   Chapter 16.7, Problem 16.15E 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

What is the pH at the equivalence point when 25 mL of 0.10 M HF is titrated by 0.15 M NaOH?

Interpretation Introduction

Interpretation:

The pH at the equivalence point when 25 mL of 0.10 M HF is titrated by 0.15 M NaOH has to be calculated.

Concept Information:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

To Calculate: The pH at the equivalence point when 25 mL of 0.10 M HF is titrated by 0.15 M NaOH

Explanation

Given data:

The volume of HF  = 25 mL

The concentration of HF = 0.10 M

The concentration of NaOH = 0.15 M

Explanation:

When HF is titrated against the NaOH , the solution will contain NaF   at the point of equivalence.

molar amount of F= molar amount of HFThe molar amount of F is calculated as,(0.10 mol HF/L)×0.025 L = 0.0025 mol F

The volume of NaOH added is calculated as follows,

Volume of NaOH Macid VacidMbase = (0.10M)(25 mL)0.15 M = 16.6 mL

Hence, the total volume is as follows,

Total volume = 25 mL + 16.6 mL = 41.6 mL =0.0416 L

The concentration of fluoride ion [F-] at the equivalence point is,

[F-] =0.0025 mol0.0416 L =0.060 M

Now, the hydrolysis of F is considered.

The base ionization constant of F can be calculated as follows,

The Ka of HF = 6.8×104 ( HF is the conjugate acid of F )

Kb =KwKa =1.0×10146

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