Question

6. Given a total dissolved lead concentration of 2.0x10^-5 M, a total dissolved NTA concentration of 2.0x10^-2 M, and a pH of 12.3 (note extreme pH)

a. Under these conditions, what is the dominant form of NTA?

b. Write an equation for total dissolved NTA and for total dissolved Pb. What assumptions can you make?

c. Calculate the predicted [Pb²⁺], considering NTA complexation only?

d. What is the [Pb²⁺] as predicted by the solubility of Pb(OH)₂(s) at pH 12.3?

e. Compare the value you calculated for part (d) to part (c). In the presence of NTA, what would happen to Pb(OH)₂(s) under these conditions?

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

a) the dissociation constants of NTA are

pK₁ = 3.03, pK₂ = 3.07, pK₃ = 10.70 at 20 °C

so the dominant for of NTA at pH 12.3 is that of NTA^3- form.

b)

Pb²⁺ + NTA^3-   $\rightleftharpoons$ Pb(NTA)^-

K_eq = [Pb(NTA)^-]/([Pb²⁺][NTA^3-] = 11.4

Since the equilibrium constant is quite high, significant amount of Pb²⁺ will be bound as PB(NTA)^- complex

c)

[Pb²⁺] = 2 x 10^-5 M

[NTA^3-] = 2 x 10^-2M

Pb²⁺ + NTA^3-   $\rightleftharpoons$ Pb(NTA)^-

Initial concentration 2x10^-5 2 x10^-2 0

At equilibrium, 2x10^-5-x 2 x 10^-2-x x

Given,

K_eq = [Pb(NTA)^-]/([Pb²⁺][NTA^3-] = 11.4

Therefore,

11.4 = x/((0.00002-x)*(0.02-x))

x = 11.4*(0.0000004-0.02x-0.00002x+x²)

=> 11.4 x² - 1.228228x + 0.00000456 = 0

=> x = (1.228228$\pm$((1.228228)²-4*11.4*0.00000456))^0.5)/(2*11.4)

=> x = 0.107 M or x = 3.71 x 10^-6 M

x cannot be 0.107 M since it would make Pb²⁺ and NTA^3- concentrations negative, therefore

x = 3.71 x 10^-6 M

Therefore at equilibrium,

[Pb(NTA)^-] = 3.71 x 10^-6 M

[Pb²⁺] = 0.000016 M

[NTA^3-] = 0.019996 M

Considering only NTA^3- complexation,

% of free Pb²⁺ = 0.000016*100/0.00002 = 81.4%

Thus most of Pb²⁺ is free after NTA^3- complexation

d)

pH = 12.3,

pOH = 14 - 12.3 (since pH + pOH = 14)

pOH = -log[OH^-] = 1.7

Therefore,

[OH^-] = 10^-1.7 = 0.019953 M

Pb²⁺ + 2OH^-   $\rightleftharpoons$ Pb(OH)₂

Ksp for Pb(OH)₂ = [Pb²⁺] [OH^-]² = 1.2 x 10^-15

Therefore,

[Pb²⁺] = 1.2 x 10^-15/(0.019953)²

[Pb²⁺] = 3.014 x 10^-12 M

Therefore considering only Pb(OH)₂ precipitation,

the concentration of free Pb²⁺ ions, [Pb²⁺] = 3.014 x 10^-12 M

e) Thus even in presence of NTA, Pb(OH)₂ will precipitate till the [Pb²⁺] reduces to 3.014 x 10^-12 M