Aqueous Equilibria. II. Acids and Bases.

Definitions:

1. Arrhenius:

An acid produces H^{+} and a base produces OH^{-}

Acid: HCl H^{+} + Cl^{-}

Base: NaOH Na^{+} + OH^{-}

2. Lowry and Brønsted:

An acid is a proton donor, a base is a proton acceptor.

HCl + H_{2}O H_{3}O^{+} + Cl^{-}

acid base

3. Lewis:

An acid is an electron pair acceptor, a base is an electron pair donor in the formation of a covalent bond between the two.

Note that Brønsted and Lewis bases are often the same.

The Brønsted acid always involves the proton accepting the electron pair. This is only one possibility for a Lewis acid.

Conjugate Acid and Bases.

In a conjugate pair, the acid has the proton, the base does not.

Some species have both a conjugate acid and a conjugate base:

conjugate acid | conjugate base | |

H_{3}O^{+} |
H_{2}O |
OH^{-} |

NH_{4}^{+} |
NH_{3} |
NH_{2}^{-} |

H_{2}SO_{4} |
HSO_{4}^{-} |
SO_{4}^{2-} |

H_{2}CO_{3} |
HCO_{3}^{-} |
CO_{3}^{2-} |

Exercises:

1 Using the Brønsted approach, pick out the acid-base conjugate pairs:

HBr + H_{2}O H_{3}O^{+} + Br^{-}

CH_{3}COOH + H_{2}O H_{3}O^{+} + CH_{3}COO^{-}

NH_{4}^{+} + H_{2}O H_{3}O^{+} + NH_{3}

S^{2-} + H_{2}O HS^{-} + OH^{-}

H_{2}O + H_{2}O H_{3}O^{+} + OH^{-}

2. Give the conjugate acid **and** conjugate base for the following species:

conjugate acid | conjugate base | |

HF | ||

H_{2}O |
||

HS^{-} |
||

H_{2}PO_{4}^{-} |
||

HPO_{3}^{2-} |

The Equilibrium Constants, K

K_{a} is the equilibrium constant for an acid reacting with water as the base:

HA + H_{2}O H_{3}O^{+} + A^{-}

K_{a} = {a(H_{3}O^{+}) * a(A^{-}) } / {a(HA) * a(H_{2}O)}

note that a(H_{2}O) = 1 in a dilute solution.

K_{b} is the equilibrium constant for a base reacting with water as the acid:

B + H_{2}O BH^{+} + OH^{-}

K_{b} = {a(BH^{+}) * a(OH^{-}) } / {a(B) * a(H_{2}O)}

and again, a(H_{2}O) = 1.

Strong and Weak Electrolytes.

An **electrolyte **is a substance whose water solution conducts electricity
due to the formation of ions.

A **strong electrolyte** produces a strong current due to high dissociation
producing many ions.

A **weak electrolyte** produces a weak current due to low dissociation
producing few ions.

Acids and bases which are highly dissociated and so produce many ions
on dissolving in water are strong electrolytes and are termed **strong acids
or bases**.

Acids and bases which are weakly dissociated produce only a few ions are
weak electrolytes and are termed **weak acids or bases.**

In terms of the equilibrium constants, K_{a}, and K_{b}, an arbitrary cut off at 1
x 10^{-2} is often used to separate strong from weak.

Important:

Do not confuse strong and weak with dilute and concentrated.

pH.

The importance of the a(H_{3}O^{+}) in living systems makes a measure of this
necessary. Because the dilute solutions involved have an a(H_{3}O^{+}) which
varies from 1.0 to 1.0 x 10^{-14}, the notion of pH was introduced.

pH = - log {a(H_{3}O^{+})}

Note that higher values of pH indicate lower values of a(H_{3}O^{+})

The use of "p" as an operature meaning "take - log of" is extended to:

pOH = - log {a(OH^{-})}

pK_{a} = - log K_{a}

pK_{b} = - log K_{b}

pK_{sp} = - log K_{sp}

Average pH values for some well known Solutions:

pH | Solution |

0 | 1 M HCl |

1 to 3 | gastric juice |

ca. 2 | limes |

2.3 | lemons |

2 to 4 | soft drinks |

2.6 to 4.4 | orange juice |

ca 3 | vinegar, wine |

ca. 3.5 | dill pickles |

4.5 to 8 | urine |

ca. 6.1 | sour milk |

ca. 6.6 | milk |

6.5 to 7.5 | saliva |

7.3 to 7.5 | blood serum |

ca. 8 | egg white |

ca. 8 | sea water |

ca. 10 | detergent solution |

ca. 10 | milk of magnesia |

ca. 11 | bar soap |

11.5 to 12 | household ammonia |

12 | washing soda |

14 | 1 M NaOH |

K

Water is one of a few substances that can react with itself as both acid and base:

2 H_{2}O H_{3}O^{+} + OH^{-}

Definition:

K_{w} = { a(H_{3}O^{+}) * a(OH^{-}) } / a(H_{2}O)^{2}

and again a(H_{2}O) = 1.

