Question

In the laboratory, a general chemistry student measured the pH of a 0.530 M aqueous solution of codeine, C18H21O3N to be 10.820. Use the information she obtained to determine the Kb for this base.

Answer 1

Comment: note that I will use B to symbolize the weak base. No one cares what the specific base is because the technique to be explained works for all weak bases. What happens is that some teachers will use the name of a specific weak base while others go the generic route.

1) Write the ionizaton equation for the base. Remember, we will use B to symbolize the base.

B + H₂O ⇌ HB⁺ + OH¯

2) Write the equilibrium expression:

K_b = ( [HB⁺] [OH¯] ) / [B]

3) Our task now is to determine the three concentrations on the right-hand side of the equilibrium expression since the K_b is our unknown.

We will use the pH to calculate the [OH¯].

We know pH = -log [H⁺], therefore [H⁺] = 10¯^pH

[H⁺] = 10¯^10.820 = 1.513 x 10¯¹¹ M

I've kept a couple guard digits; I'll round off the final answer to the proper number of significant figures.

Knowing the [H⁺] allows us to get the [OH¯]. To do this, we use K_w = [H⁺] [OH¯], so we have this:

1.00 x 10¯¹⁴ = (1.513 x 10¯¹¹) (x)

x = [OH¯] = 6.609 x 10¯⁴ M

From the ionization equation, we know there is a 1:1 molar ratio between [HB⁺] and [OH¯]. Therefore:

[HB⁺] = 6.609 x 10¯⁴ M

The final value, [B] is given in the problem. In the example being discussed, 0.530 M is the value we want. Some teachers will use 0.530 M, while others would say to first subtract the 6.609 x 10¯⁴ M value from 0.530 M. Let's do both.

c1) K_b = [(6.609 x 10¯⁴ ) (6.609 x 10¯⁴ )] / 0.530

K_b = 8.24 x 10¯⁷

c2) K_b =[(6.609 x 10¯⁴ ) (6.609 x 10¯⁴ )] / (0.530-6.609 x 10¯⁴)

K_b = 8.25 x 10¯⁷

In reality, it makes very little difference if we use the unmodified concentration of the acid (the 0.530 value) or if we do the subtration. In the above example, both values round off to 8.2 x 10¯⁷. In the end, you do what your teacher recommends. So, ask your teacher if you're not sure.