strong base approximation is not valid for S2− since only 1 of the 2 protons to be accepted ionizes 100%.

6.5.5  Weak Bases

Weak bases have reactant favored equilibrium with Kw = 1.00 × 10−14 < Kb < 1 (pedantic: KbH2O = 1.80 × 10−16 < Kb < KbOH = 55.5)

  1. amines = R3N, where R = H or C attached to other stuff
    1. ammonia (R = H) = NH3 (Kb = 1.78 × 10−5)
    2. methylamine (R1 = CH3, R2/3 = H) (Kb = 4.37 × 10−4)
    3. amino acids = the building blocks of proteins which are polymers of amino acids
  2. conjugate bases of weak acids = F, NO2, ClO, ClO2, CH3COO, HS, etc

6.5.6  Not Bases

Not bases have VERY reactant favored equilibrium such that no products are ever present with Kb < Kw = 1.00 × 10−14 (pedantic: Kb < KbH2O = 1.80 × 10−16)

  1. conjuage bases of strong acids = Cl, Br, I, ClO, NO3, HSO4
  2. CH4 (Kb ≈ 0)

Because the Kb of not bases is very reactant favored, ionization is assumed to be 0%.

6.5.7  Anhydrides

Anhydride means "without water".

Acid anhydrides are the anhydrous versions of acids.

For example, the acid anhydride of HClO is found by doubling the empirical formula to obtain an even number of H atoms that are connected to O atoms giving H2Cl2O2 and then subtracting as many H2O molecules it takes to remove all of the O atom connected H atoms to give Cl2O.

Presumably, the acid anhydride of H3PO3 would be HPO2 because only 2 of the H atoms are attached to O atoms (yielding only 2 acidic H atoms) with the 3rd H atom directly attached to the P atom.

Base anhydrides are the anhydrous versions of bases.

For example, the base anhydride of NaOH is found by doubling the empirical formula to obtain an even number of H atoms that are connected to O atoms giving Na2(OH)2 and then subtracting as many H2O molecules it takes to remove all of the O atom connected H atoms to give Na2O.

Lime, calcium oxide CaO, is a base anhydride, because when you add water to it you make calcium hydroxide, Ca(OH)2.

6.6  Adding acids and bases to water

6.6.1  Monoprotic strong acid

What is the [H3O+]eq in a 0.100 M solution of HCl (aq) at 25C? Assume the Ka of HCl is 1.00 × 107.

HCl (aq) + H2O (l) ⇋ Cl (aq) + H3O+ (aq)
RHCl (aq)+H2O (l)Cl (aq)+H3O+ (aq)
I0.100 M   0.0 M 1.00 × 10−7 ≈ 0.0 M
Cx   +x +x
E0.100 − x   x x
1.00 × 107 = 
x2
0.100−x
 
x = 0.0999999999999999999 ≈ 0.100 
[H3O+]eq = 0.100 

Since the Ka for strong acids is so large, we can assume 100% ionization, which is the strong acid approximation meaning that all of the strong acid ionizes giving [H3O+]eq = [conjugate base]eq = [strong acid]0 and [strong acid]eq ≈ 0 M. The HCl turns completely into the conjugate base of the acid, Cl in this case, and H3O+.

This is a good time to realize that Cl is such a terrible base that it won’t even react with H3O+, which is the strongest acid that persists in water due to the leveling effect (which states that all strong acids are equivalent in strength in water due to their 100% ionization allowed/leveled by the basicity of water which is a strong enough base to prevent differentiating the relative strength of the strong acids because it reacts 100% with them).

This gives a pH = −log(0.100) = 1.000.

What is the [OH]eq in a 0.100 M solution of HCl (aq) at 25C?
The concentrations of H3O+ and OH are always related through Kw.

1.00 × 10−14 = (0.100  M H3O+)[OH
[OH]eq = 1.00 × 10−13 

This solution would be labelled acidic because [H3O+]eq > [OH]eq. We see that when one concentration goes up, the other must go down so that when you multiply both, the product is still Kw.

