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Chapter 16: Acids, Bases, and Salts, Study notes of Chemistry

BOCES Practical Nursing Program - AlbanyChemistry

The stronger the acid, the weaker the conjugate base. ... weaker base. Therefore sulfuric acid is the stronger acid. ... H2SO4 is a strong acid while HSO4.

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Chapter 16: Acids, Bases, and Salts
Key topics:
pH scale; acid-base properties of water
Ka = acid ionization constant; Kb = base ionization constant
Polyprotic acids
BrØnsted Acids and Bases
Acid: proton donor
Base: proton acceptor
acid/base reaction: proton transfer
Each side of the reaction has an acid and a base !
Conjugate base: what remains after the acid donates a proton
Conjugate acid: formed when the base accepts a proton
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Chapter 16: Acids, Bases, and Salts

Key topics: pH scale; acid-base properties of water Ka = acid ionization constant; Kb = base ionization constant Polyprotic acids BrØnsted Acids and Bases Acid : proton donor Base : proton acceptor acid/base reaction: proton transfer Each side of the reaction has an acid and a base! Conjugate base: what remains after the acid donates a proton Conjugate acid: formed when the base accepts a proton

an acid and its conjugate base : conjugate acid-base pair a base and its conjugate acid : conjugate acid-base pair The two species in a conjugate pair differ only by H

There are two conjugate pairs in any acid-base reaction e.g. , HSO 3

- is the conjugate acid of what species? HSO 3 - is the conjugate base of what species? Solution: SO 3 2 – , H 2 SO 3 e.g. , (a) Write an equation in which HSO 4 - reacts (with water) to form its conjugate base. (b) Write an equation in which HSO 4 - reacts (with water) to form its conjugate acid. Solution: HSO 4 -

  • H 2 O SO 4 2 –
  • H 3 O

HSO 4 -

  • H 2 O H 2 SO 4 + OH -

Oxoacids (O—H bond): contain H, O, and a central non-metal Acidity determined by the polarity of the O—H bond:

  1. central atom from the same group of the periodic table, same oxidation number Relative strength increases with increased electronegativity of the central atom. HClO 3 > HBrO 3 H 2 SO 4 > H 2 SeO 4
  2. same central atom, different oxidation number (# of O’s) Higher oxidation number = stronger acid. e.g. , H 2 SO 4 > H 2 SO 3 e.g. , HClO 4 > HClO 3 > HClO 2 > HClO

e.g. , CH 3 COOH > CH 3 CH 2 OH

  1. different central atom, different oxidation number The stronger the acid, the weaker the conjugate base. Compare sulfuric and phosphoric acid: H 2 SO 4 HSO 4 - + H + H 3 PO 4 H 2 PO 4 - + H + The lone oxygen atoms in HSO 4

attract H

less, making it a weaker base. Therefore sulfuric acid is the stronger acid.

The Acid-Base Properties of Water Water is amphoteric : it can act as an acid or a base. Water also autoionizes to a very small degree The equilibrium constant Kw does not include the liquid, so

Kw = [H

][OH

] or Kw = [H 3 O

][OH

]

at 25

C, [H

] = 1. 0 ⇥ 10

7

= [OH

] ) Kw = 1. 0 ⇥ 10

14 For any temperature, pure water is neutral ([OH

  • ] = [H + ]) but the concentrations vary slightly which is reflected in Kw. For dilute aqueous solutions, Kw still applies. If [H

] = [OH

] the solution is neutral If [H

] > [OH

] the solution is acidic If [H

] < [OH

] the solution is basic e.g. , at 25ºC, [H

] = 4.6 x 10

  • 8 M. Find [OH

]. Is the solution acidic or basic or neutral?

(from chemwiki.ucdavis.edu) Solution: [H

][OH ] = 1. 0 ⇥ 10 14 ) [OH ] =

  1. 0 ⇥ 10 14
  2. 6 ⇥ 10 ^8 = 2. 2 ⇥ 10 7 Therefore the solution is basic. The pH and pOH Scales pH = log[H

] and pOH = log[OH ] and pX = log X

Kw = [H 3 O

][OH

] ) log Kw = log[H 3 O

] + log[OH

]

) pKw = pH + pOH (= 14 at 25 C) For pure water at 25°C, pH = pOH = 7 (neutral). When pH < 7, pOH > 7 and the solution is acidic. When pH > 7, pOH < 7 and the solution is basic.

List of strong bases. Note: the Group 2A hydroxides are completely dissociated in solutions of 0.01 M or less. These are insoluble bases which ionize 100% (the tiny amount that is soluble dissociates completely). e.g., An aqueous solution of a strong base has pH 12.24 at 25 °C. Calculate the concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH) 2. Answer: pH = 12.24 means that pOH = 14 – 12 .24 = 1. Therefore [OH

- ] = 10 - 1. = 0. With NaOH, we must have [NaOH] = 0.017 M With Ba(OH) 2 , we have [Ba(OH) 2 ] = (0.017 / 2) = 8.7 x 10 - 3 M

Weak Acids and Acid Ionization Constants Weak acids do not fully dissociate in water. The extent to which a weak acid ionizes in water depends on

  1. the concentration of the acid
  2. the equilibrium constant for the ionization reaction, Ka We will only consider monoprotic acids in this section: HA(aq) ⌦ H

(aq) + A (aq)

Ka =

[H

][A

]

[HA]

Ka is called the acid ionization constant. pKa = - log Ka The larger the value of Ka, the stronger the acid. We can calculate pH from Ka, or Ka from pH. The math is simple if: [H

]eq < (0.05) [HA]initial or [HA]initial > 400 Ka Otherwise, we will need the quadratic formula or the method of successive approximations.

