   Chapter 14, Problem 122E

Chapter
Section
Textbook Problem

# Papaverine hydrochloride (abbreviated papH+Cl−; molar mass = 378.85 g/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; Kb = 8.33 × 10−9 at 35.0°C). Calculate the pH of a 30.0-mg/mL aqueous dose of papH+Cl− prepared at 35.0°C. Kw at 35.0°C is 2.1 × 10−14.

Interpretation Introduction

Interpretation: The pH of the given 30mg/mL aqueous dose of papH+Cl prepared at 35°C to be calculated.

Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

The value of Kw is calculated by the formula,

Kw=KaKb

Explanation

Explanation

To determine: The pH of the given 30mg/mL aqueous dose of papH+Cl prepared at 35°C .

The [papH+Cl] is 0.0791M_ .

Given

The mass of papH+Cl is 30mg/mL .

The molar mass of papH+Cl is 378.85g .

The conversion of mg into g is done as,

1mg=103g

The conversion of 30mg/mL into g/mL is,

30mg/mL=0.03g/mL

The conversion of mL into L is done as,

1mL=103L

The conversion of 0.03g/mL into g/L is,

30mg/mL=30g/mL

The number of moles of a substance is calculated by the formula,

Moles=GivenmassMolarmass

The conversion of 30g/L into mol/L is,

30mg/mL=30g/L378.85g/mol=0.0791M_

The Ka value is 2.5×10-6_ .

The value, Kw=KaKb

The given value of Kb is 8.33×109 .

The given value of Kw is 2.1×1014 .

The value of Ka is calculated by the formula,

Ka=KwKb

Substitute the value of Kb and Kw in the above expression.

Kb=2.1×10148.33×109=2.5×10-6_

The equilibrium constant expression for the given reaction is, Ka=[H+][X][XH+]

Papaverine hydrochloride is assumed to be denoted by XHCl .

The dominant equilibrium reaction is,

XH+(aq)H+(aq)+X(aq)

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 