( + × University of Minnesota. Twimx Mail-ridwan abdul azeez-c ibiscms/modibis/v
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( + × University of Minnesota. Twimx Mail-ridwan abdul azeez-c ibiscms/modibis/view.php?id=45 403 42 Minnesota, Twin Cities-CHEM 2101-Fall17-BANTZ Activities and Due Dates Ch12 Gradebook 35/70 Print Caculaior-Periodic Table 10/31/2017 05:00 PM Question 3 of 7 Map Sapling Learning In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y) and metal chelate (abbreviated MY-4) can buffer the free metal ion concentration at values near the dissociation constant of e metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant This equilibriumMMY-4 MY- M EDTA isgoverned by the equation K,-4- where Ki is the association constant of the metal and Y, ar is the fraction of EDTA in the Yt form, and EDTA] is the total concentration of free (unbound) EDTA. K' is the "conditional formation constant. How many grams of Na EDTA 2H2O (FM 372.23) should be added to 1.72 g of Ba(NOs)2 (FM 261.35) in a 500-mL volumetric flask to give a buffer with pBa 7.00 at pH 10.00? (log K, for Ba-EDTA is 7.88 and -at pH 10.00 is 0.30.) Number mass Na EDTA 2H2049 O Prevous Give up &View; Solut on D check Answer 0 Next Exit about us careers privacy policy terms ofuse contact us 40 F3 F4 F5Explanation / Answer
Given Data:
K'f = alpha y4- * Kf = [MYn-4] / [Mn+][EDTA]
FM of Na2EDTA.2H2O = 372.23
FM of Ba2(NO3)2.2H2O = 261.35
Weight of FM of Ba2(NO3)2.2H2O = 1.72 g
Volume = 500 ml = 0.5L
at pH = 10
pBa2+ = 7
log Kf = 7.88
alpha y4- = 0.3
BaY2+ = [1.72 g] / {[261.35 g/mol] * [0.5L]} = 0.0132 M
Substituting in the given Kf equation, we get the following:
[0.3 * 107.88] = [0.0132]/{[10-7][X]}
Solving for X, we get: X = 0.0058 M
Total moles = [0.0058*0.5L] + [0.0132*0.05L] = 0.0095 moles
Total weight of Na2EDTA.2H2O = 0.0095*372.23 = 3.5 grams.
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