Freeze depression point, sugar in water, how do I calculate the activity?

Question

I will give some background on the concept for context and then state my question at the end:

Consider a solution containing n_A and n_B moles of solvent A and solute B, respectively. According to the Gibbs-Duhem equation:

n_Bd ln a_B = −n_Ad ln a_A (1), where a denotes the activity.

For a solution, A in this case an aqueous solution H₂O, with a molar concentration m of component B, in this case sucrose so we use the denominator S, n_S = m mol and n_H2O = 1000/18,016 = 55.51 mol (the molal concentration of pure water) which leads to

md ln a_S = −55, 51d ln aH2O, (2), which can also be written as d ln a_S = (55,51/m)*d(− ln a_H2O), (3).

If we have a set of values for 55.51/m and (− ln a_H2O) then the differential equation can be integrated between two different concentrations 1 and 2 with activities a_1,H2O, a_1,S and a_2,H2O, a_2,S,

ln a_S,2 − ln a_S,1 = ∑ (55,51/m)*d(− ln a_H2O). (4)

At a sufficiently low concentration of m_i (here ≤ 0.1 mol kg⁻¹), ideality applies, in which case a_S,1 = m_i, which means that

ln a_S,2 = ln m_i + ∑ (55,51/m)*d(− ln aH2O) = ln m_i + I (5)

where I corresponds to the integral, representing how much the activity deviates from an ideal solution (the concentration).

The question:

How do I calculate a_B,2 from that given formula (5). My first thought was a_B,2 = m_i + e^I, But that did not work well.

Answer 1

Certainly! Let's break it down without using formatted formulas.

Given Information

1. The Gibbs-Duhem equation for a solution is: nB * d(ln aB) = -nA * d(ln aA)

where:

nA and nB are the moles of solvent A and solute B, respectively.

aA and aB are the activities of the solvent and solute.

For an aqueous solution of sucrosemd(ln aS) = -55.51d(ln aH2O)

where:

1. m is the molal concentration of sucrose.
2. aS is the activity of sucrose.
3. aH2O is the activity of water.
4. Rearrange the equation:d(ln aS) = (55.51/m) * d(-ln aH2O)

Integrating the Equation

Integrate both sides from concentration 1 to concentration 2: ln aS2 - ln aS1 = ∫(55.51/m) * d(-ln aH2O)

For sufficiently low concentration mi (here ≤ 0.1 mol kg^-1), ideality applies:ln aS1 = ln mi

So the equation becomes:ln aS2 = ln mi + ∫(55.51/m) * d(-ln aH2O)

Let I represent the integral:I = ∫(55.51/m) * d(-ln aH2O)

Therefore ln aS2 = ln mi + I

Calculating aS2

1. Exponentiate both sides to solve for aS2: aS2 = exp(ln mi + I)
2. Since exp(ln mi) = mi: aS2 = mi * exp(I)

Example Calculation

Assume you have the following data points for -ln aH2O versus m:

m (mol kg^-1)-ln aH2O
0.01	0.0002
0.02	0.0004
0.05	0.0010
0.10	0.0020

1. To integrate ∫(55.51/m) * d(-ln aH2O), use numerical integration methods like the trapezoidal rule.
2. If numerical integration gives you I, you can then use the formula:aS2 = mi * exp(I)

This will give you the activity aS2 based on the initial molal concentration mi and the integral I. Make sure to perform the integration accurately to get a reliable value for I.