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 chemistry goes tougher
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Posted on 01-21-12 2:32 AM     Reply [Subscribe]
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 1. 

calculate the ratio of concentration of 2-ethylbenzonate ion to the concentration of 2-ethylbenzoic acid at?

a pH of 5.90. Ka=4.0E-5

2. 

calculate the ratio of concentration of hexanoate ion to the concentration of hexanoic acid at?

a pH of 6.09. Ka=41.3E-5

 
Posted on 01-21-12 11:50 AM     [Snapshot: 139]     Reply [Subscribe]
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So you keep posting your homeworks here and have others solve it. You will go far, specially since you don't even thank the people who solve your problems for you (check your last thread).
Am I right to be annoyed with these kind of postings?
 
Posted on 01-21-12 8:43 PM     [Snapshot: 285]     Reply [Subscribe]
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Ask ur instructor
 
Posted on 01-22-12 8:06 PM     [Snapshot: 424]     Reply [Subscribe]
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Even though I am not a chemistry major and have taken only general chemsitry, these are the some of the easiest problem realted to pH. Sove it using the :Henderson–Hasselbalch equation.
If you don't know that that is:
 http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation
p
H is given, ka is given, calculate pka and slove for the ratio of base/acid. 
 
Posted on 01-22-12 8:21 PM     [Snapshot: 446]     Reply [Subscribe]
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i know the answer but don't wanna use my brain sorry
 
Posted on 01-23-12 10:33 AM     [Snapshot: 576]     Reply [Subscribe]
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This is the easiest chemistry problem, just use HH equation

1.PH=Pka +log(B/A)  , B is the concentration of benzonate and A is the concentration of benzoic acid.
5.9=-logKa + Log(B/A)
5.9= 4.398  + log(B/A)
logB/A=1.502
B/A = 31.768


2. PH=Pka +log(B/A) , B is the concentration of hexonate ion and A is the concentration of hexanoic acid

6.09  = 4.384 + log(B/A)
Log (B/A) = 1.706

B/A = 50.815


make sure the calculation is right in case..

 


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