Question 14.13: CALCULATING THE pH AND THE EQUILIBRIUM CONCENTRATIONS IN A S...

Step 1. Let’s use Cod as an abbreviation for codeine and CodH^{+} for its conjugate acid.

The species present initially are Cod (base) and H_{2}O (acid or base).

Step 2. There are two possible proton-transfer reactions:

Step 3. Since Cod is a much stronger base than H_{2}O (K_{b}  >>  K_{w}), the principal reaction involves the protonation of codeine.

Step 4.

Step 5. The value of x is obtained from the equilibrium equation:

Assuming that (0.0012 – x) ≈ 0.0012,

x² = (1.6 × 10^{-6})(0.0012)

x = 4.4 × 10^{-5}         Approximation   (0.0012 – x) ≈ 0.0012 is justified.

Step 6. The big concentrations are

[CodH^{+}]  =  [OH^{-}]  =  x  =  4.4  ×  10^{-5} M

[Cod] = 0.0012 – x = 0.0012 – 0.000 044 = 0.0012 M

Step 7. The small concentration is obtained from the subsidiary equilibrium, the dissociation of water:

[H_{3}O^{+}]  =  \frac{K_{w}}{[OH^{-}]}  =  \frac{1.0  ×  10^{-14}}{4.4  ×  10^{-5}}  =  2.3  ×  10^{-10} M

Step 8.  pH = -log [H_{3}O^{+}]  =  -log (2.3  ×  10^{-10}) = 9.64

The pH is greater than 7, as expected for a solution of a weak base.

BALLPARK CHECK
Because is negligible compared with the initial concentration of codeine (Step 5), the [OH^{-}] equals the square root of the product of K_{b} and the initial concentration of codeine. [OH^{-}] Therefore, equals the square root of approximately (16 × 10^{-7}) × (1 × 10^{-3}), or about 4 × 10^{-5}. Since the [OH^{-}] is between 10^{-5} M and 10^{-4} M, the [H_{3}O^{+}] is between 10^{-9} M and 10^{-10} M, and so the pH is between 9 and 10. The ballpark check and the solution agree.