CA 0.4569 G s C Chegg s University of Cincinnati CX Guided S /ibiscms/mod/ibis/v

Question

CA 0.4569 G s C Chegg s University of Cincinnati CX Guided S /ibiscms/mod/ibis/view.php?id-2366009 WWW sapling learning.com Apps For quick access, place your bookmarks here on the bookmarks bar. Import bookmarks now. Attempts Score Periodic Table Print Calculator 93 Questi 4 of 7 Map Tutorial In forming a chelate with a metal ion, a mixture of free EDTA, (abbreviated Y4 and metal chelate (abbreviated MYn-4) can buffer the free metal ion concentration at values near the dissociation constant of 2.2 Tutorial the metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant. 2.3 Tutorial This equilibrium Mat Y 50 MY K'r-a K. is governed by the equation M T EDTA. where Kr is the association constant of the metal and Y ar4 is the fraction of EDTA in the Y form, and 00 DTA] is the total concentration of free (unbound) EDTA. K f is the "conditional formation constant. 67 How many grams of Na2EDTA. 2H20 (FM 372.23) should be added to 1.97 g of Mg(NO3)2 2H20 (FM 184.35) in a 500-mL volumetric flask to give a buffer with pMg 8.00 at pH 10.00 00 og Kf for Mg-EDTA is 8.79 and ay' at pH 0.00 is 0.30 Numbe mass Na2EDTA 2H20 3.98 Previous Give Up & View Solution Check Answer Next Exit Hint pyright 20 2016 Sapling Le tact 7:12 PM 3/16/2016 Available From 3/9/2016 07:00 PM 3/16/2016 11:55 PM Due Date: Points Possible 100 Grade Category: Default Description: Policies: Homework heck y up on any question You can keep trying to answer each question until you get it right or give up 5% ble to ct att eTextbook O Help With This Topic O Web Help & Videos Technical Support and Bug Reports

Explanation / Answer

Total Mg2+ = 1.97 g/184.35 g/mol x 0.5 L = 0.0214 M

pMg2+ = 8

[Mg2+] = 1 x 10^-8 M

Kf = 6.16 x 10^8

Kf' = Kf.alpha[Y4-] = 6.16 x 10^8 x 0.30 = 1.85 x 10^8

Kf' = [MgY2-]/[Mg2+][EDTA]

let x amount of EDTA remained

1.85 x 10^8 = (0.0214)/(1 x 10^-8)(x)

1.85x = 0.0214

x = 0.0116 M

Total amount of EDTA to be added = (0.0214 + 0.0116) x 0.5 x 372.23 = 6.142 g

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.