   Chapter 16, Problem 16.141QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

Ka for formic acid is 1.7 × 10−4 at 25°C. A buffer is made by mixing 529 mL of 0.465 M formic acid, HCHO2, and 494 mL of 0.524 M sodium formate, NaCHO2. Calculate the pH of this solution at 25°C after 110 mL of 0.152 M HCl has been added to this buffer.

Interpretation Introduction

Interpretation:

The pH of a given buffer solution after the addition of 110 mL of 0.152 M HCl has to be calculated

Concept Introduction:

Henderson-Hasselbalch equation:

The pH can be calculated using Henderson-Hasselbalch equation as follows,

pH = pKa + log[conjugate base][weak acid]pH = pKa + log[A-][HA]

To Calculated: The pH of a given buffer solution after the addition of 110 mL of 0.152 M HCl

Explanation

Given data:

A buffer solution made by mixing 529 mL of 0.465 M formic acid, HCHO2 with 494 mL of 0.524 M sodium formate, NaCHO2

The Ka value for formic acid is 1.7×104

The concentration of added HCl is 0.152 M

The volume of added HCl is 110 mL

pH Calculation:

A buffer solution made by mixing 529 mL of 0.465 M formic acid, HCHO2 with 494 mL of 0.524 M sodium formate, NaCHO2

Before any HCl is added:

The total volume at this point is:

Total volume  = 529 mL + 494 mL =1023 mL

NaCHO2 has Na+ and CHO2 ions.

The concentrations of HCHO2 and CHO2 after mixing together is:

[HCHO2] =M1V1V2 =(0.465 M)(529 mL)(1023 mL) =0.24045 M[CHO2] =M1V1V2 =(0.524 M)(494 mL)(1023 mL) =0.25303 M

The equilibrium reaction is:

HCHO2 + H2OH3O+ + CHO2

Use the Hendersen-Hasselbalch equation to calculate the pH.

pH  = pKa+ log[base][acid] =log(1

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