Solubility Equilibrium - Dissolution With Dissociation

Dissolution With Dissociation

Ionic compounds normally dissociate into their constituent ions when they dissolve in water. For example, for calcium sulfate:

As for the previous example, the equilibrium expression is:

$K^\ominus = \frac{\left\{\mbox{Ca} ^{2+}(aq)\right\}\left\{\mbox{SO}_4^{2-}(aq)\right\}}{ \left\{\mbox{CaSO}_4(s)\right\}} =\left\{\mbox{Ca} ^{2+}(aq)\right\}\left\{\mbox{SO}_4^{2-}(aq)\right\}$

where K is the thermodynamic equilibrium constant and braces indicate activity. The activity of a pure solid is, by definition, equal to one.

When the solubility of the salt is very low the activity coefficients of the ions in solution are nearly equal to one. By setting them to be actually equal to one this expression reduces to the solubility product expression:

The solubility product for a general binary compound A_pB_q is given by

A_pB_q pAq+ + qBp-

K_sp = pq (electrical charges omitted for simplicity of notation)

When the product dissociates the concentration of B is equal to q/p times the concentration of A.

= q/p

Therefore

K_sp = p (q/p)q q

=(q/p)q × p+q

The solubility, S is 1/p . One may incorporate 1/p and insert it under the root to obtain

$S = { \over p} = { \over q} = \sqrt{K_{\mathrm{sp}} \over {(q/p)^q} p^{p+q}} = \sqrt{K_{\mathrm{sp}} \over {q^q} p^p}$

Examples

CaSO₄: p=1, q=1,

Na₂SO₄: p=2, q=1,

Al₂(SO₄)₃: p=2, q=3,

Solubility products are often expressed in logarithmic form. Thus, for calcium sulfate, K_sp = 4.93×10−5, log K_sp = -4.32. The smaller the value, or the more negative the log value, the lower the solubility.

Some salts are not fully dissociated in solution. Examples include MgSO₄, famously discovered by Manfred Eigen to be present in seawater as both an inner sphere complex and an outer sphere complex. The solubility of such salts is calculated by the method outlined in dissolution with reaction.

Read more about this topic: Solubility Equilibrium

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