6.6.2  Monoprotic weak acid

What is the [H3O+]eq in a 0.100 M solution of HF (aq) at 25C?
Using the acid ionization reaction with its associated Ka = 6.76 × 10−4:

RHF (aq)+H2O (l)F (aq)+H3O+ (aq)
I0.100 M   0.0 M 1.00 × 10−7 ≈ 0.0 M
Cx   +x +x
E0.100 − x   x x
6.76 × 10−4 = 
x2
0.100−x
 
quadratic formula:  x = 
7.891 × 10−3
  or  −8.567 × 10−3 

Instead of solving this quadratic equation with the quadratic formula, you can iterate. x, how far the reaction needs to go to get to equilibrium, is going to be small because Ka is small compared to 0.100 M. If x is very small, x ≈ 0.0, then 0.100 - x ≈ 0.100.

6.76 × 10−4 = 
x2
0.100
 
x = 8.22 × 10−3 

Now that we have a good guess for x, plug it in instead of 0.0 for the second iteration:

6.76 × 10−4 = 
x2
0.100−8.22 × 10−3
 
x = 7.88 × 10−3 

Now that we have a better guess for x, plug it in for the third iteration:

6.76 × 10−4 = 
x2
0.100−7.88 × 10−3
 
x = 7.89 × 10−3 

Fourth iteration:

6.76 × 10−4 = 
x2
0.100−7.89 × 10−3
 
x = 7.89 × 10−3 

Now we stop because the x we put in is the same x that was returned; the iterations have converged.

Regardless of the mathematical method used to solve the equation:

[H3O+]eq = 7.89 × 10−3 M 

This leads to a percent ionization of:

7.89 × 10−3
0.100
 × 100 = 7.89% 

This leads to a pH = −log(7.89 × 10−3) = 2.103.

6.6.3  Dibasic strong base

What is the [H3O+]eq in a 0.100 M solution of Ca(OH)2 (aq) at 25C?
Ca(OH)2 (s) → Ca2+ (aq) + 2 OH (aq)

Since this compound dissociates upon dissolution directly creating OH, we just use the Kw relation to get [H3O+]:

[OH] = 0.100 M Ca(OH)2 


OH
Ca(OH)2
 


 = 0.200  M OH 


The concentrations of H3O+ and OH are always related through Kw.

1.00 × 10−14 = [H3O+](0.200 M OH
[H3O+] = 5.00 × 10−14 

This solution would be labelled basic because [H3O+] < [OH].

The pH = −log(5.00 × 10−14) = 13.301.

6.6.4  Monobasic weak base

What is the pH in a 0.100 M solution of NH3 (aq) at 25C?
Using the base ionization reaction with its associated Kb = 1.78 × 10−5:

RNH3 (aq)+H2O (l)NH4+ (aq)+OH (aq)
I0.100 M   0.0 M 1.00 × 10−7 ≈ 0.0 M
Cx   +x +x
E0.100 − x   x x
1.78 × 10−5 = 
x2
0.100−x
 
quadratic formula:  x = 
1.33 × 10−3
  or  −1.34 × 10−3 
[OH]eq = 1.33 × 10−3 M 

This leads to a percent ionization of:

1.33 × 10−3
0.100
 × 100 = 1.33% 

The pOH = −log(1.33 × 10−3) = 2.876 which allows you to get the pH = 14 − pOH = 14 − 2.876 = 11.124.

6.6.5  Ionic Compounds

Problem

What is the pH in a 0.100 M solution of NaF (aq) at 25C?
NaF (s) ⇋ Na+ (aq) + F (aq)

What can act as an acid or base?

Na+ doesn’t make a hydrated metal cation with any meaningful acidity and it has no electrons to form a bond with H+ so it can’t be a base either.

F is the conjugate base of HF, so perhaps it can act as a base.

F (aq) + H2O (l) ⇋ HF (aq) + OH (aq)Kb = ?
HF (aq) + H2O (l) ⇋ F (aq) + H3O+ (aq)Ka = 6.76 × 10−4
F (aq) + HF (aq) + 2 H2O (l) ⇋ HF (aq) + F (aq) + H3O+ (aq) + OH (aq) 
2 H2O (l) ⇋ H3O+ (aq) + OH (aq)Ka × Kb = Kw


For conjugate acid/base pairs only:

Ka × Kb = Kw     (10)

So the Kb for F is:

