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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: SweetAsC12H22O11 on November 09, 2014, 09:37:31 AM

Title: dipotassium phtalate with perchloric acid
Post by: SweetAsC12H22O11 on November 09, 2014, 09:37:31 AM
How many grams of dipotassium phtalate (MW=242.31 should be added to 30.00mL of 0.400M HClO4 to give a pH of 5.10 when the solution is diluted to 500mL?


I started by taking care of the dilution (C1V1=C2V2) to get the concentration of the diluted perchloric acid solution (0.024M). Assuming that is the solution that was diluted...

Now, because the problem deals with a strong acid, the perchloric acid must dissociate completely (right?).

     HClO4  :rarrow: H+ + ClO4-

Usually with a strong acid, use of systematic treatment is required. How can it be applied to this problem? Can anyone point me in the right direction please?

Thank you

     
Title: Re: dipotassium phtalate with perchloric acid
Post by: Borek on November 09, 2014, 03:31:34 PM
This is a buffer problem.
Title: Re: dipotassium phtalate with perchloric acid
Post by: SweetAsC12H22O11 on November 10, 2014, 04:59:18 PM
Great, thanks! I've attempted the problem.

I used the concentration of the acid to find the number of moles:

     0.024M x 0.5L = 0.012moles of acid initially

Then, I used the HH equation to determine the number of moles of added dipotassium phthalate:

     pH = pKa2 + log [A-]/[HA]

     (I used the pka2 for phthalic acid since we are adding dipotassium phthalate)

     5.10 = 5.408 + log x/(0.012-x)

     x = [H+] = 0.00396M x 0.5L = 0.00198 moles of dipotassium phthalate added

     Multiply by MW of 242.31 gives a mass of 0.479 g of added dipotassium phthalate.

Does this look right?
Title: Re: dipotassium phtalate with perchloric acid
Post by: Borek on November 10, 2014, 06:22:52 PM
I am afraid it is wrong.

Buffer contains an acid and its conjugate base. What is the acid and what is its conjugate base in this problem?
Title: Re: dipotassium phtalate with perchloric acid
Post by: SweetAsC12H22O11 on November 10, 2014, 09:55:47 PM
Okay, thanks!

The acid is HClO4 and its conjugate base is ClO4-.

So the HH equation should be:

     pH = pKa + log [ClO4-]/[HClO4]

     Right?

What pKa value should be used? If I look up perchloric acid in my textbook, I can't find it. Is that because it's a strong acid?
Title: Re: dipotassium phtalate with perchloric acid
Post by: Borek on November 11, 2014, 03:53:57 AM
Quote from: SweetAsC12H22O11 on November 10, 2014, 09:55:47 PM
The acid is HClO4 and its conjugate base is ClO4-.

No, that's why you can't get the right answer. Think it over - what do you put in the solution? How will it react with a strong acid present?
Title: Re: dipotassium phtalate with perchloric acid
Post by: SweetAsC12H22O11 on November 11, 2014, 07:29:48 AM
Okay. So dipotassium phthalate is added to the acid solution, and the acid is a proton donor so it will protonate the diphth so that it become monopotassium phthalate.

So (ignoring K since salts dissociate...):

     C8H4O42- + H+  ::equil:: HC8H4O4-

     pH = pKa + log (C8H4O42-/HC8H4O4-)

Is this correct?
Title: Re: dipotassium phtalate with perchloric acid
Post by: Hunter2 on November 11, 2014, 08:02:35 AM
This looks good.
Title: Re: dipotassium phtalate with perchloric acid
Post by: SweetAsC12H22O11 on November 11, 2014, 09:21:08 AM
Great! Now I am trying to figure out what values to use for the ratio of:

     log [c8H4O42-]/[Hc8H4O4-]

The concentration of acid is 0.024M. So if I use an ICE table, the concentration of diphth should be F-0.024, and the concentration of monophth should be 0.024M, is that right?

Plugging into HH:

     5.10 = 5.408 + log (F-0.024)/0.024

     F = 0.0358M x 0.5L = 0.0179moles diphth x 242.31g/mol = 4.338g





     
Title: Re: dipotassium phtalate with perchloric acid
Post by: Borek on November 11, 2014, 04:22:12 PM
I got around 4.3g as well.
Title: Re: dipotassium phtalate with perchloric acid
Post by: SweetAsC12H22O11 on November 11, 2014, 06:01:12 PM
Thank you!