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.
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.
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
Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 107AE: A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution...
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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
pKa 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.
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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