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Positive Behavioral.. Positive Behavioral...Professional R.... S Teaching Clas... E Tardiness Beh.... Search ResultsExtra Cred... X C Chemistry que....+ Students www.clayton.edu/Portals/273/chem1212_extraquiz3.pdf C Q Search Page: - + Automatic Zoom Document3 Word HOME INSER DESIG PAGE L REFER MALI REVIE VIEW Foxit P Joel Lymon x Arial FILE b) 100, Styles Editing Paste Clipboard Font Paragraph Styles G It's one of the perks of having Office 365 GET THE NEW OFFICE See what's new Update Office 1) Estimate the pH that results when the following two solutions are mixed. You may need to look up a constant from your textbook to answer part(s) of this question c) 200 00 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2 PAGE 1 OF 1 43 WORDS +9096 9:39 PM 7/12/2016 Search the web and Windows )Explanation / Answer
1)
we know that
moles = molarity x volume (ml) / 1000
so
moles of CH3COOH taken = 0.3 x 200 / 1000 = 0.06
moles of Ba(OH)2 taken = 0.3 x 100 / 1000 = 0.03
now
given CH3COOH and Ba(OH)2
the reaction is given by
2CH3COOH + Ba(OH)2 ---> Ba(CH3COO)2 + 2H20
we can see that
moles of CH3COOH reacted = 2 x moles of Ba(OH)2 taken = 2 x 0.03 = 0.06
so
all the Ba(OH)2 and CH3COOH are completely used up
now
moles of Ba(CH3COO)2 formed = moles of Ba(OH)2 taken = 0.03
now
final volume= 200 + 100 = 300 ml
now
only Ba(CH3COO)2 remains in the solution
now
moles oof CH3COO- = 2 x moles of Ba(CH3COO)2 = 2 x 0.03 = 0.06
final volume = 300 ml
so
[CH3COO-] = 0.06 x 1000 / 300 = 0.2 M
now
as it is a salt of weak acid CH3COOH and strong base Ba(OH)2
it will undergo anionic hydrolysis
CH3COO- + H20 ---> CH3COOH + OH-
Kb = [CH3COOH] [OH-] /[CH3C00-]
using ICE table
initial conc of CH3COO- , CH3COOH ,OH- are 0.2 , 0 , 0
change in conc of CH3COO- , CH3COOH , OH- are -x , +x , +x
final conc of CH3COO- , CH3COOH , OH- are 0.2 -x , x , x
now
Kb = [x] [x] / [0.2-x]
Kb value for acetic acid is 5.5 * 10-10
as Kb value is very small , x will be very small
so
0.2-x ---> 0.2
so
Kb = [x]^2 / 0.2
[x]^2 = 0.2 * 5.5 * 10-10
[x] = 1.048 * 10-5
now
pOH = -log [OH-]
pOH = -log 1.048 * 10-5
pOH = 4.98
now
pH = 14 - pOH
pH = 14 - 4.98
pH = 9.02
so
pH of the resulting solution is 9.02