KbF = 
Kw
KaHF
 = 
1.00 × 10−14
6.76 × 10−4
 = 1.48 × 10−11 

Using the base ionization reaction with its associated Kb = 1.48 × 10−11:

RF (aq)+H2O (l)HF (aq)+OH (aq)
I0.100 M   0.0 M 1.00 × 10−7 ≈ 0.0 M
Cx   +x +x
E0.100 − x   x x
1.48 × 10−11 = 
x2
0.100−x
 
quadratic formula:  x = 
1.22 × 10−6
 
[OH]eq = 1.22 × 10−6  M 

This leads to a percent ionization of:

1.22 × 10−6
0.100
 × 100 = 0.00122% 

The pOH = −log(1.22 × 10−6) = 5.914 which allows you to get the pH = 14 − pOH = 14 − 5.914 = 8.086.

This pH is not too far from neutral pH = 7 because F is a very weak base.

———————————————————————–

Problem

What is the pH of a 0.100 M solution of FeCl3 (aq)?
FeCl3 (s) ⇋ Fe3+ (aq) + 3 Cl (aq)

Is Cl a base just like F? Technically, yes. Actually, no.

KbCl = 
Kw
KaHCl
 = 
1.00 × 10−14
10+7
 = 1.00 × 10−21 

The Kb of Cl is so low that the auto-ionization of H2O generates more hydroxide than Cl would, so Cl has negligible basicity.

The Fes+ attracts 6 H2O to make a hydrated metal cation, [Fe(H2O)6]3+, which has a Ka = 6.3 × 10−3 which leads to [H3O+] = 0.022 M with a pH = 1.66.

———————————————————————–

6.6.6  Diprotic sulfuric acid

Problem

What is the [H3O+]eq in a 0.200 M solution of H2SO4 (aq) at 25C?
Since the Ka1 for sulfuric acid is so large, we assume 100% ionization for the first ionization. It turns completely into the conjugate base of the acid, HSO4, and H3O+:

H2SO4 (aq) + H2O (l) ⇋ HSO4 (aq) + H3O+ (aq)

[H3O+] = 0.200 M and [HSO4] = 0.200 M 

However, this is only the H3O+ generated from the first ionization. There is still another H+ to lose in the conjugate base molecule generated, HSO4. Using the acid ionization reaction with its associated Ka2:

RHSO4 (aq)+H2O (l)SO42− (aq)+H3O+ (aq)
I0.200 M   0.0 M 0.200 M
Cx   +x +x
E0.200 − x   x 0.200 + x
0.012 = 
(x)(0.200+x)
0.200−x
 
quadratic formula:  x = 
0.01077
  or  −0.2228 

Instead of solving this quadratic equation with the quadratic formula, you can iterate. x, how far the reaction needs to go to get to equilibrium, is going to be small because Ka is small compared to 0.200 M. If x is very small, x ≈ 0.0, then 0.200 ± x ≈ 0.200.

0.012 = 
0.200x
0.200
 
x = 0.012 

Now that we have a good guess for x, plug it in instead of 0.0 for the second iteration:

0.012 = 
(x)(0.200+0.012)
0.200−0.012
 
x = 0.0106 

Third iteration:

0.012 = 
(x)(0.200+0.0106)
0.200−0.0106
 
x = 0.0108 

Fourth iteration:

0.012 = 
(x)(0.200+0.0108)
0.200−0.0108
 
x = 0.0108 

Now we stop because the x we put in is the same x that was returned; the iterations have converged.

Regardless of the mathematical method used to solve the equation:

[H3O+]eq = 0.200 + x = 0.200 + 0.0108 = 0.211  M 

This leads to a percent ionization of:

0.211
(2 H+/H2SO4)(0.200)
 × 100 = 52.8% 

———————————————————————–

6.6.7  Periodic trends in acidity of binary acids and basicity of the halides

Acidity increases from left to right following electronegativity.

Electronegativity is the ability to draw bonding electrons to itself as a result of an atom’s size and effective nuclear charge.

Due to the EN of the central atom, the H becomes more and more partially positive (δ+), thus making it easier and easier to remove with the electrons from the B-L base.

acidCH4NH3H2OHF
pKa563315.73.17
ΔEN2.5−2.1=0.43.0−2.1=0.93.5−2.1=1.44.0−2.1=1.9
Hδ+-0.044+0.371+0.636+0.782
Xδ+0.176-1.113-1.272-0.782

Acidity increases from top to bottom NOT following electronegativity, but instead the ionic radius of the central atom.

