Morphine, C 17H 19NO 3, is often used to control severe post-operative pain. What is the pH of the solution made by dissolving 25.0 mg of morphine in 100. mL of water? (For morphine, K b = 1.62 × 10 –6.)

pH = 9.58

Explanation:

First of all, we need to determine the molarity of the solution.

We determine the molar mass of morphine:

12g/m . 17 + 1 g/m . 19 + 14 g/m + 16 g/m . 3 = 285.34 g/m

molar mass g/m, is the same as mg/mm

25 mg . 1 mmol / 285.34 mg = 0.0876 mmoles / 100 mL = 8.76×10⁻⁴ M

In diltuted solution, we must consider water.

Mass balance for morphine = [Morphine] + [Protonated Morphine]

8.76×10⁻⁴ M = [Morphine] + [Protonated Morphine]

As Kb is too small, I can skipped, the [Protonated Morphine]

8.76×10⁻⁴ M = [Morphine]

In the charge balance I will have:

[OH⁻] = [H⁺ morphine] + [H⁺]

Let's go to the Kb expression

Morphine + H₂O  ⇄  MorphineH⁺  +  OH⁻        Kb

Kb = [MorphineH⁺]  [OH⁻] / [Morphine]

Kb = [MorphineH⁺]  [OH⁻] / 8.76×10⁻⁴ M

So now, we need to clear [MorphineH⁺] to replace it in the charge balance

Kb  . 8.76×10⁻⁴ M / [OH⁻] = [MorphineH⁺]

Now, the only unknown value is the [OH⁻]

[OH⁻] = Kb .  8.76×10⁻⁴ M / [OH⁻]  + Kw/[OH⁻]

Remember that Kw = [H⁺] . [OH⁻]

[H⁺] = Kw/[OH⁻]

[OH⁻]² = 1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴

[OH⁻] = √(1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴)

[OH⁻] = 3.76×10⁻⁵  →  - log [OH⁻] = pOH = 4.42

pH = 14 - pOH  →  14 - 4.42 = 9.58