An analytical chemist is titrating 225.8 mL of a 0.5300 M solution of butanoic acid (HC,H,CO,) with a 0.8200 M solution of KOH. The p K, of butanoic acidis 4.82. Calculate the pH of the acid solution after the chemist has added 40.67 mL of the KOH solution to it.Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.Round your answer to 2 decimal places.pH = ||

Answered: An analytical chemist is titrating…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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An analytical chemist is titrating 225.8 mL of a 0.5300 M solution of butanoic acid (HC,H,CO,) with a 0.8200 M solution of KOH. The p K, of butanoic acid
is 4.82. Calculate the pH of the acid solution after the chemist has added 40.67 mL of the KOH solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.
Round your answer to 2 decimal places.
pH = ||

Expert Solution

Step 1

Given that:

Concentration of butanoic acid = 0.5300 M
Concentration of KOH base = 0.8200 M
Volume of the butanoic acid = 225.8 mL
pK_a of the butanoic acid = 4.82

To find: pH of the solution when 40.67 mL of the KOH is added to the solution.

Step 2

Calculate number of moles of butanoic acid and KOH.

$\begin{array}{l} Moles of butanoic acid = (Concentration) (Volume in ml \times 10^{- 3}) \\ Moles of butanoic acid = (0.5300) (225.8 \times 10^{- 3}) \\ Moles of butanoic acid = 0.119674 moles \\ Moles of KOH base = (Concentration) (Volume in ml \times 10^{- 3}) \\ Moles of KOH base = (0.8200) (40.67 \times 10^{- 3}) \\ Moles of KOH base = 0.0333494 moles \end{array}$

Step 3

KOH is a strong base and butanoic acid is a weak acid. Here KOH is in smaller amount. It reacts with butanoic acid in equal amounts to give salt of butanoic acid and water.

CH₃CH₂CH₂COOH + KOH → CH₃CH₂CH₂COO^- K⁺ + H₂O

KOH neutralize 0.0333494 moles of butanoic acid to give 0.0333494 moles of salt

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