Question

3. A 25.00 mL 0.0250 M calcium carbonate sample is titrated with 14.3 mM EDTA solution. Both sample and titrant are buffered at pH = 10.0 a. What is the titration reaction and what is the value of the conditional formation constant? b. What is the calcium ion concentration when 12.35 mL of titrant has been added? c. What is pCa2+ at the equivalence volume?

Answer 1

Given that Volume of Ca+2 solution = 25.00 mL Conc. of Cat2 solution = 0.0250 M Conc. of EDTA = 14.3 mm = 0.0143 M. logk = 10.65 So Kf = 4.47 x 1010 Formation constant of Ca(EDTA)]-2 = 4.47 x 1010 (a) The reaction between Ca+2 and EDTA-4 (Y-4) is At pH =10, the fraction of EDTA that exists Y-4 is Oyne = 0.30 So the Conditional Formation constant = K = 0.30x 4.47 *100 = 1.34x1010

(b) Addition of 12.35 mL of EDTA mmols of Ca+2 = M*V = 0.0250 M* 25.00 mL = 0.625 mmol mmols of EDTA = M*V = 0.0143 M* 12.35 mL = 0.176605 mmol Before the equivalence point, as Cat2 is present in excess. (Ca+2] is calculated by the concentration of unreacted Ca+2 Net moles of Ca+2 = 0.625 - 0.176605 = 0.448395 mmol moles 0.448395 mmol ["" } Total Volume (25.00+12.35) mL [ca"] = Techno -=0.012005M Concentration of Cat2 after addition of 12.35 mL = 0.0120 M

(c) At Equivalence point: Ca+2 forms 1:1 complex with EDTA At equivalence point, Number of moles of Ca+2 = Number of moles of EDTA Number of moles of Ca+2 = M*V = 0.0250 M* 25.00 mL = 0.625 mmol Number of moles of EDTA = 0.625 mmol Volume of EDTA required = moles/Molarity = = 0.625 mmol / 0.0143 M = 43.71 mL V = 43.71 mL At equivalence point, all Ca+2 is converted to (CaY-2] complex. So the concentration of Cat2 is determined by the dissociation of (CaY-2] complex. -27 initial moles of Ca+2 [Car]= Total Volume 0.625 mmol -= 0.009097M (25.00+43.71)mL ki Cara 0.009097 Initial Change 0 +* 0 + x -X Equilibrium X 0.009097 - x 0.009097 [Cay?] 0.009097 - x [Ca+2][Y-41 20.009097 0.009097 .x 3X=-K 1610 = 6.79x10-13 1.34x1 x=8.24 x 10-7 [Ca+]=8.24x10-?M pCa =- log (8.24 x10-7)=6.08