Sciencemadness Discussion Board

Calculation of refractive index of binary aqueous solutions

DistractionGrating - 19-12-2014 at 11:53

I am attempting to calculate the refractive index of various salt water mixes. My understanding is that according to Moelwyn-Hughes (1961), the molal refraction R of either a pure substance or a homogenous mixture of molal volume V and refractive index n is given by R = V(n^2-1)/(n^2+2) (equation #1), and that for a binary solution of solute mole fraction y1 and solvent mole fraction y2, with partial molal refractions R1 and R2 respectively, the molal refraction of the solution is R = y1R1+y2R2 = R2+(R1-R2)y1 (equation #2). In the case of a pure substance, molal volume V = M/d, where M is the molecular weight and d is density (equation #3). In the case of a binary solution, V = 1/d * (y1M1+y2M2) = 1/d * (M2+(M1-M2)y1 (equation #4).

If you have R, M, and d, then you can calculate n = ((M+2d)/(M-d))^(1/2) (equation #5).

Attempting this calculation for the simple case of a 5% NaCl solution at 20C, I get the following:

H2O MW = 18.01528
H2O nD20 = 1.333
H2O d @ 20C = 0.9982

NaCl MW = 58.442
NaCl nD20 = 1.5442
NaCl d = 2.163

Using equation #3, V of H2O = 18.0477
Using equation #1, R of H2O = 3.7132

Using equation #3, V of NaCl = 27.0190
Using equation #1, R of NaCl = 8.5320

The mole fraction y1 of the solute in a 5% NaCl solution is (5/58.442)/(5/58.442+95/18.01528) = 0.015965. The mole fraction y2 of the solvent is 1 - 0.015965 = 0.948035.

Using equation #2, I calculate the R of the solution = 3.7901.

Using equation #4, I calculate the V of the solution = 18.04668.

Now, consulting the CRC Handbook of Chemistry and Physics to check my work, I see that a 5% solution of NaCl at 20C has a density (d) of 1.0340, and a refractive index (n) of 1.3418. Using formula #4, I calculate the molal weight M = 18.66070, and V = 18.04668, which agrees with the value I arrived at above. However, when I use formula #1 to calculate R, I get 3.8009, which differs from the value of 3.7901 I arrived at above. Furthermore, if I use equation #5 with R = 3.8009, M = 18.66070, and d = 1.0340, I get n = 1.3418, but if I substitute my calculated R = 3.7901, I get n = 1.3407 instead.

If I repeat the exercise for another solution, e.g., a 10% solution of KCL instead, I get similar results, where my answer is near the correct answer, but still quite imprecise.

So, my question is: Is the formula taken from Moelwyn-Hughes merely a crude approximation, or am I doing something wrong in my calculations somewhere?

[Edited on 19-12-2014 by DistractionGrating]

[Edited on 19-12-2014 by DistractionGrating]

DistractionGrating - 19-12-2014 at 19:47

After doing some more searching, I suspect that I may find my answer in this paper: http://www.ecd.bnl.gov/pubs/BNL-79546-2007-JA.pdf

unionised - 20-12-2014 at 03:59

The equation is pretty good, and your maths seems to be right.
But there's a difference between a sodium ion surrounded by water and a sodium ion surrounded by chloride ions.
The calculation has no way to allow for that.

DistractionGrating - 20-12-2014 at 20:27

Digging deeper into this, apparently you need empirically derived data for each solute compound in order to determine the partial molar refraction at a given concentration. Fortunately, for most of the solute compounds I'm interested in, the necessary data is in the CRC Handbook, and I can derive the necessary polynomial approximations from that data. Thanks!