Question

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=\) \(378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0 \mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

Short answer

The pH of the 30.0 mg/mL aqueous dose of PapH+Cl- prepared at 35.0°C is approximately 3.31.

Step-by-step solution

1. Calculate the concentration of PapH+Cl- in mol/L

First, convert the 30.0 mg/mL concentration to moles per liter (mol/L). The molar mass of PapH+Cl- is 378.85 g/mol. \[ \text{Concentration} = \frac{(30.0~\text{mg} / \text{mL}) \times (1~\text{g} / 1000~\text{mg}) \times (1~\text{mol} / 378.85~\text{g})}{1~\text{mL} / 0.001~\text{L}} \] \[ \text{Concentration} = 0.0793 ~\text{mol/L} \]

2. Calculate the Ka using the provided Kb

To find the Ka value of the conjugate acid PapH+Cl-, we can use the relationship between Kb, Ka, and Kw for a conjugate acid-base pair: \[ K_\text{a} = \frac{K_\text{w}}{K_\text{b}} \] Plug in the provided values for Kb and Kw: \[ K_\text{a} = \frac{2.1 \times 10^{-14}}{8.33 \times 10^{-9}} \] \[ K_\text{a} = 2.52 \times 10^{-6} \]

3. Create the ICE table and determine equilibrium concentrations

Set up the ICE table for the reaction, where PapH+Cl- dissociates into Pap and H+: \[ \text{PapH}^{+}\text{Cl}^{-} \rightleftharpoons \text{Pap} + \text{H}^{+} \] Initial concentrations: \[ \text{PapH}^{+}\text{Cl}^{-}: 0.0793~\text{M} \] \[ \text{Pap}: 0 \] \[ \text{H}^{+}: 0 \] Changes in concentration: \[ \text{PapH}^{+}\text{Cl}^{-} \rightarrow -x \] \[ \text{Pap} \rightarrow +x \] \[ \text{H}^{+} \rightarrow +x \] Equilibrium concentrations: \[ \text{PapH}^{+}\text{Cl}^{-} \rightarrow 0.0793-x \] \[ \text{Pap} \rightarrow x \] \[ \text{H}^{+} \rightarrow x \]

4. Use the Ka expression to solve for x

Now, apply the Ka expression using the equilibrium concentrations: \[ K_\text{a} = \frac{[\text{Pap}] \times [\text{H}^{+}]}{[\text{PapH}^{+}\text{Cl}^{-}]} = \frac{x \times x}{0.0793 - x} \] Plugging the Ka value we found: \[ 2.52 \times 10^{-6} = \frac{x^2}{0.0793 - x} \] Since Ka is small, we can assume x is small compared to the initial concentration, so 0.0793 - x ≈ 0.0793. Thus, we can solve for x: \[ 2.52 \times 10^{-6} \approx \frac{x^2}{0.0793} \] \[ x \approx \sqrt{2.52 \times 10^{-6} \times 0.0793} \approx 4.88 \times 10^{-4} \]

5. Calculate the pH from the H+ concentration

Remember, H+ concentration equals x. Use this to find the pH of the solution: \[ \text{pH} = -\log ([\text{H}^{+}]) \] \[ \text{pH} = -\log (4.88 \times 10^{-4}) \approx 3.31 \] The pH of the 30.0 mg/mL aqueous dose of PapH+Cl- prepared at 35.0°C is approximately 3.31.

Key concepts

Acid-Base Equilibrium

Acid-base equilibrium refers to the balance that exists when an acid and its conjugate base coexist in a solution. When an acid donates a proton (H\(^+\)), it forms its conjugate base. Similarly, when a base accepts a proton, it turns into its conjugate acid. This dynamic interplay is crucial in understanding how solutions stabilize their pH.

In the context of our exercise, Papaverine hydrochloride (PapH\(^+\)Cl\(^{-}\)) acts as a conjugate acid. It can dissociate into Pap, the base, and H\(^+\) ions, adjusting the pH of the solution. The efficient conversion between these forms ensures that the solution's pH remains stable until an equilibrium is reached.

• A solution reaches equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
• At equilibrium, the concentrations of acids and bases do not change, although their reactions continue to occur.

Vasodilators

Vasodilators are a class of drugs that help expand blood vessels, leading to increased blood flow. This is critical in treating conditions like hypertension, where improved blood flow can alleviate symptoms and lower heart strain. Papaverine hydrochloride is an example of such a medication.

By acting on the smooth muscles of blood vessels, vasodilators reduce resistance in the vascular system. This not only helps enhance circulation but can also ensure that organs receive enough oxygen and nutrients. The use of vasodilators must be precisely calibrated to avoid lowering blood pressure too much.

• Vasodilators play a key role in managing cardiovascular diseases.
• Their mechanism often involves influencing molecular pathways to relax blood vessels.

Understanding the chemistry behind drugs like Papaverine hydrochloride, including their acid-base properties, is fundamental for determining their effectiveness and safety in medical applications.

Conjugate Acid-Base Pair

A conjugate acid-base pair consists of two species that transform into each other by gain or loss of a proton. In the exercise, Papaverine hydrochloride (PapH\(^+\)Cl\(^{-}\)) and Pap serve as the conjugate acid-base pair.

This pair illustrates the basic concept of proton exchange that underpins pH calculations:

• When PapH\(^+\) loses a proton, it becomes Pap, a weak base.
• Conversely, Pap can gain a proton to regenerate PapH\(^+\).

Conjugate pairs are crucial in buffering solutions, allowing them to resist significant pH changes. The relationship between a conjugate acid and base is depicted by their dissociation constant (Ka for acids and Kb for bases), indicating how readily the pair donates or accepts protons.

Equilibrium Expression

The equilibrium expression is a mathematical representation of the concentrations of reactants and products at equilibrium. In the context of the exercise, we use this to describe the dissociation of PapH\(^+\)Cl\(^{-}\) into Pap and H\(^+\).

For any given reaction, the equilibrium constant \( K_a \) is derived from:\[ K_a = \frac{[ ext{products}]}{[ ext{reactants}]} \]
In our example,
\[ K_a = \frac{[ ext{Pap}][ ext{H}^+]}{[ ext{PapH}^+]} \]
This equation illustrates how shifting the concentration of one component affects the others, striving to maintain equilibrium.

• If too much H\(^+\) is produced, the solution becomes more acidic, decreasing the concentration of PapH\(^+\).
• If PapH\(^+\) doesn’t fully dissociate, fewer H\(^+\) ions are available, slowing pH changes.

Achieving a balance in these reactions allows for accurate pH determination and offers insights into how chemical changes affect solution properties.