Basicity decreases with increasing size of the anion because the bond to the H+ becomes weaker as the negative charge holding onto the H+ becomes less concentrated (don’t forget Vsphere = (4/3) π r3) and because the proton is farther away due to the increasing size.

F is a stronger base than I, therefore HF is a weaker acid than HI.

acidpKaΔENIR (pm)V (pm3)Factor
HF3.174.0−2.1=1.91197.06 × 1061.000
HCl-73.0−2.1=0.91671.95 × 1072.762
HBr-92.8−2.1=0.71822.53 × 1073.584
HI-102.5−2.1=0.42063.66 × 1075.184

Proof of concept can be had when looking at the solubility of sodium or lithium halides.

The solubility of NaF and LiF should be much lower than NaI and LiI because the F is holding onto the Na+ and Li+ more tightly than an I can because F is a better base because it’s 1− charge is much more concentrated.

NaXSolubility (mol/100 g H2O)LiXSolubility (mol/100 mL H2O)
NaF0.09669LiF0.005166
NaCl0.6141LiCl1.9873
NaBr0.8825LiBr1.92
NaI1.1875LiI1.25

6.6.8  Periodic trends in acidity of oxoacids and basicity of polyatomic ions

An oxoacid is the protonated form of an oxygen containing polyatomic ion.

Sulfuric acid, H2SO4, is the oxoacid of the sulfate polyatomic ion, SO42−.

Oxoacid strength increases with (listed in order of importance):

  1. The number of O atoms (also represented by an increase in the oxidation state of the central atom)
  2. Electronegativity of the central atom (e.g. S in H2SO4)
acidHClO4HClO3HClO2HClO
Kastrong, ≈ 107 >>> 1strong, ≈ 103 > 11.1 × 10−22.9 × 10−8
acid   HBrO
Ka   2.8 × 10−9
acid HIO3 HIO
Ka 1.7 × 10−1 2.3 × 10−11

Other examples include: H2SO4 > H2SO3 > H2SeO3 and HNO3 > HNO2.

6.7  Lewis Definition

Consistent with the Brønsted-Lowry definition; Lewis acid = Brønsted-Lowry acid; Lewis base = Brønsted-Lowry base.

Formation of hydrated metal cations (and other complex ions) involves the metal cation acting as the Lewis acid and the water (or other ligand) donating a lone pair as a Lewis base to make what is called a coordinate covalent bond (coordinate indicates that the 2 bonding electrons come from the same molecule).

Cations (with missing electrons) not only attract negative electrons, but have empty orbitals to put the bonding pair of electrons from the Lewis base into.

Atoms with incomplete octets also have an empty orbital to put new, bonding electrons into, e.g. compounds of boron and aluminum (BX3 or AlX3).

The carbon atom in CO2 acts as a Lewis acid while H2O acts as a Lewis base to form carbonic acid, H2CO3.

CO2 (aq) + H2O (l) ⇋ H2CO3 (aq)

6.8  Buffers

Buffers are solutions that resist change in pH, [H3O+]

  1. Must neutralize any added acid or base
  2. Must contain both an acid AND a base (2 compounds) ≠ amphoteric/amphiprotic (1 compound)
  3. Must not neutralize itself

A buffer is a solution of a weak conjugate acid-base pair present in roughly equal concentrations.
The percent ionization of a weak acid or base is not enough to create a roughly equal concentration of its conjugate.

The conjugate is often introduced by dissolving an ionic compound containing the conjugate into the solution of the weak acid/base.

Examples: 0.1 M HF & 0.1 M NaF, 0.1 M NH3 & 0.2 M NH4Cl, 0.5 M CH3COOH & 0.3 M NaCH3COO
How does a buffer work?

If a strong base is added, instead of creating OH (affecting [H3O+]), it is neutralized by the weak acid component to produce the weak base.

HF (aq) + NaOH (aq) ⇋ NaF (aq) + H2O (l) = F (aq) + H2O (l) + Na+ (aq)
If a strong acid is added, instead of creating H3O+, it is neutralized by the weak base component to produce the weak acid.

