May 31, 2020, 12:30:07 AM
Forum Rules: Read This Before Posting

### Topic: Calculating [HY3-] in a solution prepared with EDTA, why do we use pK6?  (Read 4422 times)

0 Members and 1 Guest are viewing this topic.

#### BatMiteDimension

• New Member
• Posts: 6
• Mole Snacks: +0/-0
##### Calculating [HY3-] in a solution prepared with EDTA, why do we use pK6?
« on: November 18, 2012, 07:56:41 PM »
The question is "Calculate [HY3-] in a solution prepared by mixing 10.0 mL of 0.01 M VOSO4, 9.90 mL of 0.01 EDTA, and 10.0 mL of buffer with a pH of 4.00".

The answer is to first find [VO2+], [VOY2-], and then [Y4-] (= [VOY2-]/([VO2+] x K6)) and use that to find [HY3-] which is through the equation ([H+][Y4-])/K6

I just don't understand why we are using K6 (derived from pK6 of H6Y2+) and using that equation to find HY3-. Is this what we do if we want to find any of the fractional compositions of EDTA or just for HY3-? What made us use Y4- instead of anything else?

Thank you!

P.S.: I posted this question yesterday but I did not format the post properly. I reposted it here now with proper formatting

#### Borek

• Mr. pH
• Deity Member
• Posts: 25773
• Mole Snacks: +1686/-400
• Gender:
• I am known to be occasionally wrong.
##### Re: Calculating [HY3-] in a solution prepared with EDTA, why do we use pK6?
« Reply #1 on: November 19, 2012, 04:43:41 AM »
[Y4-] (= [VOY2-]/([VO2+] x K6))

K6 or Kf?

Somehow I am not convinced method you outlined will give a correct result (but I can be easily wrong, something is not clicking). Can you list Ka6 and Kf?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### BatMiteDimension

• New Member
• Posts: 6
• Mole Snacks: +0/-0
##### Re: Calculating [HY3-] in a solution prepared with EDTA, why do we use pK6?
« Reply #2 on: November 19, 2012, 11:10:03 AM »
Oh, I'm sorry! You're correct, it is Kf in [Y4-] (= [VOY2-]/([VO2+] x Kf))

Kf for VOY2-= 1018.7
pK6 for H6Y2+=10.37

I still don't understand why we use K6 for the rest of the question though.