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Chapter 14: Problem 122

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 aqueous solution containing \(30.0 \, \mathrm{mg / mL}\) of papaverine hydrochloride at \(35.0 ^\circ \mathrm{C}\) is approximately 1.82.

Step by step solution

01

Find the molar concentration of papH+ Cl-

First, we need to convert the concentration from mg/mL to mol/L. Using the given concentration and molar mass: \[\frac{30.0 \, \mathrm{mg}}{\mathrm{mL}} \times \frac{1 \, \mathrm{mol}}{378.85 \, \mathrm{g}} \times \frac{1 \, \mathrm{g}}{1000 \, \mathrm{mg}} = 7.93 \times 10^{-5} \, \mathrm{M}\] So, the molar concentration of papH+ Cl- is \(7.93 \times 10^{-5} \mathrm{M}\).
02

Calculate the concentration of the weak base (pap) and conjugate acid (papH+).

Since papH+ Cl- is a conjugate acid of the weak base pap, their concentrations in the solution will be equal. Thus, the initial concentration of both pap and papH+ Cl- is \(7.93 \times 10^{-5} \mathrm{M}\). We will assume that the concentration of Cl- does not affect the pH of the solution as Cl- is the conjugate base of a strong acid (HCl) and will not hydrolyze in the solution.
03

Write the equilibrium expression for the reaction between the weak base (pap) and water.

When the weak base pap reacts with water, it forms papH+ and OH-. The equilibrium expression is: \[K_{\mathrm{b}} = \frac{[\mathrm{papH}^+][\mathrm{OH}^-]}{[\mathrm{pap}]}\]
04

Calculate the equilibrium concentration of OH-.

Let x be the change in the concentration of pap, papH+, and OH-. At equilibrium, we have: \[[\mathrm{papH}^+] = 7.93 \times 10^{-5} - x\] \[[\mathrm{OH}^-] = x\] \[[\mathrm{pap}] = 7.93 \times 10^{-5} + x\] Plugging these values into the Kb expression, we get: \[8.33 \times 10^{-9} = \frac{x(7.93 \times 10^{-5} - x)}{7.93 \times 10^{-5} + x}\] Since the value of Kb is very small, we can assume that x is negligible compared to \(7.93 \times 10^{-5}\). This simplifies the equation: \[x = 8.33 \times 10^{-9} \times 7.93 \times 10^{-5}\] Solving for x, we get: \[x = 6.60 \times 10^{-13}\] Thus, the equilibrium concentration of OH- is \(6.60 \times 10^{-13} \mathrm{M}\).
05

Find the pOH.

To find the pOH, we take the negative logarithm of the hydroxide ion concentration: \[pOH = -\log(6.60 \times 10^{-13}) = 12.18\]
06

Find the pH.

Finally, subtract the pOH from 14 to find the pH: \[pH = 14 - 12.18 = 1.82\] So, the pH of the aqueous solution containing \(30.0 \, \mathrm{mg / mL}\) of papaverine hydrochloride at \(35.0 ^\circ \mathrm{C}\) is approximately 1.82.

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