Effects of Solutes on Water

Just knowing how compounds dissolve in water doesn’t really tell you that much practical information. This section will cover what you might expect to see happen as a result of water containing a solute, or perhaps many solutes.

Aqueous means dissolved in water. Water solvates ions or molecules according to their charge distribution and intermolecular interactions; the solute need not have started as a solid. Dissolving a salt can separate its lattice ions, while dissolving a molecular substance can disperse intact molecules. State the species instead of calling every solute a molecule.

You have covered that different molecules have differing solubilities, but all those solubilities are for a pure solution. That is, a solution of only water and that compound. What happens, then, when you have multiple compounds dissolved in water?

Adding another dissolved salt can change an equilibrium, but not because water molecules are simply used up. In a concentration model, a common ion changes the reaction quotient and can reduce the dissolution of a sparingly soluble salt. A spectator electrolyte that supplies no common ion can change activities through ionic-strength effects; that is a distinct, more advanced issue. Do not infer a universal decrease from any arbitrary added salt.

Ionic strength weights each dissolved ion’s concentration by the square of its charge. It helps describe how interactions change activities, rather than providing a universal “more ions means less soluble” rule. Under some conditions an inert electrolyte can increase molar solubility. For AP concentration calculations, state the ideal approximation and analyze common-ion mass action separately.

Ionic strength actually goes past just the number of ions, and it also looks at the charge on the ions. Therefore, higher charges result in higher Ionic Strengths. The formula for determining Ionic Strength is as follows:

$I = \frac12 \cdot\sum_{i=1}^Nc_i\cdot z_i^2$

The ci refers to the ion’s concentration, and the zi refers to the ion’s charge number (the absolute value of the charge). The summation function just shows that you add up the effects of all the ions in the water, for N ions.

Keep two effects separate. A common ion alters the concentration quotient in the simple Ksp model. A salt without a common ion can alter activity coefficients, and the direction is not determined merely by counting total ions. Quantitative activity corrections are enrichment; the required AP task is to explain and calculate the stated ideal common-ion equilibrium.

Reviewed enrichment calculation: compare the formal ionic strengths of (a) 1.00 M NaCl plus 0.750 M Mg(NO₃)₂ and (b) 2.00 M KOH, assuming complete dissociation and stated final concentrations. This asks only for ionic strength, not a solubility ranking. Neither solution by itself specifies enough chemical equilibrium information to rank the solubility of the original incorrectly written copper sulfate.

The salt effect isn’t the only way a compound’s solubility can be affected. There’s one more major effect known as the common ion effect. This occurs when you have two compounds in a solution which both contain the same ion. One example would be if you were to dissolve NaCl in a solution of CuCl. Both Compounds contain Chlorine, therefore you should see the common ion effect.

The common ion effect serves to lower the solubility of a compound depending on the concentration of the ion already present in the solution.The reason this happens stems from Le Chatelier’s Principle. To figure out why, let’s look at the aforementioned case.

For a clear common-ion example, compare NaCl(aq), which supplies Cl⁻, with sparingly soluble CuCl(s), copper(I) chloride. Cupric chloride means copper(II) chloride, CuCl₂, and is a different formula. The common-ion calculation must identify the correct salt and oxidation state.

$NaCl_{(s)}\rightarrow Na^+_{(aq)}+Cl^-_{(aq)}$

$CuCl_{(s)}\rightarrow Cu^+_{(aq)}+Cl^-_{(aq)}$

If you imagine a solution of just NaCl, you notice that it contains water, Na+ and Cl-. Now if you were to add in some CuCl, it would also dissolve. However, there is already some Cl- in the solution. This means that the equilibrium of that dissolution equation will be pushed towards the CuCl(s) side, meaning that the solubility will decrease.

For a sparingly soluble salt, a balanced dissolution equation and itsKsp can connect ion concentrations to saturation under stated conditions. For A_aB_b(s)⇌aA(aq)+bB(aq), the powers inKsp are the stoichiometric coefficientsaandb. They are not defined simply as the ion charge numbers. Identify the actual ion formulas, charge balance and any initial common-ion concentration before calculating a molar solubility.

For a sparingly soluble salt A_aB_b(s)⇌a A(aq)+b B(aq), the ideal concentration expression is Ksp=[A]^a[B]^b. The exponents are the balanced stoichiometric coefficients, not automatically the charge numbers. Specify the ion charges in the balanced dissolution equation and omit the pure solid.

In this formula, $[A]$ and $[B]$ are the concentrations at saturation for the salt.

The dissolved ion concentrations are equal only for a 1:1 salt in a solution that has no other sources of those ions. For CaF₂, dissolving s mol L⁻¹ supplies [Ca²⁺]=s and [F⁻]=2s in the simple model. With a common ion already present, add its initial concentration to the stoichiometrically produced amount.

Reviewed common-ion replacement: find the molar solubility of AgCl at a temperature where Ksp=1.80×10⁻¹⁰ in 0.0100 M NaCl. Assume complete NaCl dissociation, ideal concentrations, unchanged final volume and no other silver complexes. Then s(0.0100+s)=Ksp; because s≪0.0100, s≈1.80×10⁻⁸ M.

One other topic we have yet to cover is the effect that solutes have on the boiling point and freezing point of their solvent. You might be surprised to know that while water has a boiling point of 100ºC, salt water has a boiling point above 100ºC. How far above depends on how salty the water is. A simple relationship exists for the calculation of the boiling point elevation and freezing point depression.

$\Delta T_{FP} = K_F\cdot b \cdot i$

$\Delta T_{BP} = K_B\cdot b \cdot i$

In these equations, b refers to the molality of the salt and i refers to the number of ions in the salt. If you don’t remember, molality is the moles of the salt per the mass of water (in kg). The KB and KF values refer to constants of the solvent. For water, KB is 0.512 and KF is 1.853. Both have units of $\text{º}C\cdot kg/mol$. It might be useful to remember that water has a density of 1kg/L.

The original 50%-by-mass NaOH colligative calculation is superseded as a real-solution prediction. Dilute ideal freezing/boiling relations cannot be applied quantitatively at that loading without major nonideality and phase information. Colligative properties and molality are beyond required AP coverage. A defensible enrichment exercise must supply a dilute model, solute amount, solvent mass and an effective van’t Hoff factor with its assumptions.