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i Chrome File Edit View History Bookmarks People Window Help 26% D } 'I us, Tue 7:49 PM a ??? T. Blackboard @ Texas Tech Blackboard @ Texas Techl × × ?Home-201877-CHEM-13 ?.? Cengage xMindTap - Cengage Learnin x GKp benzoie acid - Google S x Ahmed i Secure https://ng.cengage.com/static/nb/ui/evo/index.html?eISBN-9781305657571&id;=324758682&nbld-843147;&snapshotld-8431478; ?aw' ? MyTech Raiderlink G chegg ? Blackboard Texas.. a The originals ?Synonyms, PaperHelp ? Calculate Your GPA GPA calculator Y uTube to mp3 Co ::: Apps 01 ~ 5% MINDTAP 17 Challenge -u s a Search this course ? ? Ahmed From Cengage Question 1 Use the References to access important values if needed for this question A 48.1 mL sample of a 0.327 M aqueous acetic acid solution is titrated with a 0.406 M aqueous solution of potassium hydroxide. How many milliliters of potassium hydroxide must be added to reach a pH of 4.485? mL Submit Answer 4 question attempts remaining A-Z Question 12 Current score: 7.6/12 pts (63.33 %) Autosaved at 7:49 PM 10

Explanation / Answer

CH3COOH + KOH <-----> CH3COOK + H3O

the Henderson–Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

Acetic acid pka = 4.76

required pH = 4.485

Lets assume we need a L of 0.406 M KOH to reach pH of 4.485

Since, the pKa is more than than the required pH or before reaching the equivalence point (pH = pKa)

Therefore,
concentration of unreacted acetic acid [CH3COOH] = (initial moles CH3COOH - moles KOH added)/total volume
                                       = (0.0481 L x 0.327 moles/L - a L x 0.406 moles/L)/( 0.0481 L + a L)
                                       = (0.01573 - a x 0.406)/(0.0481 +a)
Concetration of acetate [CH3COO-] = moles of KOH added/ toal volume
                                  = (a L x 0.406 moles/L)/(a + 0.0481 L)
                              
the Henderson–Hasselbalch equation:

pH = pKa + log ([CH3COO-]/[CH3COOH])
                              
4.485 = 4.76 + log {(0.01573 - a x 0.406)/(0.0481 +a)/(a L x 0.406 moles/L)/(a + 0.0481 L)}
-0.275 = log { (0.01573 - a x 0.406)/(a L x 0.406 moles/L))
0.5309 = { (0.01573 - a x 0.406)/(a L x 0.406 moles/L))
0.21554 a =(0.01573 - a x 0.406)
0.62154 a = 0.01573
a = 0.0253 L
a = 25.3 mL

therefore, we need 25.3 mL of 0.406 m of KOH required to reach the pH = 4.485

Hence the answerr = 25.3 mL

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