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pH Calculation (WILL RATE LIFESAVER)

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Date Posted: 7/22/2008 2:27:31 PM  Status: Live
pH Calculation (WILL RATE LIFESAVER)
Course Textbook Chapter Problem
General Chemistry Chemistry: The Central Science (10th) by Brown, Lemay, Bursten 16&17 N/A
Question Details:
This is the almost same kind of homework question I posted below.


Q.
pH of 20.0mL of 0.100M acetic acid (Ka=1.8*10-5) solution is _____.
a(2.3) b(2.5) c(2.7) d(2.9)

5.0mL of 0.200M NaOH is added to (1) solution above, pH is ______.
a(2.7) b(3.7) c(4.7) d(5.7)
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Answers:

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Karma Points: 4,258
Date Posted: 7/23/2008 1:29:48 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
1) only weak acid is present initially
                    CH3COOH  -----> H+    +     CH3COO-
at equilibrium:   (0.1-x)                  x                  x
Ka = [H+][CH3COO-]/[CH3COOH] = 1.8x10-5
(x*x)/(0.1-x) =  1.8x10-5
on solving we get, x= 0.00133M
[H+] = 0.00133M
pH = - log (0.00133M) = 2.876
 
 
After adding 5ml of 0.2M NaOH base
moles of CH3COOH = 0.1M*20ml = 2mmol CH3COOH
moles of NaOH = 0.20M*5ml = 1mmol NaOH
              CH3COOH  +  NaOH ------> CH3COONa +H2O
I                2mmol            1mmol                   0
C               -1mmol         -1mmol                1mmol
E                 1mmol              0                     1mmol
weak acid and its salt - buffer
need two equations one for salt, one for weak acid
[CH3COONa] = 1mmol/25ml = 0.04M
after adding base
              CH3COOH  -----> H+    +     CH3COO-
at equilibrium:   (0.04-x)         x                  x
              CH3COONa-----> Na+    +     CH3COO-
                     0.04                   0.04            0.04
Ka = [H+][CH3COO-]/[CH3COOH] = 1.8x10-5
(x*(0.04+x))/(0.04-x) =  1.8x10-5
on solving we get, x=  1.79x10-5
[H+] =1.79x10-5
pH = 4.75 



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