# Why is rate of dissolution constant?

1. May 20, 2012

### sodium.dioxid

A(s) <---> B(aq)
For the dissociation of A into ions, the rate law predicts a constant rate:
rate = k [solid], where the concentration of the solid is constant.
But this doesn't make sense because as the solid is becoming smaller, surface area is decreasing. Less surface area means less molecules breaking off at any interval of time (lower rate). What gives?

2. May 20, 2012

### Mike H

Hmmm. Could be a few possibilities, right off the top of my head -

1.) There are more sophisticated expressions for the rate of dissolution that take surface area into account - where did you find this one?;

2.) one could be working with some sort of normalized expression for the rate of dissolution which already has taken surface area into account; or

3.) one is dealing with a solid that dissolves very slowly, so the change in surface area is negligible under the time scales involved of the measurements.

3. May 20, 2012

### sodium.dioxid

The rate law in question is not something I got from anywhere. I just used my previous knowledge of formulating rate laws and applied it in this situation. From consulting many textbooks, they introduce equilibrium constant and it's relation to the rate law for gases, but solubility product constant is randomly presented without it's relation to solution-related rates. Since equilibrium constants and rate laws were introduces in context with gases, one would think that the same relationship further applies to solutes, especially when there is a lack of warning against this exact practice on the book's part. In short, I have realized that the rate law is not what I have written.

4. May 21, 2012

### Staff: Mentor

Dissolution before saturation is not an equilibrium process, so the solubility products doesn't apply. Once the solution is saturated amount of solid (or rather its surface) doesn't matter, as both dissolution and precipitation occur at the same surface - so for both processes surface is identical.

5. May 22, 2012

### sodium.dioxid

Thank you Borek. I just looked up a textbook by Zumdahl and it had the same explanation.