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Author: Subject: Buffer solution problem/HCl gas
SoluBull
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[*] posted on 11-2-2005 at 21:11
Buffer solution problem/HCl gas


This question was on a past test in my class and a group of my friends at 2 TAs weren't able to get it. Any help would be appreciated :cool:

"If a buffer solution is prepared by mixing 180.0 mL of 0.35 M NaAc with 70.0 mL of 0.60 M HAc, the amount of HCl (g) which must be bubbled into this buffer to double [H3O+] is closest to:

a) 0.025 moles
b) 0.018 moles
c) 0.035 moles
d) 0.012 moles
e) 0.028 moles

The correct answer was b) 0.018 moles.

How do you solve this problem?
Thanks in advance to all those who help!
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Magpie
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[*] posted on 12-2-2005 at 17:13


original solution:

moles NaAc =(180/1000)0.35 = 0.063
moles HAc = (70/1000)0.60 = 0.042

HAc --> <---------- H+ + Ac-
NaAc ----> Na+ Ac-

[H+][Ac-]/[HAc] = 1.75 x 10^-5 = Ka for acetic acid (eqn 1)

Let x = moles H+ at equilibrium

x(0.063 + x)/(0.048-x) = 1.75x10^-5 from eqn 1

(0.063x + x^2)/(0.042-x) = 1.75x10^-5
x^2 is negligible so disregard
x=1.17 x 10^-5 (pH = 4.93)

After bubbling HCl through:

[H+] = 2.34 x 10^-5 (pH = 4.63)

Let y = H+ added that ties up with Ac-

(2.34x10^-5)(0.063-y)/(0.042 +y) =
1.75 x 10^-5 (from eqn 1)

y=0.0181

The extra H+ contributed to supply the free H+ is negligible compared to that soaked up by the Ac- from the NaAc. ;)

Note: I have made an error in setting this up and will submit a correction soon.
Instead of using "moles" in the equations I should have use "molarity." - Magpie

[Edited on 13-2-2005 by Magpie]




The single most important condition for a successful synthesis is good mixing - Nicodem
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SoluBull
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[*] posted on 12-2-2005 at 20:18


Thank you so much for your help. It never ceases to amaze me how little I know.

:o

Thanks again for all your help!
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Magpie
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[*] posted on 13-2-2005 at 14:05


Here's the correction to my earlier post:

original solution:

volume = 180mL + 70mL = 250mL = 0.25L

moles NaAc =(180/1000)0.35 = 0.063; [NaAc] = 0.063/0.25 = 0.252M
moles HAc = (70/1000)0.60 = 0.042
[HAc] = 0.042/0.25 = 0.168M

HAc --> <---------- H+ + Ac-
NaAc ----> Na+ Ac-

[H+][Ac-]/[HAc] = 1.75E-5 = Ka for acetic acid (eqn 1)

Let x = [H+] at equilibrium

x(0.252 + x)/(0.168-x) = 1.75E-5 from eqn 1

0.252x + x^2 = (0.294E-5) - (1.75E-5) x

x^2 and (1.7510E-5)x terms are negligible so drop.

x= 0.294E-5/0.252 = 1.17E-5
(pH = 4.93)

After bubbling HCl through:

[H+] = 2.34E-5 (pH = 4.63)

Let y = [H+] added via HCl gas:

(2.333E-5)[0.252-(y-2.333E-5)]/[0.168 +(y - 2.333E-5)] =
1.75 x 10^-5 (from eqn 1)

dropping negligible terms:

(2.333E-5)(0.252-y)/(0.168+y) =1.75E-5

y=0.2939/4.083 = 0.07198M (moles/L)

for 0.25L: moles HCl = 0.07198/4 = 0.018

[Edited on 14-2-2005 by Magpie]




The single most important condition for a successful synthesis is good mixing - Nicodem
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