Acids, Bases, and Buffers—H+ in Chemistry

Chapter Story: respiratory alkalosis due to faster removal of CO2 from the lungs than it arrives.

You need to know how to use your calculators to find "log" and do calculations on exponents.

General descriptions of acids and bases when dissolved in water:
Acids:
a) Produce hydrogen ion(s) (i.e., proton, H+).
    e.g., HCl, H2SO4 (sulfuric acid, the culprit of acid rain (click here to see the Acid Rain Program at EPA), a main ingredient in acid-lead battery), HNO3 (nitric acid), H3PO4 (phosphoric acid), H2CO3 (carbonic acid)
    In fact, hydronium ion(s) (H3O+) is formed.
        AH + H2O ® A– + H3O+

b) Neutralize "bases" to produce water and salts.
    HCl + NaOH ® H2O + NaCl
    2HCl + Ca(OH)2 ® 2H2O + CaCl2

c) Turn litmus red.

Bases:
a) Produce hydroxide ion(s) (i.e., OH–).
    e.g., NaOH, Ca(OH)4, KOH, Mg(OH)2, NH3
        NH3 + H2O ® NH4+ + OH–

b) Neutralize acids to produce water and salts.
    See examples in Acids-b above.

c) Turn litmus blue.

"Reaction" between two water molecules
A water molecule interacts with another water molecule via a H-bond. An equilibrium is established in which a hydrogen ion (proton, H+) is transferred from one water molecule to another to produce basic hydroxide ion (OH–) and acidic hydronium ion (H3O+). Water thus can serve as both an acid and a base!
        H2O + H2O h H3O+ + OH–

In pure water at 25 °C, [H3O+] = [OH–] = 1 ´ 10–7 M and the ion-product [H3O+] ´ [OH–] is a constant Kw.
        Kw = [H3O +][OH–] = (1 ´ 10–7 M)(1 ´ 10–7 M) = 1 ´ 10–14 M2

       <Excercise: What is the concentration of H+ and OH– in ppm at neutral water?>

Acidic: [H3O +] > 10–7 M > [OH–]
Basic: [OH–] > 10–7 M > [H3O+]

BrÆnsted-Lowry acids and bases
Acids: giving H+ in aqueous solution
    For example: H—Cl + H2O ® H3O+ + Cl–

As a general form:
    A—Hn + H2O ® n H3O+ + An–

Bases: giving OH– in aqueous solution (H+ acceptor)
    For example: NaOH in water ® Na+ + OH–

As a general form:
    B—(OH)n in water ® Bn+ + n OH–

Strong acids: 100% dissociated in water (into H3O+ and An–)
    e.g., HCl, HNO3, H2SO4 (both protons!!)---delete the footnote on P. 226!

Strong bases: 100% dissociated in water (into OH– and Bn+)
    e.g., LiOH, NaOH, Ca(OH)2, etc. (cf. Table 9.2)

The pH scale---power of [negative] H+ concentration
    [H+] = 10–pH pH = –log[H+]

A= 10B Þ logA = log10B = B

pH:       0 ------------------> 14
[H+]:     1 ´ 100 ----------> 1 ´ 10–14 M
[OH–]: 1 ´ 10–14---------> 1 ´ 100 M

Example: pHs of various substances and some body fluids on P. 230.

Exercises
(1) What is the pH of 3.00 mM [H+] solution?
        3.00 mM = 0.00300 M
        pH = –log(0.00300) = 3.00

(2) [H+] = 2.00 ´ 10–10 M, what is the pH? What is [OH–]?
        pH = –log(2.00 ´ 10–10) = 9.70
        [OH–] = Kw/[H+] = 10–14/2.00 ´ 10–10 = 5.00 ´ 10–5 M

(3) [H+] = 7.71 ´ 10–11 M, what is the pH? What is [OH–]?
        See #2!

