Change of temperature upon dilution
At such low concentrations, can’t you just assume an ideal solution? Then the  change in entropy from mixing would be $\Delta S = -R(x_A\ln x_A + x_B \ln x_B)$. But because the solution is so dilute, you can see that both terms are going to be essentially zero. (For a 10 nM solution, the contribution from entropy of mixing is on the order of $10^{-8}$ J K$^{-1}$ mol$^{-1}$.)

The dilution process would thus lower the total entropy, since there’s less uncertainty in the system. (You’d get the value you compute at 10 nM, the final state, minus the value at 10 µM, the initial state.)