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Equivalence point (WILL RATE LIFESAVER)

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Date Posted: 7/22/2008 2:19:06 PM  Status: Live
Equivalence point (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 one big question that consists of 4; please give me a solution.
You can also separate answers; thank you.


Question.
 
pH of 20.0mL of 0.100M HCl solution is ______
a(-1) b(0) c(1) d(2) e(3)

5.0mL of 0.200M NaOH is added to (1) solution above, pH is _____
a(1.2) b(1.4) c(1.6) d(1.8)
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Sage
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Date Posted: 7/23/2008 12:54:45 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
1) only strong acid is present
                        HCl -----> H+  +  Cl-
                        0.1M        0.1M    0.1M
[H+] = 0.1M
pH = - log [H+]
pH =   -(-1) =1
2)after adding 5.0ml of 0.2M NaOH solution
moles of HCl = 0.1M*20ml = 2mmol HCl
moles of NaOH = 0.2M*5.0ml = 1mmol NaOH
              HCl  +   NaOH -----> NaCl +  H2O
initial:    2mmol      1mmol             0
change:  -1mmol    -1mmol          1mmol
final:        1mmol        0                 1mmol
strong acid and its salt are treated as strong acid
HCl ---->H+  +  Cl-
[H+] =1mmol/25ml = 0.04M
pH = -log (  0.04M) = 1.4



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