I have a question regarding what happens to the solubility of $\ce{CaSO_4}$ or the rate that $\ce{CaSO_4}$ will dissolve when we add $\ce{NaCl}$ to the solution:

$\ce{CaSO_4}$ will dissolve into $\ce{Ca^{2+}}$ and $\ce{SO_4^{2+}}$:

$$\ce{CaSO_4$ $<=>$ $Ca^{2+}$+$SO_4^{2+}}$$

The ionic strength has the expression:

$$I = \frac{1}{2} \sum (m_i z_i^2)$$

where $m_i$ is the molar concentration of species $i$, and $z_i$ is the charge of species $i$.

If we add $\ce{NaCl}$ to the solution, it follows that we get more ions, and therefore the ionic strength will increase.

The expression for activity coefficients for ionic strength up to about 0.5 is:

$$\log\gamma_{i} = -A z_{i}^{2} \left({\sqrt{I}\over 1 + \sqrt{I}}\right) - 0.3 I$$

where $A$ is a constant, $\gamma_{i}$ is the activity coefficient for species $i$, and $I$ is the ionic strength. If I then use $A = 0.5$, I get:

$\gamma_{Ca_{2+}}$=0.296 at $I=0.5$

$\gamma_{Ca_{2+}}$=0.291 at $I=0.4$

$\gamma_{Ca_{2+}}$=0.297 at $I=0.3$

$\gamma_{Ca_{2+}}$=0.318 at $I=0.2$

Since we increase the ionic strength by adding $\ce{NaCl}$, is it correct that the activity product of calcium and sulfate would decrease (even though I did not get that in the above calculations), and that because of this calcium sulfate would dissolve more?

  • $\begingroup$ LaTeX tips: use \$\ce{Ca^{2+}}\$ for better formatting! $\endgroup$
    – Kenny Lau
    Commented May 14, 2016 at 9:30
  • $\begingroup$ Yes your conclusions are right. Increasing ionic strength tends to favour dissolution $\endgroup$ Commented May 14, 2016 at 10:43


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