|
[VIEWED 9222
TIMES]
|
SAVE! for ease of future access.
|
|
|
|
blackheaven123
Please log in to subscribe to blackheaven123's postings.
Posted on 01-21-12 2:32
AM
Reply
[Subscribe]
|
Login in to Rate this Post:
0 
? 
|
| |
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
|
| |
|
|
|
|
Kiddo
Please log in to subscribe to Kiddo's postings.
Posted on 01-21-12 11:50
AM [Snapshot: 139]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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?
|
| |
|
|
resham32
Please log in to subscribe to resham32's postings.
Posted on 01-21-12 8:43
PM [Snapshot: 285]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
|
|
| |
|
|
Tenacious
Please log in to subscribe to Tenacious's postings.
Posted on 01-22-12 8:06
PM [Snapshot: 424]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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
pH is given, ka is given, calculate pka and slove for the ratio of base/acid.
|
| |
|
|
intelligentguy
Please log in to subscribe to intelligentguy's postings.
Posted on 01-22-12 8:21
PM [Snapshot: 446]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
i know the answer but don't wanna use my brain sorry
|
| |
|
|
bhustigre
Please log in to subscribe to bhustigre's postings.
Posted on 01-23-12 10:33
AM [Snapshot: 576]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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..
|
| |
