Chapter 18

Acid-Base Equilibria

Homework:

Auto Ionization of Water

 

H₂O + H₂O ---> H₃O⁺ + OH^-

 

Note: For purposes of brevity the hydronium ion, H₃O⁺ , will be represented by H⁺.

In neutral water there are equal amounts of H⁺  and OH^-.  The quantities of H+ and  OH- in the solution are equal at a concentration of 1x10^-7 M each.  The equilibrium constant for this is:

 

      K_eq = [H⁺][OH^-]/[H₂O]  

 

where the concentration of water is 55.5 M and is essentially a constant.  You should verify that the concentration of water in water is 55.5M.  If we include the [H₂O] constant into the K_eq, the constant becomes K_w or the autoionization constant for water and has the value of 1x10^-14,  that is:    K_w=[H⁺][OH^-]

 

In solutions of acids and bases the equilibrium of the autoionization is shifted to the left when either an acid or base are added to water.

 

Addition of a base increases the concentration of hydroxide and decreases the concentration of hydronium.  Addition of an acid increases the concentration of hydronium and decreases the concentration of the hydroxide ion.  A system used to measure the concentration of hydronium ion is called pH.

 

     pH = - log[H⁺]

or

     [H⁺] = 10^-pH

From the point of view of the hydroxide:

 

     pOH = - log[OH^-]

 

or

 

     [OH^-] = 10^-pOH

 

 [H⁺]           pH               [OH^-]            pOH           K_w    

1x10⁺¹          -1                1x10^-15         15                1x10^-14

1x10^-0          0                 1x10^-14            14                1x10^-14

1x10^-1          1                 1x10^-13         13                1x10^-14

1x10^-2          2                 1x10^-12         12                1x10^-14

1x10^-3          3                 1x10^-11         11                1x10^-14

1x10^-4          4                 1x10^-10         10                1x10^-14

1x10^-5            5                 1x10^-9              9                 1x10^-14

1x10^-6          6                 1x10^-8              8                 1x10^-14

1x10^-7          7                 1x10^-7            7                 1x10^-14

1x10^-8          8                 1x10^-6               6                 1x10^-14

1x10^-9          9                 1x10^-5            5                 1x10^-14

1x10^-10          10                1x10^-4              4                 1x10^-14

1x10^-11          11                1x10^-3            3                 1x10^-14

1x10^-12        12                1x10^-2            2                 1x10^-14

1x10^-13        13                1x10^-1            1                 1x10^-14

1x10^-14        14                1x10^-0            0                 1x10^-14

1x10^-15        15                1x10⁺¹          -1                 1x10^-14

From this table it is clear that 

 

 pH + pOH = 14    and    K_w = [H⁺][OH^-] = 1x10^-14.

 

Solutions of Strong Acids and Bases

 

Strong acids and bases are 100% ionized.  The concentration of H⁺ is always equal to the concentration of strong acid present in a solution and the concentration of OH^- is always equal to the concentration of strong base present in solution.

 

Problem:

Calculate the pH and pOH for a 2M solution of HCl.

 

Problem:

Calculate the pH, and pOH for a 6M solution of NaOH.

 

Solutions of Weak Acids and Bases

 

Weak acids and bases are only partially dissociated and give comparatively low concentrations of H⁺  and OH^- respectively.

 

When weak acid dissociates:

 

      HX ↔ H⁺ + X^-

 

we can write an equilibrium constant for the dissociation.

 

      K_a = [H⁺][X^-]/[HX]

 

These equilibrium constants can be found in the book or in references such as the Handbook of Chemistry and Physics, CRC Press. 

 

Problem:

Calculate the pH of a 1M solution of HOAc.  K_a = 1.8x10^-5

 

Calculate the pH of a 1M solution of NH₃  K_b=1.8x10^-5

 

Polyprotic Acids

 

A polyprotic acid is an acid with more than one acidic hydrogen; H₂SO₄, H₃PO₄ and H₂CO₃ are three common examples.

 

The dissociation of these acids takes place in a stepwise fashion and each step has its own K value; K_a1, K_a2, and K_a3 for the first three sequential dissociations.

 

Consider the dissociation of phosphoric acid:

 

      H₃PO₄    ---> H⁺ + H₂PO₄^-   K_a1 = 7.2x10^-3

 

      H₂PO₄^- ---> H⁺ + HPO₄^-2     K_a2 = 6.3x10^-8

 

      HPO₄^-2   ---> H⁺ + PO₄^-3      K_a3  = 4.2x10^-13

 

We can see that the successive dissociations of the acid have successively smaller K values (by about 10^-5 for each successive dissociation).

Problem:

Calculate the concentration of PO₄^-3 in a 1 M solution of phosphoric acid.

 

Is there an easier way?

 

Simply add the reactions and you get

 

  K_a1,2,3 = K_a1K_a2K_a3 = [H⁺]³[PO₄^-3]/[H₃PO₄]

 

IONS OF WEAK ACIDS AND BASES

 

From the Brønstead Lowrey theory of acids and bases, we know that an acid gives a conjugate base and a base gives a conjugate acid. Salts of weak acids and bases have their conjugate effect on solutions made from them.

For example: Sodium acetate contains the acetate ion, which is the conjugate base of acetic acid.  We therefore expect acetate to have less acidic character than acetic acid (which is another way of saying more basic character).  The acetate ion reacts with water to form acetic acid and hydroxide.

 

      OAc^-  +  H₂O ---> HOAc  +  OH^-

 

The equilibrium expression for this is :

     Kb = [OH^-][HOAc]/[OAc^-]

 

The question is how can we find the K_b for this reaction,  it is not published data.

 

If we examine the K_w for water and the K_a for the acid, it is easy to see how we could derive the numeric quantity for the K_b of the ion of a weak base. 

 

Problem:

Calculate the pH of a solution made by adding 1.00 moles of NaOAc to 500 ml of water. 

 

The same goes for the ions of weak bases:

 

Problem:

Calculate the pH of a solution made by adding 2 M of NH₄Cl to 500 ml of water.

 

Problem:

Calculate the product of the K_axK_b for any acid-conjugate base system.

 

Review the Brønstead-Lowrey theory and Lewis Acid-Base Theory.