NaF (aq) + HCl (aq) ⇋ HF (aq) + NaCl (aq) = HF (aq) + Na+ (aq) + Cl (aq)
Recognize the product of neutralization is the other component of the buffer.
It can’t neutralize itself because the products are the same as the reactants (why we must use a conjugate acid-base pair).

HF (aq) + NaF (aq) ⇋ NaF (aq) + HF (aq)
Professor’s tricky way of making a buffer: react one component of buffer with a limiting reactant amount of strong acid/base to create the second component.

RHF (aq)+OH (aq)F (aq)+H2O (l)
I0.2 mol 0.1 mol 0 mol  
C-0.1 mol -0.1 mol +0.1 mol  
F0.1 mol 0.0 mol 0.1 mol  

The "F" row is the "final" result of the limiting reactant problem which isn’t necessarily the solution to the equilibrium problem.

6.8.1  pH of a Buffer

Problem

What is the pH of a buffer solution that is 0.100 M CH3COOH & 0.120 M NaCH3COO?

CH3COO = Ac for simplicity.

Buffer = 0.100 HAc & 0.120 M NaAc

RAc (aq)+H2O (l)HAc (aq)+OH (aq)Kb
I0.120   0.100 ≈0.0 
RHAc (aq)+H2O (l)Ac (aq)+H3O+ (aq)Ka
I0.100   0.120 ≈0.0 
Cx   +x +x 
E0.100 − x   0.120 + x x 
Ka = 
[Ac]eq[H3O+]eq
[HAc]eq
 
1.8 × 10−5 = 
(0.120 + x)(x)
(0.100 − x)
 

Ka is small and so x is going to be small, even more so since a bunch of product already exists at equilibrium!

[Ac]eq = (0.120 + x) ≈ 0.120 = [Ac]0 
[HAc]eq = (0.100 − x) ≈ 0.100 = [HAc]0 
1.8 × 10−5 = 
(0.120)(x)
(0.100)
 
x = 1.8 × 10−5 


0.100
0.120
 


= 1.5 × 10−5 
[H3O+]eq = 1.5 × 10−5 M → pH = 4.82 

What we did is replace the equilibrium concentrations of HAc and Ac with their initial concentrations under the approximation that the change toward equilibrium, x, was going to be too small to change their initial concentration values.

For a generic acid, we get:

Ka = 
[A]eq[H3O+]eq
[HA]eq
 ≈ 
[A]0[H3O+]eq
[HA]0
 

Solving for [H3O+]eq, we get:

[H3O+]eq = Ka 


[HA]0
[A]0
 


Taking the -log of both sides gives the Henderson-Hasselbalch equation:

−log
[H3O+]eq 
= −log(Ka) + −log


[HA]0
[A]0
 


→ pH = pKa + log


[A]0
[HA]0
 


    (11)

Applied to our previous problem where Ka = 1.8 × 10−5pKa = 4.74:

pH = 4.74 + log


0.120
0.100
 


= 4.74 + 0.08 = 4.82 

The pH is higher than the pKa of HAc because [Ac]0 > [HAc]0.

———————————————————————–

Problem

What is the pH if I add 500 mL of 0.100 M HCl to 1.0 L of the 0.100 M HAc & 0.120 M NaAc buffer?

R1HCl (aq)+Ac (aq)Cl (aq)+HAc (aq)
R2H3O+ (aq)+Ac (aq)H2O (l)+HAc (aq)
I(0.5 L)(0.1 M) = 0.05 mol (1 L)(0.12 M) = 0.12 mol   (1 L)(0.1 M) = 0.1 mol
C-0.05 -0.05   +0.05
F0.0 mol 0.07 mol   0.15 mol


While we don’t know K1 for R1, we do know K2 for R2 which is such a large K that we can justify treating the reaction as a limiting reactant problem instead of an equilibrium problem.

K2 = Ka−1 = (1.8 × 10−5)−1 = 5.6 × 104 

The limiting reactant was HCl (or rather H3O+ made from HCl) and so we’re left with a buffer.

pH = 4.74 + log





0.07 mol
Vtotal
0.15 mol
Vtotal
 





= 4.74 + log


0.07 mol
0.15 mol
 


= 4.41  

The pH went down because we added a strong acid which converted Ac into HAc, so having more of the acid component results in a pH lower than the pKa.

