Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: 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 of papaverine hydrochloride prepared at \(35.0^{\circ} \mathrm{C}\) can be calculated using the given values for \(K_b\), \(K_w\), concentration, and molar mass. Convert the concentration to mol/L, calculate the \(K_a\) from the \(K_b\), and solve the quadratic equation for the H+ ion concentration. Finally, use the -log formula to determine the pH of the solution.

Step by step solution

01

Calculate concentration of papH+Cl- in mol/L

First, we need to convert the given concentration of the drug (30.0 mg/mL) from mg/mL to mol/L. Use the molar mass of papH+Cl- (378.85 g/mol) to convert the concentration: \(Concentration (mol/L) = \frac{30.0 (mg/mL) \times \frac{1 g}{1000 mg} \times \frac{1 L}{1000 mL}}{378.85 (g/mol)}\)
02

Calculate the concentration of the conjugate base (pap) in the solution

Since we are given the Kb value for the weak base pap in the problem statement, use the relationship between Ka and Kb to calculate the concentration of the conjugate base (pap) in the solution: \(K_a = \frac{K_w}{K_b}\) \(K_a = \frac{2.1 \times 10^{-14}}{8.33 \times 10^{-9}}\) Now that we have the Ka value, set up an equation using the ionization constant Ka: \(K_a = \frac{[H^+][pap]}{[papH^+Cl^-]}\) We can assume that [H+] and [pap] are the same since they are produced in a 1:1 ratio. Thus, the equation becomes: \(K_a = \frac{x^2}{([papH^+Cl^-]-x)}\)
03

Solve for x, the concentration of H+ ions in the solution

Since we have a quadratic equation for x, use the quadratic formula to solve for x: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where K_a = a, -[papH^+Cl^-] = b, and -[papH^+Cl^-]x = c. Substitute the values into the quadratic formula and solve for x: \(x = \frac{[papH^+Cl^-] \pm \sqrt{([papH^+Cl^-])^2 - 4(K_a)([papH^+Cl^-])}}{2(K_a)}\) Since x is very small compared to the original concentration of papH+Cl-, we can simplify the equation by using the approximation [papH+Cl-] ≈ [papH+Cl-] - x: \(x \approx \frac{[papH^+Cl^-] \pm \sqrt{([papH^+Cl^-])^2}}{2(K_a)}\)
04

Calculate the pH of the solution

Now that we have the concentration of H+ ions (represented by x) in the solution, we can calculate the pH using the formula: pH = -log[H+] pH = -log(x) Substitute the x value found in step 3 into the equation to get the pH of the aqueous solution of papaverine hydrochloride.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 M \mathrm{KNO}_{2}\) b. \(0.45 M\) NaOCl c. \(0.40 M \mathrm{NH}_{4} \mathrm{ClO}_{4}\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a \(3.0 \times 10^{-7}-M\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

An unknown salt is either \(\mathrm{NaCN}, \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{NaF}, \mathrm{NaCl},\) or NaOCl. When 0.100 mole of the salt is dissolved in 1.00 L of solution, the \(\mathrm{pH}\) of the solution is \(8.07 .\) What is the identity of the salt?

The pH of a 0.063-M solution of hypobromous acid (HOBr but usually written HBrO) is 4.95. Calculate \(K_{\mathrm{a}}\).

Consider \(1000 .\) mL of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4} .\) How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of \(\mathrm{HA}\) is dissociated at equilibrium? Assume that HA is nonvolatile.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free