   Chapter 15, Problem 41E

Chapter
Section
Textbook Problem

Calculate the pH of a buffered solution prepared by dissolving 21.5 g benzoic acid (HC7H5O2) and 37.7 g sodium benzoate in 200.0 mL of solution.

Interpretation Introduction

Interpretation:

The mass of benzoic acid, sodium benzoate and volume of their solution is given. The pH value of this solution is to be calculated.

Concept introduction:

The pH value is the measure of H+ ions. The relation between pH and pKa is given by Henderson-Hassel Bach equation. According to this equation,

pH=pKa+log[Base][Acid]

Explanation

Explanation

To determine the number of moles of benzoic acid and sodium benzoate

Mass of benzoic acid is 21.5g .

Mass of sodium benzoate is 37.7g .

The molar mass of benzoic acid is 122.12g/mol .

The molar mass of sodium benzoate is 144.10g/mol .

Formula

The number of moles is calculated using the formula,

Numberofmoles=MassofcomoundMolarmassofcompound (1)

Substitute the values of mass and molar mass of benzoic acid in the above equation.

Numberofmoles=MassofcomoundMolarmassofcompound=21.46122.12g/mol=0.176mol_

Substitute the values of mass and molar mass of sodium benzoate in equation (1).

Numberofmoles=MassofcomoundMolarmassofcompound=37.68144.10g/mol=0.261mol_

To determine the concentration of benzoic acid and of sodium benzoate

Moles of benzoic acid is 0.176mol .

Moles of sodium benzoate is 0.261mol .

Volume of solution is 200mL .

The conversion of milliliter (mL) into liter (L) is done as,

1mL=103L

Hence,

The conversion of 200mL into liter is,

200mL=(200×103)L=0.2L

The concentration of compound in a solution is calculated using the formula,

Concentrationofcompound=MolesofcompoundVolumeofsolution (2)

Substitute the values of moles of benzoic acid and volume of solution in the above equation.

Concentrationofcompound=MolesofcompoundVolumeofsolution=0

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