———————————————————————–

Problem

What is the pH if I add 1.20 L of 0.100 M HCl to 1.0 L of the 0.100 M HAc & 0.120 M NaAc buffer?

R1HCl (aq)+Ac (aq)Cl (aq)+HAc (aq)
R2H3O+ (aq)+Ac (aq)H2O (l)+HAc (aq)
I(1.2 L)(0.1 M) = 0.12 mol (1 L)(0.12 M) = 0.12 mol   (1 L)(0.1 M) = 0.1 mol
C-0.12 -0.12   +0.12
F0.0 mol 0.0 mol   0.22 mol


Here there is no limiting reactant, HCl/H3O+ and Ac are present in stoichiometric amounts.

What’s left is not a buffer, but a solution of HAc, and that’s it.

Ka = 
[Ac][H3O+]
[HAc]
 = 
(x)(x)
0.22 mol
Vtotal
x
  where  Vtotal = 1.0  L buffer  + 1.2  L HCl  = 2.2  L  
1.8 × 10−5 = 
x2
0.100 − x
 → x = 1.33 × 10−3 = [H3O+] → pH = 2.88 

Having destroyed the buffer, the pH is solely dictated by the final concentration of the remaining buffer component, in this case it was the acid component HAc that remained.

———————————————————————–

Problem

What is the pH if I add 2.00 L of 0.100 M HCl to 1.0 L of the 0.100 M HAc & 0.120 M NaAc buffer?

R1HCl (aq)+Ac (aq)Cl (aq)+HAc (aq)
R2H3O+ (aq)+Ac (aq)H2O (l)+HAc (aq)
I(2.0 L)(0.1 M) = 0.2 mol (1 L)(0.12 M) = 0.12 mol   (1 L)(0.1 M) = 0.1 mol
C-0.12 -0.12   +0.12
F0.08 mol 0.0 mol   0.22 mol


The limiting reactant is Ac so there is excess HCl/H3O+ still floating around.

Vtotal = 1.0  L buffer  + 2.0  L HCl  = 3.0  L solution 
[HAc] = 
0.22 mol
3.0 L
 = 0.0733  M HAc 
[HCl] = 
0.08 mol
3.0 L
 = 0.0267  M HCl = 0.0267  M H3O+ 

This mixture of acids can be solved using a RICE table for the Ka of HAc in the presence of one of its products, H3O+ from the excess HCl.

RHAc (aq)+H2O (l)Ac (aq)+H3O+ (aq)Ka=1.8 × 10−5
I0.0733   0 0.0267 
Cx   +x +x 
E0.0733 − x   x 0.0267 + x


Ka = 
[Ac][H3O+]
[HAc]
 
1.8 × 10−5 = 
(x)(0.0267 + x)
0.0733 − x
 → x = 4.94 × 10−5 
[H3O+] = 0.0267 + 0.0000494 = 0.0267494 = 0.0267  M → pH = 1.57 

Having destroyed the buffer and left with an excess of HCl, the pH is solely dictated by the final concentration of HCl and has nothing to do with the remaining HAc floating around.

———————————————————————–

6.8.2  Designing a Buffer

Picking which acid to use

The Henderson-Hasselbalch equation is only valid when the further ionization of either buffer component when mixed together is negligible compared to their initial concentrations.

For example, dissolving HA doesn’t make enough A from acid ionization to significantly change the amount of A that was directly dissolved when the buffer was prepared.
This means the Henderson-Hasselbalch equation is only valid for describing buffer solutions where the two components, the acid and base = HA and A, are in roughly equal concentrations that are at least within a factor of 10.

0.1 < 
[A]
[HA]
 < 10 

According to the Henderson-Hasselbalch equation:

pH = pKa + log(0.1) = pKa − 1 
pH = pKa + log(10) = pKa + 1 

This allows us to understand that a buffer will only resist large changes in pH within 1 pH unit of the acid component’s pKa.

pH = pKa ± 1 

When designing a buffer you must first pick the pH you want your solution buffered at and then you pick the appropriate acid with the pKa that most closely matches that pH.