At 25ºC, K_{w} = 1 x 10^{-14}. (Experimental measurement)

So, in pure water at 25ºC:

1 x 10^{-14} = a(H_{3}O^{+}) * a(OH^{-})

and:

a(H_{3}O^{+}) = a(OH^{-}) = 1 x 10^{-7}.

Taking logs of the equation for K_{w}:

- log K_{w} = - log{a(H_{3}O^{+})} - log {a(OH^{-})}

pK_{w} = pH + pOH

14 = pH + pOH (at 25ºC)

In pure water at 25ºC, pH = pOH = 7.

Because of the fixed relationship between pH and pOH there is only need for one scale, the pH scale.

Limits of the pH scale.

Note that the relationship:

pH = - log {a(H_{3}O^{+})}

can be measured experimentally only for ideal (read dilute) solutions
when a(H_{3}O^{+}) = {[H_{3}O^{+}] / 1 mol/L}.

When either the [H_{3}O^{+}] or the [OH^{-}] becomes greater than 1 M the
activities can no longer be measured and so neither can pH or pOH.

An acid in water donates a proton to the water:

HA + H_{2}O H_{3}O^{+} + A^{-}

and the [H_{3}O^{+}] increases. Consequently the pH goes down from 7 (the
pH of pure water.)

Calculation of the [H_{3}O^{+}] (acid solutions):

1. **For a strong acid, assume 100% reaction to H _{3}O^{+}.**

Example: calculate the pH of a 0.0100 M solution of HBr, a strong acid.

Assume that all HBr reacts to gibe H_{3}O^{+}, the [H_{3}O^{+}] is equal the initial
[HBr] = 0.0100 M.

So pH = - log a(H_{3}O^{+}) = - log 0.0100 = 2.0

In this problem the [H_{3}O^{+}] from the water itself is ignored.

If the **acid concentration is very low**, this may not be possible:

Example: calculate the pH of a 1 x 10^{-8} M solution of HBr.

If the [H_{3}O^{+}] from water is ignored, this will give a pH of 8, which is
basic!

2 H_{2}O |
H_{3}O^{+} |
+ OH^{-} | ||

I | 1 x 10-8 | 0 | ||

C | x | x | ||

E | 1 x 10-8 + x | x |

^{}K_{w} = 1 x 10^{-14} = (1 x 10^{-8} + x)x

x = 9.5 x 10^{-8}

[H_{3}O^{+}] = 1.05 x 10^{-7}

pH = 6.98

- for a weak acid, use K
_{a}to find [H_{3}O^{+}].

**Example: calculate the pH of 0.0100 M hyperchlorous acid, HOCl, pK _{a} =
7.2**

**Step A: Analysis**

HOCl | + H_{2}O |
H_{3}O^{+} |
+ ClO^{-} | ||

I (mol/L) | 0.0100 | 0 | 0 | ||

C | -x | x | x | ||

E | 0.0100 - x | x | x |

K_{a} = {a(H_{3}O^{+}) * a(ClO^{-})} / a(HOCl)

6.3 x 10^{-8} = x * x /(0.0100-x)

unknown: pH, calculated from x.

**Step B: Brainstorm (trivial)**

**Step C: Calculate**

If it is suspected that x is very small compared with the value to which it is added or subtracted (K is very small), the an approximate solution can be quickly obtained by ignoring the x in such sums or differences.

approx:

6.3 x 10^{-8} = x * x /(0.0100)

so an approximate value for x is found to be 2.53 x 10^{-5}.

**Validate this approximation by checking that x < 5% of the value to
which it is added or subtracted:**

{x / 0.0100}*100% = 2.53 x 10^{-5}*100% / 0.01

= 0.25%

so the approximation is valid.

x = 2.53 x 10-5

pH = 4.6

What to do if the Approximation is not Valid.

Example:

Calculate the pH of a 0.0100 M solution of HF. pK_{a} = 3.17

Step A: Analysis.

