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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: unsavedhero on March 16, 2014, 06:53:46 PM

Title: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: unsavedhero on March 16, 2014, 06:53:46 PM: Calculate the pH after 35.0mL of .150M NaOH has been added to 35mL of .150M acetic acid. Ka for acetic acid (CH3COOH) is 1.8 x 10^-5
The answer is apparently pH = 8.81 but I have no idea how they got it. I mean when you convert the acetic acid and NaOH into moles you get the same exact number of moles so they cancel each other out. I would assume the pH to be 7 but apparently its not. someone please help me my brain is about to explode
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: Borek on March 16, 2014, 07:13:30 PM: In other words: calculate pH of the ~~0.085M~~ 0.075M sodium acetate solution.

Does it help?
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: unsavedhero on March 16, 2014, 07:16:30 PM: mehh not really. i understand that sodium acetate forms but how'd you get .085M. I tried doing .085 x 1.8x10^-5 and taking -log of that and it was close. 8.19 but not 8.81
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: unsavedhero on March 16, 2014, 07:24:51 PM: OMG never mind i got the answer using .085M but I have no idea how you got .085M. Kb= (1.0 x 10^-14)/(1.8 x 10^-5) = 5.555 x 10^-10 (.085)M = 4.722 x 10^-11. Square root of 4.72 x 10^-11 = 6.871 x 10^-6. -log of that gives like 5.163. 14 - 5.163 = approximatelly 8.81. if you could tell me how you got .085M that would be fantastic
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: Borek on March 16, 2014, 07:29:12 PM: Sorry, typo - should be 0.075M. But the difference between 0.075 and 0.085 is low enough to not make large effect on teh pH.

0.075M just because of dilution - you start with 0.15M and dilute it twice.
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: unsavedhero on March 16, 2014, 07:32:50 PM: Quote from: Borek on March 16, 2014, 07:29:12 PM
Sorry, typo - should be 0.075M. But the difference between 0.075 and 0.085 is low enough to not make large effect on teh pH.

0.075M just because of dilution - you start with 0.15M and dilute it twice.
ironic cause "I am known to be occasionally wrong." lol hmm i see. you did .00525/.070L which equals .075M. thanks a lot. you really helped me. should have realized that when all the H+ from the acetic acid forms water that there will be a base left over.
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: Borek on March 16, 2014, 07:42:30 PM: Quote from: unsavedhero on March 16, 2014, 07:32:50 PM
hmm i see. you did .00525/.070L which equals .075M

No, I did 0.15/2. Much faster.
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: unsavedhero on March 16, 2014, 07:45:38 PM: Quote from: Borek on March 16, 2014, 07:42:30 PM
Quote from: unsavedhero on March 16, 2014, 07:32:50 PM
hmm i see. you did .00525/.070L which equals .075M

No, I did 0.15/2. Much faster.
ugh your methods are unorthodox... f&#$ it! I'm making meth!
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: Borek on March 17, 2014, 03:45:25 AM: Quote from: unsavedhero on March 16, 2014, 07:45:38 PM
ugh your methods are unorthodox...

Shortcut - yes, unorthodox - I don't think so. You mix two identical volumes, so the final concentration of any ion must be half of the original.
Title: Re: A 35.0mL sample of 0.150M Acetic Acid is titrated with 0.150M NaOH solution.
Post by: DrCMS on March 17, 2014, 05:14:55 AM: Quote from: unsavedhero on March 16, 2014, 07:45:38 PM
I'm making meth!

Do not joke about stuff like that or we'll have to stop helping you in case it is actually true.