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Chapter 14: Problem 146

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

Expert verified
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

01

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. \[ \text{Concentration} = \frac{(30.0~\text{mg} / \text{mL}) \times (1~\text{g} / 1000~\text{mg}) \times (1~\text{mol} / 378.85~\text{g})}{1~\text{mL} / 0.001~\text{L}} \] \[ \text{Concentration} = 0.0793 ~\text{mol/L} \]
02

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: \[ K_\text{a} = \frac{2.1 \times 10^{-14}}{8.33 \times 10^{-9}} \] \[ K_\text{a} = 2.52 \times 10^{-6} \]
03

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: \[ \text{PapH}^{+}\text{Cl}^{-} \rightarrow 0.0793-x \] \[ \text{Pap} \rightarrow x \] \[ \text{H}^{+} \rightarrow x \]
04

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: \[ 2.52 \times 10^{-6} \approx \frac{x^2}{0.0793} \] \[ x \approx \sqrt{2.52 \times 10^{-6} \times 0.0793} \approx 4.88 \times 10^{-4} \]
05

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.

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Most popular questions from this chapter

A solution is made by adding \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to \(50.0 \mathrm{~mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the \(\mathrm{pH}\) of the solution. b. Calculate the acetate ion concentration.

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