Problem

How many grams of the sodium salt does one need to add to 1.00 L of a 0.200 M acid solution to have a buffer with pH = 4.50?

acidKapKa
HIO31.7 × 10−10.77
HClO2.9 × 10−87.54
CH3COOH1.8 × 10−54.74
HC9H7O43.3 × 10−43.48

We can see acetic acid, CH3COOH = HAc, is the best choice having the closest pKa = 4.74 to the desired pH = 4.50.

This means the sodium salt to our acid is sodium acetate, NaCH3COO = NaAc.

We can use the Henderson-Hasselbalch equation to tell us how much NaAc we need by setting the pH = 4.50 and remembering that our HAc (pKa = 4.74) concentration is 0.200 M as defined in the problem.

4.50 = 4.74 + log


[Ac]
0.200



[Ac] = 0.200 × 104.50−4.74 = 0.11509  M 
mol Ac = 0.11509  M  × 1.00 L = 0.11509  mol Ac 
0.11509 
 mol Ac × 
mol NaAc
mol Ac
 = 0.11509 mol NaAc
 
0.11509 
 mol NaAc × 
82.0343 g NaAc
mol NaAc
 = 9.44 g NaAc = 9.44 g NaCH3COO
 

One would need to add 9.44 g NaCH3COO to 1.00 L of 0.200 M CH3COOH to obtain a buffer with pH = 4.50.

The volume change from the addition of the 9.44 g NaAc doesn’t matter because it is really the mole ratio that determines the pH since the molar concentration ratio is the same quantity because the buffer is all in the same volume of solvent.

———————————————————————–

Picking concentrations

The higher the initial concentrations of the components of the buffer, the higher the capacity of the buffer because the ratio of HA and A won’t change as much allowing it to stay within a factor of 10 more easily, but also the resulting pH changes will be smaller.

Adding an acid to a buffer with a lower concentration of buffer components would have the following effect on the change in pH, Δ pH, away from the acid’s pKa if we view the Henderson-Hasselbalch equation like this:

pH = pKa + Δ pH 
Δ pH = log(
0.1−0.05
0.1+0.05
) = log(
0.05
0.15
) = log(0.333) = −0.477 

The buffer component ratio not only dropped from 0.1/0.1 = 1 to 0.333 but the pH went down 0.477 units.

Adding the same amount of acid to a higher concentration buffer results in a smaller change in the ratio, from 0.2/0.2 = 1 to 0.6, but also a smaller pH change, 0.222.

Δ pH = log(
0.2−0.05
0.2+0.05
) = log(
0.15
0.25
) = log(0.600) = −0.222 

6.9  Titrations

Titration = analytic method to determine the unknown concentration of an analyte using a standardized solution of titrant (standardized = the exact concentration has been determined).

This is accomplished by volumetric addition of titrant solution to the analyte solution until a stoichiometric amount of titrant has been added (stoichiometric = consistent with the balanced chemical equation resulting in no limiting or excess reactants).

Considerations:

  1. Need the reaction between titrant and analyte to go 100% so that all titrant and analyte are consumed stoichiometrically.
  2. Need some way to indicate that a stoichiometric amount of titrant has been added.

Titration Curve (for acid/base titrations) = A plot of the pH of the analyte solution (y-axis) versus volume of added titrant solution (x-axis).

The goal of this section is to be able to calculate the pH of the analyte solution provided the volume of titrant solution added so far.

In order to do this, you not only need to know the identity and concentration of the titrant solution (which you always will in the lab because this is determined by you) but you’ll also need to know the identity and concentration of the analyte solution (which is often NOT something you know because the whole point of the titration is to determine the concentration of the analyte solution where you may never know the identity of the analyte!).

Depending on how much titrant solution is added to the analyte solution, the pH of the resulting solution will be calculated any number of ways.

Examples with relevant reactions will be given for the titration of a acetic acid, HAc, analyte with potassium hydroxide, KOH, titrant.

  1. Initial: Vtitrant = 0.0 mL
    HAc (aq) + H2O (l) ⇋ Ac (aq) + H3O+ (aq)
  2. Buffer Region: Vtitrant < Vequivalence point
  3. Equivalence point: Vtitrant = Vequivalence point
  4. Error Region: Vtitrant > Vequivalence point
Problem

What is the pH at the equivalence point of the titration of 0.100 M NaOH into an analyte solution of 12.5 mL 0.243 M butyric acid (Ka = 1.5 × 10−5).