(4) What is [H+] and [OH–] of a solution at pH 5.3?
        [H+] = 10–pH = 10–5.3 = 5.0 ´ 10–6 M <<Don’t forget the unit!!>>
        [OH–] = Kw/[H+] = 10–14/5.0 ´ 10–6 = 2.0 ´ 10–9 M
Alternatively,
        [OH–] = 10–(14 – pH) = 10–(14 – 5.3) = 10–8.7 = 2.0 ´ 10–9 M

A basic logarithmatic calculation:

A = B ´ C Þ logA = logB + logC
Since Kw = [H+] ´ [OH–],
thus, logKw = log[H+] + log[OH–]
and – logKw = –log[H+] + (–log[OH–]).
We thus obtain 14 = pH + pOH
    or pOH = 14 – pH.

Therefore, [OH–] = 10–pOH = 10–(14 – pH) as in [H+]-pH conversion.

(5) What is [H+] (and [OH–]) at pH 12.2?
    See #4!

(6) What is the pH of a 2.3 ´ 10–4 M [OH–] solution?
    Find [H+] first!
    [H+] = 10–14/2.3 ´ 10–4 = 4.3 ´ 10–11 M
    pH = –log(4.3 ´ 10–11) = 10.4

(7) What is the pH of [OH–] = 4.0 ´ 10–11 M solution?
    See question #6

What are the concentrations of H+ and OH– in ppm at neutral water?
    Ans.: pH = 7 at neutral, i.e., [H+] = [OH–] = 10–7 M
    For H+: 10–7 M = 10–7 mol/L = 10–7 g/L = 10–4 mg/L = 10–4 ppm
        Remember that for dilute aqueous solutions, 1 L = 1 kg.  Thus, mg/L is equivalent to mg/106mg, i.e., ppm.
    For OH–: 10–7 M = 10–7 mol/L = 17 ´ 10–7 g/L = 17 ´ 10–4 mg/L = 1.7´ 10–3 ppm

What are the concentrations of H+ and OH– in ppm at pH 5.5?

Some more exercises:
(8) What is the pH of a 2.3 ´ 10–4 M Ca(OH)2 solution?
    Ca(OH)2 is a strong base, which gives 2OH– in water.
    [H+] = 10–14/(2 ´ 2.3 ´ 10–4) = 2.2 ´ 10–11 M
    pH = –log(2.2 ´ 10–11) = 10.7

(9) What is the pH of 4.0 ´ 10–11 M Ca(OH)2 solution?
    See question #8

Neutralization—to produce different "salts" and water
Examples
(1) HCl + NaOH ® NaCl + H2O
(2) H2SO4 + 2NaOH ® Na2SO4 + 2H2O
(3) HCl + NH3 ® NH4Cl
(4) HCl + Mg(OH)2 ® MgCl2 + 2H2O
(5) H3PO4 + 3NaOH ® Na3PO4 + 3H2O
(6) H2CO3 + 2NH3 ® (NH4)2CO3

Weak acids and bases: <100% dissociation in water
For example: acetic acid CH3COOH
    CH3COOH + H2O ® H3O+ + CH3COO–

When it reaches equilibrium, the ionization constant is
    Keq = Ka = [H3O+][CH3COO–]/[CH3COOH] = 1.8 ´ 10–5 M

For any weak acid AHn
    Ka = [A–][H3O+]/[AHn]
    pKa = –logKa, like in the case of pH

Similarly, for weak bases such as ammonia,
    NH3 + H2O ® NH4+ + OH–
    Keq = Kb = [NH4+][OH–]/[NH3]

What is the relationship of Ka and acidity, and Kb and basicity?
    HCl, Ka ~ 103 M; H3PO4, Ka = 7.5 ´ 10–3 M; H2CO3, 4.3 ´ 10–7 M

    Ammonia, Kb = 1.75 ´ 10–5 M; morphine, Kb = 1.41 ´ 10–10 M

Conjugate Acids and Bases
The acids/bases counterparts of bases/acids that are formed upon neutralization (upon dissolving in water).
        CH3COOH + H2O ® H3O+ + CH3COO–
        (Water serves as a base above.)

