[SweetAsC12H22O11]

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  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

     

[Borek]

This is a buffer problem.

[SweetAsC12H22O11]

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?

[Borek]

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?

[SweetAsC12H22O11]

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?

[Borek]

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?

[SweetAsC12H22O11]

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...):

     C₈H₄O₄^2- + H⁺  HC₈H₄O₄^-

     pH = pKa + log (C₈H₄O₄^2-/HC₈H₄O₄^-)

Is this correct?

[Hunter2]

This looks good.

[SweetAsC12H22O11]

Great! Now I am trying to figure out what values to use for the ratio of:

     log [c₈H₄O₄^2-]/[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





     

[Borek]

I got around 4.3g as well.

[SweetAsC12H22O11]

Thank you!