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Topic: Titration Error  (Read 11022 times)

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Offline zeshkani

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Titration Error
« on: October 02, 2006, 10:06:49 PM »
i have very quick question.

What would be the titration error if you used phenolphthalein as the indicator for the titration of a 25.00 mL solution of 0.08943 M ammonia with 0.1000 M HCl?

Offline Borek

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Re: Titration Error
« Reply #1 on: October 03, 2006, 03:22:13 AM »
Huge. Show us what you think and we will tell you why.
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Offline zeshkani

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Re: Titration Error
« Reply #2 on: October 03, 2006, 10:49:18 AM »
here

 .08943M* .025L / .1 * 1000 = 22.357mL

THEN

.08943-x  =x2

x= square root of the Kb(1.8e-5)*.08943
x=1.6e-6
-log(1.6e-6)
=5.79 pOH
 = 14-5.79
=8.21 pH

then  from here i dont know where to go ?


Offline Borek

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Re: Titration Error
« Reply #3 on: October 03, 2006, 11:00:18 AM »
You must describe what you are doing. I can help, but I don't want to waste my time guessing what you did and why.

here

 .08943M* .025L / .1 * 1000 = 22.357mL

THEN

.08943-x  =x2

Are these two connected in some way, or just thrown at random?

Quote
8.21 pH

If I guess correctly what you trying to do - you are wrong.
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Offline zeshkani

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Re: Titration Error
« Reply #4 on: October 03, 2006, 11:16:13 AM »
well dont we have to multiply the [.08943] by the (.025L) then divide by .1 then multiply by 1000ml = to give the amount of ml need to reach the equ.pt. which cam out to be 22.357

then
do the

Initial =  .08943                     0         0
Equ   =  .08943-x                  x          x

x2= Kb of ammonia * .08943M
x = square root of Kb * .08943M
 which gives us the [OH] CONCENTRATION
take the -log(of the[OH]
which is the pOH ?
then just take 14- pOH
which gives pH

and from here on iam trying to do this but i dont know if iam right or wrong

([OH] - .08943M) - .08943 / ([OH]-.08943)*100 = .001789% ???

Offline Borek

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Re: Titration Error
« Reply #5 on: October 03, 2006, 01:17:41 PM »
well dont we have to multiply the [.08943] by the (.025L) then divide by .1 then multiply by 1000ml = to give the amount of ml need to reach the equ.pt. which cam out to be 22.357

Are you using it in further calculations, or have you calculated it just to calculate it?

Quote
Initial =  .08943                     0         0
Equ   =  .08943-x                  x          x

x2= Kb of ammonia * .08943M
x = square root of Kb * .08943M
 which gives us the [OH] CONCENTRATION
take the -log(of the[OH]
which is the pOH ?
then just take 14- pOH
which gives pH

and from here on iam trying to do this but i dont know if iam right or wrong

([OH] - .08943M) - .08943 / ([OH]-.08943)*100 = .001789% ???

You are calculating some pH. pH of what? What for?

What is definition of titration error?
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