        CH3COO–+ H2O ® OH– + CH3COOH
        (Water serves as an acid above.)

        HCN + H2O ® H3O+ + CN–

        NH3 + H2O ® OH– + NH4+

        NH4+ + H2O ® NH3 + H3O+

        CH3COOH + NH3 ® CH3COO– + NH4+

Weak acids (bases) produce strong conjugate bases (acids).

In the equilibrium CH3COOH + H2O h H3O+ + CH3COO–
    Ka: "forward" equilibrium; Kb: "backward" equilibrium
    Ka ´ Kb = [H3O+][OH–] = Kw

Read Sections 9.6 and 9.7!!

Buffers: A compound which can react with both acids and bases can serve as a buffer when dissolved in water.

Buffer solutions: Solution that undergo very slight change upon addition of acids or bases, e.g., H2PO3–, HCO3– (the buffer system in the blood), etc.
    HCO3– + OH– ® H2O + CO32–
    HCO3– + H3O+ ® H2O + H2CO3

Upon hydration of CO2, H2CO3 is formed. However, CO2 is constantly removed which lowers H2CO3 concentration. If CO2 is removed too fast or too slow, the pH of blood thus becomes more basic (alkalosis) or more acidic (acidosis), respectively.

Henderson-Hasselbalch equation
    (For derivation of the equation, see P. 245.)

    pKa = pH – log ([proton acceptor]/[proton donor])
    The pH value can be calculated when pKa and the concentrations of the proton donor and acceptor are known.

Example
Calculate the pH of an aqueous solution consisting of 0.0080 M acetic acid and 0.0060 M sodium acetate. The Ka value of acetic acid is 1.74 ´ 10–5 M.
    pH = 4.76 + log (0.0060/0.0080) = 4.63

Titration
Quantitatively adding "one stuff" (like an acid solution) to "another stuff" (like a base solution), e.g., to determine the amount of an unknown acid or base using a standard base or acid in neutralization reactions

A reaction that is stoichiometrically complete is said to be at its equivalence point, and the experimentally determined point of completion is the end point which can be determined with indicators.

Acid-base indicators
H-indicator + OH– a indicator– + H2O
    The two forms of indicators have different colors.

In a neutralization reaction,
    number of OH– ions = number of H+ ions

Exercise: How many moles (1) sulfuric acid, (2) nitric acid, (3) hydrochloric acid, (4) phosphoric acid, (5) acetic acid, respectively, are required to neutralize 1 mol sodium hydroxide?

Equivalent: the mass of acid (base) donating (accepting) 1 mol H+
examples: 1 equivalent HNO3 (= MW), Ca(OH)2 (= MW/2), H2SO4 (= MW/2), H3PO4 (= MW/3), etc.

Normality (N)— equivalent of an acid or a base per liter
    (= n ´ M, n = # proton donated or accepted)

example:
    1) 1 M HNO3 = 1 N
    2) 1 M Ca(OH)2 = 2 N
    3) 1 M H2SO4 = 2 N
    4) 1 M H3PO4 = 3 N

equivalent of acid or base = volume (L) ´ normality (N)
During dilution, the number of equivalents of the acids and bases are not changed.
        i.e., Ni ´ Vi = Nf ´ Vf

In neutralization reactions, the equivalents of the acid and base are the same.
        i.e., Na ´ Va = Nb ´ Vb

Examples:
1) What is the final concentration of 200 mL 2.0 N HCl when it is added into 500 mL water?
    final volume = 700 mL
    2.0 ´ 200 = Nf ´ 700 Þ Nf = 0.57 N

2) How much 2.00 M sulfuric acid is needed to neutralize 50.0 mL 0.50 N sodium hydroxide?
    2.00 M H2SO4 = 4.00 N
    4.00 ´ Va = 0.50 ´ 50.0 Þ Va = 6.25 mL