Question

Iminodiacetic acid (IDA) is often used in ion exchange resins as a way to remove heavy metals from industrial waste waters. It forms a 2:1 complex with many heavy metals, with a high selectivity for divalent over monovalent ions. A 32.50 mL sample of 0.0483 M Ni2+ was added to 32.50 mL of 0.198 M iminodiacetic acid (the completely protonated form is abbreviated as H3A+ ) buffered at pH 7.00. Given that ?A2− α A 2 − is 4.6×10−3 4.6 × 10 − 3 at pH 7.00, what is the final concentration of Ni2+ Ni 2 + remaining in solution? Ni2+(aq)+2A2−(aq)↽−−⇀NiA2−2(aq)?f=2.00×1014 Ni 2 + ( aq ) + 2 A 2 − ( aq ) ↽ − − ⇀ NiA 2 2 − ( aq ) K f = 2.00 × 10 14

Answer 1

Ni 2+ (aq) + 2 A2 (ar) - Nitrag) The equilibrium constant (formation constant): x [Ni A2-] in At = [N; 2+] CAR mine diactic acid has following forme (HAN) Here reaction occurs auth A2-form. At any pt the man balance on ininodi- -acetic and CIDAD = [H 3A +] + [A] + [HA] + [42-] Fraction of A2- at any in r where to IDAA- Sinino diactic Did] PH : [A2-) ) LA2- = CADA : [A 2-] & LA2 X CIDAA Now, plugging the value of tar into equations & NA 2 - - EN; 2 x [LAZ-X CADAA]

CMA 27 F IN1 24 x 27 x Cape - INARI rat par KEXKAR IN; Com & 12:00) =) 2:0 X10%Yx (4*6*10*)" (NAF) =) 20X10 X TN 24 Capas TN;+) CEDA - 498 2x109.... (ü) Here total volume of the solution = (32.50+ 3250 mita 65 m. Initial (N; 2+] = Mwist & VN; 2 + VNiat + VIDAL bulu Ma Molarily - volume 0.8483 m X 9250 (3215 +325) = 0.02415M MIDAR XVIDAR Initias, (torr - 65 mm 6198 M X 32.5 m A 0099 M = Ni 2+(ar) + 2A²-(ar) Initial: 0.02415 0.092 0.899 M & Ni a "sand on 14 iguu change: x -20 Familim (0.0245 ) (0.09 ) here, an 0.02415m because all the 2+ auill be converted N; A,2- solution Nire comer due to dinociation of NjA 22- NH

[Ni A22] = 0.02415 m. CIDAAT 1099 m. plugging the values into eanation (iii): 0.02415 IN; 2+] X To 09972 2 = 41232 x109 0.02415 To 09972 [n 1 2 3 2x109) in 5.88 10-10 Concentration of Niet remaining in solution in 5.881010 m. .