At equilibrium we have

Ka =

x

2

0. 50 x This is a quadratic equation in x.

The “simple” math is to put (0. 50 x) ⇡ 0. 50 Then we have Ka = x 2

  1. 50 ) x 2 = (0.50)(7. 1 ⇥ 10 4 ) ) x = 0. 019 The pH is then = - log(0.019) = 1. (redo with the) Quadratic formula approach: We rewrite Ka = x 2
  2. 50 x (^) as x 2
  • Kax (0.50)Ka = 0 and use the quadratic formula (page A-4) to write x = (Ka ± p K a^2 + 4(0.5)Ka)/2 = 0. 0185 This is more accurate because it gives

Ka =

2

4 whereas if we use x = 0.019 we get Ka = 7.5 x 10

  • 4 (redo with the) Successive Approximations approach (page A-4): This approach is more accurate than the simple approach but less complicated than using the quadratic formula. We begin like the simple approach: Starting from Ka = x 2
  1. 50 x (^) we put (0.^50 ^ x)^ ⇡^0.^50 to get Ka = x 2
  2. 50 ) x 2 = (0.50)(7. 1 ⇥ 10 4 ) ) x = 0. 019

A weak acid ionizes more the weaker its concentration. (data for acetic acid) Next, we use this value of x in the (0.5 0 – x) part of Ka, giving Ka = x 2

  1. 50 0. 019 = x 2
  2. 481 ) x = 0. 0185 Then we do it again: Ka = x 2
  3. 50 0. 0185 = x 2
  4. 4815 ) x = 0. 0185 We stop when the answer stops changing. (The simple approach was good enough in this case because the simple math assumptions held.) Percent Ionization For a monoprotic acid (HA),

percent ionization =

[H

]eq

[HA] 0

First we set 0.^0015 ^ x^ ⇡^0.^0015 to get

x 2 = (0.0015)Ka ) x = 1. 64 ⇥ 10 4 M Then we put this value of x in the denominator:

Ka =

x

2

0. 0015 1. 64 ⇥ 10 ^4

x

2

which gives x = 1.55 x 10

  • 4 . We repeat to get Ka = x 2
  1. 0015 1. 55 ⇥ 10 ^4

x 2

  1. 001345 which gives x = 1.56 x 10
  • 4 The answer has stopped changing so we are done. This results in x = 1.56 x 10
  • 4 M with pH = 3. and % ionization = 10.4 % Using pH to Determine Ka We can use the pH of a weak acid to determine the equilibrium concentrations, which gives us the equilibrium constant. e.g. , Calculate the Ka of a weak acid if a 0.015 M solution of the acid has a pH of 5.03 at 25ºC. Solution: pH = 5.03 = log[H

] ) [H

] = 10 5. 03 = 9. 33 ⇥ 10 6 M HA H

  • A

Initial 0.015 0 0 Change - 9.33 x 10

  • 6
  • 9.33 x 10
  • 6
  • 9.33 x 10
  • 6 Equilibrium 0.01499 9.33 x 10
  • 6 9.33 x 10
  • 6

Ka =

6

2

9

All these Bronsted bases have a lone pair of electrons on a nitrogen atom which can accept a proton. Weak Bases and Base Ionization Constants To define the base ionization constant Kb we write: B( aq ) + H2O( l ) HB

( aq ) + OH

  • ( aq )

Kb =

[HB

][OH

]

[B] where HB+^ is the conjugate acid of base B

Conjugate Acid-Base Pairs Consider dissolving NaCl in water. Since Cl

  • is the conjugate base of a strong acid (HCl), it is called a weak conjugate base and it will not generate HCl: Cl
  • H 2 O 6! HCl + OH Next consider dissolving NaF in water. Since F

is the conjugate base of a weak acid (HF), it is called a strong conjugate base and it will generate HF:

F

+ H 2 O ⌦ HF + OH

Let us examine HF in more detail: F

  • H 2 O ⌦ HF + OH Kb = [HF][OH ] [F ] HF + H 2 O ⌦ F
  • H 3 O

Ka = [F ][H 3 O

] [HF] Add these two together: 2H 2 O ⌦ OH

  • H 3 O

K = Ka ⇥ Kb = [OH ][H 3 O

] = Kw Therefore pKa + pKb = pKw = 14 (at 25 C) This is the conjugate acid-base pair relationship. e.g. , Determine (a) Kb of the weak base B whose conjugate acid HB

has Ka = 8.9 x 10

  • 4 and (b) Ka of the weak acid HA whose conjugate base has Kb = 2.1 x 10
  • 8 . Solution: We just use Ka x Kb = Kw = 1.00 x 10
  • 14 (a) Kb = Kw / Ka = 1.00 x 10
  • 14 / 8.9 x 10
  • 4 = 1.12 x 10
  • 11 (b) Ka = Kw / Kb = 1.00 x 10
  • 14 / 2.1 x 10
  • 8 = 4.76 x 10
  • 7

(from www.chem.wisc.edu) Diprotic and Polyprotic Acids Some acids can undergo two or three ionizations: Note that the conjugate base in the first ionization serves as the acid in the second ionization. The second ionization constant is very small compared to the first one (usually at least 1000x smaller). This is because it is harder to remove a proton (H

) from a negatively charged species. This lets us make some simple approximations. e.g. , Calculate the concentrations of H 2 SO 4 , HSO 4

  • , SO 4 2 – , and H

ions in a 0.14 M sulfuric acid solution at 25 ºC. Ka1 = very large ; Ka2 = 1.3 x 10

  • 2 = 0.