Question

7. A low concentration of NTA has equilibrated with PbCO₃(s) at pH 7.00 in a medium having a final HCO₃^- concentration of 7.50x10^-4 M.

a. Determine [Pb²⁺].

b. Write out K_T and the overall reaction for K_T.

c. What is the ratio of [Pb(NTA)^- ] to [H(NTA)^2-] in this solution?

Potentially useful data: Reaction Hg 2 +2e = Hgº(s) Feste = Fe2 Cu 2 +2e = Cuº(s) 2H + 2e = H2(g) Pb2 +2e=Pbº(s) Ca*2 + 2e = Cdº(s) 1/4 O2(g) + H + = 1/2 H2O 2 02(g) + 2H+ + 2e = H,O2 1/2 NO3 + H + = 1/2 NO,' + 1/2 H2O 1/5 NO3 + 6/5 H +=1/10 N2(g) + 3/5 H2O 1/8 NO3 + 5/4 H+ + = 1/8 NH. + 3/8 H2O 1/8 SO. +9/8 H + e + 1/8 HS + 1/2 H2O 1/4 CO2 + H + = 1/4 {CH,0} + 1/4 H2O 1/4 {CHO} +Hp+ee 1/4 CH4 +1/4 H20 1/8 CO2 + H+ = 1/8 CH4 + 1/4 H2O 1/2 MnO2(s) + 2H+ + = 1/2Mn+2 + H2O Reaction Alt + OH = Al(OH)2 Alt + 2OH = Al(OH)2 A1** +30H = Al(OH)3(s) A1+3 +40H = Al(OH)4 Fe(OH)3(s) + 3H* = Fe*3 + 3H2O Fe(OH)(s) + 2H+ = Fe+2 + 2H2O Pb(OH)2(s) = Pb*2 + 2OH pe Reaction 13.35 CO2(aq) + H2O = HCO3 +H 13.2 HCO3 = C0,2 +H 5.71 CO2(g) = CO2(aq) 0.00 H S(g) = H,S(aq) -2.13 H2S(aq) = HS+H -6.81 | HS=S? +H 20.75 H3PO, = H2PO4 + 11.5 H PO, 2 HPO, 2 +H* 14.15 HPO 2 = P04 +H 21.05 AIPO.() = A1*' + PO, 14.9 PbCO3(s) = Pb*2 + CO2 4.25 CaCO3(s) = Ca*+ Co, 2 -1.20 MnCO3(s) = Mn+2 +CO, 2 6.94 FeCO3(s) = Fe2+ CO2 2.86 H(NTA) = H + H2(NTA) 20.42 H (NTA) = H + H(NTA) 2 PK H(NTA)?= H+ +(NTA) -9.0 Cat2 + (NTA)' = Ca(NTA) -18.7 Mnº? + (NTA)' = Mn(NTA) -27.0 Fe 2 + (NTA) * = Fe(NTA) -33.0 | Pb2+ (NTA) = Pb(NTA) -3.96 Alt + (NTA) = A(NTA) -12.9 Fe* + (NTA) = Fe(NTA) 19.79 HOCI = H + OCI pK 6.35 10.33 1.47 0.99 7.0 12.9 2.17 7.21 12.32 21.0 12.82 8.35 9.3 10.46 1.66 2.95 10.28 -8.2 -8.7 -9.6 -11.4 -13.4 -17.9 7.52 For the reaction: aA + bB =cC + DD, K= [C][D]d at equilibrium, Q = [ C D]d at any given moment [A]a[B]b [A]a[B]" Henry's Law: [X(aq)] = K Px [X(aq)] = concentration of gas X in water KH Henry's Law constant at specific T.P.S Px-partial pressure of gas X in solution at equilibrium Nernst Equation: E=E°+ 2.303RT log (1/0) or pe = pɛ° + 1/n log (1/0) nF pε = E/(2.303RT/F) and pɛ° = E°/(2.303RT/F), where 2.303RT/F = 0.0591 at 25°C At Equilibrium, pe = 0 = pɛ° + (1/n) log(1/K) and therefore, npɛ° = logK AGⓇ = -2.303RT log K = -2.303nRT(pɛ°) and AG -- FE = -2.303nRT (pɛ) In ([N]/[N]) = kt therefore [N]/[N]=ck and In ([X]/[X].) = at therefore [X]/[X]=e**

Answer 1

ANSWER

NTA equilibrium with PbCO3 at pH 7.00,

​Pka = -logKa

Pbcoz (s) As poet & co? Ksp = [Pb 2+ ] [ cor- . - - lig HINTA) 2- Kal H+ + (NTA) Ra= [H * ] [INTAI? Si -_ [H (NTA)2) Pb²+ +INTAY PLANTA) KA = Pb(NTA)" ippets in this fiind THt't come / Hcou 14CO3] cc (iv) Overall Reaction Pblog(s) + HINITA PENTAJfheon K : Pb(NTA) [Htcom HINTA?- from ean (1) (i), (i) and in k = Kep. Kazike - 4.06x102 - Kaze K = 4.06 x 102/ her

(a) from ean (i) Ksp = [Pb2+) [ron? os s²= Ksp 5 sakse } for Pb CO₂ (s) = 51.5lX90 13 ) Ksp = 1.57X10 = 3.89 x 107 concentration, [Pb ²+] = 3.89% 10 M concentration q Hooi = 7.50 xiom from ean (us К. (РЬ (УТА) [нсо, ) HIPOTAJ?- LPb INTA- JK HINTAO2- (H což) 4.068 10 2 7.50 x 104 .0.54 x 102 IP6TTUTAJ 597 1 HINTA ) 2 - Teacher's Signature.............