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THoneymoon x Ga Blackboard L x D bboard2010 x E-Mail Accesx E Sign into yo x C Chemistr q My GRE My x GILove Kanye x CHEM 1215 X C WWW sapling learning.com /ibiscms/mod/ibis/view.php?id-2501029 Sapling Learning Sapling Learning My Assignment Resources 4/25/2016 11:55 PM 61.9/100 Gradebook O Assignment Information Attempts Score Print A Calculator Periodic Table rom: 4 10/2016 06:00 PM 4/25/2016 11:55 PM Points Possible: 100 Map Dab General Chemistr Grade Category: Graded Description Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). A list of ionization constants can be found here. You can check your answers You can view solutions when you complete or give Number up on any question. 0.05 You can keep trying to answer each question until ApH you get it right or give up Calculate the change in pH when 6.00 mL of 0.100 M NaOH (aq) is added to the original buffer solution. Number O Help With This Topic 0.15 ApH Web Help & Videos O Technical Support and Bug Reports There is additional feedback available! ew this feedbac bottom divider bar. Click on the divider bar again to hide the additional feedback. Previous 3 Give Up & View Solution 8 Try Again Next Exit Explanation careers partner prhag polly terms ofuse contact us m Cortana. Ask me anything Logout 4/25/2016

Explanation / Answer

pH = pka + log [conjugate base] /[acid] is formula

NH4+ is acid then NH3 is conjugate base , pka of NH4+ = 9.25

initially pH = 9.25 + log ( 0.1/0.1)

     = 9.25

NH3 moles = M x V = 0.1 x ( 100/1000) = 0.01

NH4+ moles = M x V = 0.1 x ( 100/1000) = 0.01

HCl moles added = M x V = 0.1 x ( 6/1000) = 0.0006

HCl gives H+ which adds to NH3 to form NH4+    ( solution volume now = 106 ml = 0.106 L)

hence NH3 moles = 0.01-0.0006 = 0.0094 ,   [NH3] = moles/vol = (0.0094) / ( 0.106) =0.08868

NH4+ moles = 0.01+0.0006 =0.0106 , [NH4+] = (0.0106/0.106) = 0.1 M

pH = 9.25 + log ( 0.08868/0.1)

= 9.2

pH change = 9.2-9.25 = -0.05

now NaOH moels added = M x V =0.1 x ( 6/1000) = 0.0006

NaOH gives OH- which reacts with NH4+ to give Nh3 and H2O

NH4+ moles = 0.01-0.0006 = 0.0094 , [NH4+] =0.0094/0.106 = 0.08868

NH3 moles = 0.01+0.0006 = 0.0106 , [NH3] = 0.0106/0.106 = 0.1

pH = 9.25 + log ( 0.1/0.08868)

= 9.3

pH change = 9.3-9.25 = 0.05

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