For simplicity, the analyte butyric acid = HA and its analyte conjugate = A.

Since we’re at the equivalence point, the moles of titrant, OH (Na+ is just a spectator ion), must be equal to the moles of analyte, HA, so that there is no limiting or excess reactants.

RHA+OHA+H2O
I0.0125 L × 0.243 M = 0.0030375 mol 0.0030375 mol 0 mol 
C-0.0030375 mol -0.0030375 mol +0.0030375 mol  
F0.0 mol 0.0 mol 0.0030375 mol  
[A] = 
0.0030375 mol A
0.0125 L HA + ? L NaOH
 

We do not know the volume of titrant NaOH that was added to the analyte solution to reach the equivalence point.

What volume of the 0.100 M NaOH solution contains the needed 0.0030375 mol NaOH necessary to get to the equivalence point?

Vequivalence point = 
0.0030375 mol NaOH
0.100 M NaOH
 = 0.030375 L NaOH
 
[A] = 
0.0030375 mol A
0.0125 L HA + 0.030375 L NaOH
 = 0.0708455 M A
 

We can now use the base properties of the created analyte conjugate, A, to calculate the pH of the analyte solution at the equivalence point.

But first, we need to calculate the Kb of butyrate, A, from the Ka of butyric acid, HA.

Kb = 
Kw
Ka
 = 
1 × 10−14
1.5 × 10−5
 = 6.7 × 10−10 

Now we can solve the equilibrium of the A floating in water.

RA+H2OHA+OH
I0.0708455   0 ≈ 0
Cx   +x +x
E0.0708455 − x   x x
x2
0.0708455 − x
 = 6.7 × 10−10 
x = 6.87 × 10−6 = [OH
pH = −log


1 × 10−14
6.87 × 10−6
 


= −log(1.46 × 10−9) = 8.84 

We’ll notice that the analyte solution at the equivalence point contains only the analyte conjugate, A, and as such should have a basic pH > 7.

———————————————————————–

6.9.1  Distinguishing features of Titration Curves

6.9.2  Color Indicators

Color indicators are often acids/bases themselves, so HIn will be used for the acidic form of the indicator and In will be used for the basic form of the indicator.

HIn (aq)+H2O (l)In (aq)+H3O+ (aq)
color 1   color 2 

A very small amount of color indicator is added to the analyte solution.

By Le Chatelier’s Principle, the predominant color will be determined by the pH of the analyte solution.

If the analyte is an acid, then at initial time the solution will be color 1 because the H3O+ present from ionization of the weak acid analyte prevents the formation of In.

As the weak acid analyte is titrated by the addition of strong base titrant, the pH of the analyte solution increases.

At the equivalence point, which we’ve already established will have a basic pH > 7, the concentration of H3O+ will be very low allowing the indicator’s equilibrium to shift to the product side creating In and making the solution start to appear color 2.

The transition from color 1 to color 2 occurs when the concentrations of HIn and In are roughly equal (consistent with the idea of a buffer) as they switch which one is present in the higher concentration.

Using the Henderson-Hasselbalch equation to model the behavior of the indicator during the color change at the point that HIn and In have equal concentrations, we see that the analyte solution will change from color 1 to color 2 when the pH of the titration is near the pKa of the acid form of the indicator, HIn.

pHequivalence point = pKaHIn + log


[In]
[HIn]
 


pKaHIn + log(1) = pKaHIn 

It is very important to select an indicator with a pKa within ± 1 of the pH at the equivalence point, otherwise the indicator will change color at the wrong time (e.g. in the Buffer Region or the Error Region, or if very poorly selected at a pH never exhibited in the particular titration you’re doing).

The equivalence point indicated by a color indicator is called the end point due to its lack of precision.

Because the pKa of the indicator is unlikely to correspond to the exact equivalence point pH, the color change will occur slightly before or after the real equivalence point pH; however the change in pH during the transition from the Buffer Region to the Error Region is so large and so sudden (occuring with only minute addition of titrant solution) that the equivalence point in the middle is often well approximated by the color indicated end point.


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