HF | + H_{2}O |
H_{3}O^{+} |
+ F^{-} | ||

I | 0.0100 | 0 | 0 | ||

C | (-1)x | (+1)x | (+1)x | ||

E | 0.0100 - x | > | x |

K_{a} = a(H_{3}O^{+}) * a(F^{-}) / a(HF) = 6.76 x 10^{-4}

= x * x / (0.0100 - x) = 6.76 x 10^{-4}

Use the approximation x << 0.01

6.76 x 10^{-4} = x * x / 0.0100

from which x_{1} = 0.00260.

Test the approximation: 0.0026 * 100% / 0.01 = 26%.

The approximation is not valid.

Use the method of successive approximations:

Substitute into the quadratic expression the approximate value obtained for x, and resolve to obtain a second approximation.

Repeat this to get a third approximation, and then continue until **two
successive** approximations are the same.

Second approximation:

6.76 x 10^{-4} = x * x / (0.0100 - 0.00260)

from which x_{2} = 0.00224

Third approximation (you must do this one!)

6.76 x 10^{-4} = x * x / (0.0100 - 0.00224)

from which x_{3} = 0.00229

Check successive approximations from now on:

x_{2} and x_{3} not same, so find x_{4}:

Fourth approximation:

6.76 x 10^{-4} = x * x / (0.0100 - 0.00229)

from which x_{4} = 0.00228.

x_{4} is (to < ½%) the same as x_{3}.

Thus we take x = 0.00228 as the correct result.

From which: [H_{3}O^{+}] = 0.00228 M = a(H_{3}O^{+})

and pH = 2.64.

Note: you could also solve the quadratic equation.

Answer:

**Strong acids are:**

**HCl, HBr, HI, HNO _{3}, H_{2}SO_{4}, HClO_{4}**

**Memorize this list.**

Weak acids are:

every other acid.

If the acid is not on the above list it is weak.

Note about calculations in general.

When a relationship can be developed between several variables, knowledge of all but one can allow the calculation of that one.

In the case of acid/base equilibria:

HA | + H_{2}O |
H_{3}O^{+} |
+ A^{-} | ||

I (mol/L) | a | 0 | 0 | ||

C | (-1)x | x | x | ||

E | (a - x) | x | x |

K_{a} = x * x /(a - x)

This is a relationship in three variables, given any two the other can be calculated.

Problems can then:

give K_{a} and starting acid concentration, calculate pH (x)

give K_{a} and pH, calculate starting acid concentration.

give pH and starting acid concentration, calculate K_{a}.

This is a reminder and, of course, it applies to any relationship we have met or will meet in this course.

1. Strong bases.

This term base here is a hang over from the Arrhenius definition of a base
(a substance that produces OH^{-}.) A Brønsted base is a proton acceptor,
strong bases in fact just dissociate in water:

NaOH(s) Na^{+} + OH^{-}

We will assume that strong bases are 100% dissociated, so that the [OH^{-}]
from them is equal to the original [MOH].

Calculate the pH of 0.01 M Mg(OH)_{2}.

Answer: [OH^{-}] = 0.02,

pOH = 1.7

pH = 12.3 (pH + pOH = 14)

**CARE: The calculated concentration is the [OH ^{-}], and - log of this is
the pOH.**

2. Very dilute solutions of strong bases.

Use same technique as for very dilute acid solutions.

3.

Weak bases do react with water to give OH^{-}:

B + H_{2}O BH_{}^{+} + OH^{-}

To find the [OH^{-}] and hence the pH, requires the use of an equilibrium
calculation.

Example: calculate the pH of a 0.0100 M solution of ammonia.

**Step A: Analysis.**

NH_{3} |
+ H_{2}O |
NH_{4}^{+} |
+ OH^{-} | ||

I (mol/L) | 0.0100 | 0 | 0 | ||

C | (-1)x | x | x | ||

E | 0.0100 - x | x | x |

Kb = {a(NH_{4}^{+}) * a(OH^{-})} / a(NH_{3})

1.8 x 10^{-5} = x * x / (0.0100 - x)

Unknown, pH, obtainable from x.

**Step B: Brainstorm (trivial)**

**Step C: Calculate**

Try the approximation

1.8 x 10^{-5} = x * x / 0.0100

x = 4.24 x 10^{-4}

Check the approximation: 4.24 x 10^{-4} * 100% / 0.01 = 4.24%

the approximation is valid.

x = [OH-] = 4.24 x 10^{-4} M

pOH = 3.4 (don't forget this!!!)

pH = 10.6