[Club33]

I am trying to do a lab where it is asking me to convert the mass of potassium hydrogen phthalate to moles so I need to calculate the molar mass of KHPh.

However, we haven't covered phthalates in class and I am not sure how to take that into consideration when I am calculating the molar mass.  Is phthalate just a name and not representative of a substance?  Do I just add the molar mass of potassium and hydrogen or do I need to take anything else into consideration?

Thanks for your help and have a great day!

Janelle

[Arkcon]

KHPh is a shorthand for a common weak acid used in titrations, called potassium hydrogen phthalate, as you know.  Phthalate is a common name for an organic anion  -- and yes, it does really exist.  The bottle you used has it's formula written on the side, and you could sum the atom's weights.  Or a quick web look up of "potassium hydrogen phthalate" will give you the answer.

Phthalates are also a general class of organic chemicals used as plasticizers, now out  of favor due to health complications, so a search for "phthalates" is going to drive you quietly crazy without helping your problem.

[Club33]

Thanks!  I had done a web search of phthalate but not potassium hydrogen phthalate.  I found it on Wikipedia with a formula of KHC₈H₄O₄.  Does that sound right?  I know wikipedia is not always a reliable source...

Thanks for your help, I really appreciate it!!

Janelle

[ARGOS++]

Dear Club33;  Dear Janelle;

Precisely!,   ─   And what gives that for the MW?

Good Luck!
                    ARGOS⁺⁺

[Arkcon]

I like the text block associated with the wikipedia, it has a clear molecular diagram, and it counts the atoms correctly, you can also double check the atomic mass's sums.  But if you're really wondering  about the internet, remember to check the bottle next time while you're in the lab.

[Club33]

Good point...  I will next time...   

Anyway, I used my class periodic table and calculated a MM of 204.23 g/mol of KHC₈H₄O₄.  Based on that and the mass of my sample (1.2064g) I got 5.9071x10^-3 mol KHC₈H₄O₄.

So now...  how do I convert this to equivalents??   

[ARGOS++]

Dear Club33;  Dear Janelle;

Moles are a kind of “Equivalents”.
But I think you forgot something in your Questions, if I read correctly:

“Equivalents” for what, or for what Reaction?

We know only that your MW is very correctly calculated, and also your translation into Moles looks very precise! 
But for what is it used?


Good Luck!
                    ARGOS⁺⁺

[Club33]

Yeah, it would help if I gave the equation that we worked with...

KHC₈H₄O_{4 (aq)} + NaOH _(aq) --> KNaC₈H₄O_{4 (aq)} + H₂O _(l)

I understand the basic definition of equivalents and that Eq_{acid reacted} = Eq_{base reacted}, but I do not understand how to convert the moles to equivalents...

Thanks for your help.

Janelle

[ARGOS++]

Dear Club33;  Dear Janelle;

That looks pretty much better!,   ─   But I have now got some doubts!

Did you real stop the Reaction/Titration after the “Half Reaction”?
Did you not realise, that there is a second acidic group?:  "Phthalic Acid”

What Indicator did you use to detect the “End Point”?


Good Luck!
                    ARGOS⁺⁺

[Arkcon]

Did you not realise, that there is a second acidic group?:  "Phthalic Acid”

For phthalic acid, yes, but potassium hydrogen phthalate has one ionizable hydrogen.  'Tho back in my early days on this forum, I'd seen some general chem students try to react all 4 hydrogens in potassium hydrogen phthalate with base.  He. He.  Benzene ring hydrogens reacting with NaOH.  Funny.

[ARGOS++]

Dear Club33;  Dear Janelle;

Sorry!,   ─    Now I did a Mistake!, because I read not correct!

If you know that the Reaction is done by “Equivalents”, and you know how many moles of KHC₈H₄O₄ you have, and you use your “Equivalents” Balance:
 

Can you tell how many moles^*) of NaOH are required for finish the Reaction (for 100%)?

^*) That's your “Equivalents” in NaOH!

Sorry! once again for my mistake!

Good Luck!
                    ARGOS⁺⁺

Dear Arkcon : Yes I realised my Mistake by myself, that's why I'm already back!

[Club33]

OK, the molar ratio of KHPh to NaOH is 1:1 so for a 100% yield, I would need 5.9071x10^-3 mol NaOH to neutralize.

So since there is one ionizable H in the KHPh and one ionizable OH in the NaOH (so this ratio is also 1:1), would my equivalency be the same as the number of moles?

5.9071x10^-3 Eq KHPh

Thanks!!

Janelle

[ARGOS++]

Dear Club33;  Dear Janelle;

If you really have a pure KHC₈H₄O₄ (KHPh) then you are right in the sense, that 5.9071 * 10^-3 mole KHPh is “equivalent” to 5.9071 * 10^-3 mole NaOH for your Reaction.

That allows you now to determine the %-age of your real KHPh, in case your Indicator told you that less NaOH was required to reach the “Equivalents Point”.

(So you don’t need a Unit: [Eq], or how ever you would like to call it.)

Hope anyway to have been finally of some help to you.

Good Luck!
                    ARGOS⁺⁺