A 20.00-ml sample of a 0.100 M solution of butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10-6), is titrated with 0.100 M KOH

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A 20.00-ml sample of a 0.100 M solution of butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10-6), is titrated with 0.100 M KOH

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A 20 00 Ml Sample Of A 0 100 M Solution Of Butanoic Acid Ch3ch2ch2cooh Ka 1 54 X 10 6 Is Titrated With 0 100 M Koh 1
A 20 00 Ml Sample Of A 0 100 M Solution Of Butanoic Acid Ch3ch2ch2cooh Ka 1 54 X 10 6 Is Titrated With 0 100 M Koh 1 (114.08 KiB) Viewed 74 times
A 20.00-ml sample of a 0.100 M solution of butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10-6), is titrated with 0.100 M KOH solution. Answer the following questions (1 thru 8): 1. Calculate the volume of KOH (in mL) required to reach the equivalence point? 2. Calculate the pH before adding any KOH 3. Calculate the pH after addition of 10.00 mL of KOH 4. Calculate the pH at the equivalence 5. Calculate the pH at the half equivalence point. Show that the titration is at the half equivalence point 6. Calculate the pH after addition of 25.00 mL of KOH 7. Use the calculated pH values, obtained from questions 1 thru 7, to carefully draw the resulting titration curve. Clearly label the axes. Refer to lectures notes and textbook to ensure the proper shape of the titration curve. 8. Choose an appropriate indicator for this titration.
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