SCH 4U - Weak Bases

Weak Bases

A weak base dissociates only partially in solution.  The most common weak base is ammonia:

NH₃(aq)   +  H₂O(l)  ↔  NH₄⁺(aq)  +  OH^-(aq)         

K_b = [NH₄⁺][ OH^-]

        -----------------

             [NH₃]

Many bases are the salts of strong acids (Na₃PO₄ ↝ PO₄^3-, CaSO₄ ↝ SO₄^2-, KCN ↝ CN^-):

Relationship Between K_a & K_b

Recall that the stronger the acid, the weaker the conjugate base.  Consider the NH₃/NH₄⁺ conjugate acid-base pair:

NH₃(aq)   +  H₂O(l)  ↔  NH₄⁺(aq)  +  OH^-(aq)    K_b = [NH₄⁺][ OH^-]

                                                                                     ----------------

                                                                                         [NH₃]

                NH₄⁺(aq)  ↔  NH₃(aq)   +  H⁺(aq)      K_a = [NH₃][H⁺]

                                                                                    --------------

                                                                                        [NH₄⁺]

__________________________________________________________

                    H₂O(l)  ↔  H⁺(aq)  +  OH^-(aq)       K_a x K_b = [H⁺][OH^-] = K_W

                                                                                       

Using K_a x K_b =  K_W we can calculate K_b if we know the K_a of the conjugate acid.

Finding K_b, pK_b or pH for Weak Bases

Example: What is the pH of a 0.25 M solution of sodium phosphate?

Answer:

• Create a balanced equilibrium equation using the salt’s anion.  (For right now, trust me that it is only the anion portion that we use.  In the next lesson, I will show you why.)  Notice that the anion acts as a base, since it accepts a proton from water.
• Create an ICE table, as usual and fill in the table with the data given and the variable x, as appropriate.  Notice that liquid water is crossed out.
• Calculate the K_b value for PO₄^3-, using the K_a value of the conjugate acid, HPO₄^2-(obtained from the Ka table).
• Set up the K_b expression and sub in the eq [ ]s.
• Do a bunch of algebra and ultimately find the roots.  Determine which one is correct.  This root is equivalent to the eq[OH^-].
• Determine pOH.
• Determine pH. 

     PO₄^3-(aq)  +  H₂O(l)  ↔  HPO₄^2-(aq)  +  OH^-(aq)   

I        0.25               --                 0                    0
C        - x                --               + x                 + x
E     0.25 – x           --                 x                    x

K_a(HPO₄^2-) = 4.2 x 10^-13

K_a x K_b = 1.00 x 10^-14 
K_b = 1.00 x 10^-14/4.2 x 10^-13 
K_b = 2.4 x 10^-2

                                                                                                                          

K_b = [HPO₄^2-][OH^-]/[PO₄^3-]
2.4 x 10^-2 = x²/(0.25 –x) 
x = 6.4 x 10^-2 or –8.6 x 10^-2        

pOH = -log[OH^-]
        = -log(6.4 x 10^-2)
        = 1.19

pH = 14.00 - pOH 
pH = 14.00 - 1.19
pH = 12.81

Homework # 21, 22, 26