Solution Equilibria and the Common Ion Effect

From MyMCAT

Introduction

While Solution Equilibria discusses the concepts of dissociation equilibria and its related Ksp, it does not provide a complete picture (or the complete understanding necessary for the MCAT). One last topic remains, and this is the concept of the common ion effect and how it relates to solubility.

Consider the scientist in the first example dissolving NaCl. How would the example compare if the scientist did a second solubility experiment, this time in hydrochloric acid instead of water? If one looks at the the solubility expression, there is no change between them as it is the same reaction, NaCl dissolving but assuming then that both solutions will dissolve the same amount of salt is false!

Why is this the case if there equilibrium expressions and Ksp expressions are the same? First lets consider the solutions themselves and then lets lets examine the Ksp expressions more closely. In the water we have a LOT of H₂O and a negligible amount of H⁺ and Cl^-. In the acid solution on the other hand, the HCl produces a LOT of Cl^- but Cl^- also comes from the salt, so right away we should be able to recognize that there is a complex of interactions occuring here. It is no longer a simple dissociation between salt and ions as one of the ions is also coming from somewhere else. Le Chatelier's Principle states that an equilibrium will counteract any shifts in the equilibrium position, seeing as how the dissociation reaction of the salt IS an equilibrium, we can consider the extra Cl^- (from the HCl) a shift the equilibrium wants to oppose. In this direction, the equilibrium will go towards the reactants, or the solid NaCl. The common ion effect will ALWAYS reduce the solubility of the solid in question.

Now let us consider the Ksp expression for the above example,

Again, this expression will be the same for both the salt in pure water and salt in HCl solutions, but what about the actual numbers? If we were to measure the concentrations of the ions involved, are they the same? No. Clearly the Cl^- must be different (as we added a lot of Cl^- in the HCl case) and as such, the equilibriums will shift resulting in a change to the other ions involved.

An example in action

Let's consider the dissociation of AgCl in the presence of a common ion. Firstly, the Ksp for AgCl (at 25C) is 1.77 x 10¯¹⁰, and if we wanted we could solve for what the concentration of the ions in the dissolved solution would be (try it yourself). But more importantly, lets say that solution is already 0.01M in [Cl^-] (from some previously dissolved NaCl). As usual, the Ksp expression is still as follows,

but this time, what should we put for numbers if we are solving for the molar solubility? Normally, we reason that if x moles of AgCl dissolves we get x moles of Ag⁺ and x moles of Cl^-, but this time there is already some Cl^- so we will have to take this into account and include it,

Do you notice the change? its no longer a simple x², but rather a complex quadratic (that may be quite difficult to solve). Because the MCAT does not allow calculators we must make a rather bold simplification. Lets assume that the x terms is much smaller than the 0.01. So small that we could just ignore x in that addition and still get close to the same answer when we multiply it all out.

Solving this expression with the known Ksp value we can derive x to be x = 1.77 x 10^-8 M. If one actually did the original case of that in pure water they would see that in that case the solubility was 1.33 x 10^-5. 10^-8 compared to 10^-5 is clearly a huge drop in the molar solubility and again to reemphasize, the common ion effect will always reduce solubility.

1. The Ksp of PbI₂ is 8.7 x 10^-9 at 25°C. What would the molar solubility of PbI₂ be at the same temperature if it were dissolved in 0.1M NaI?

	8.7 x 10-7
	1.7 x 10-10
	34 x 10-8
	9.3 x 10-5
→	As with all Ksp questions, we should first write out the dissociation reaction and from that determine the Ksp expression.
→
→	From this it follows then that the Ksp will be
→
→	In a basic problem, we would assume that one mole of PbI2 forms one mole of Pb and two mole of I thus we would have equal concentrations of "x" to put into the Ksp expression, however in this case we must keep in mind that there is already a known amount of I- present. Therefore we will get
→
→	notice the iodide terms is "2x+0.1". This is because we get TWO moles of I- for every mole dissolved AND we are starting with 0.1 already present.
→	If we then make the simplification that 0.1 is much bigger than 2x, we can reduce it to something easily solvable without a calculator
→
→	Thus x = 8.7 x 10-7

2. What is the molar solubility of barium flouride (Ksp = 1.0 x 10^-6) in a solution that is 150mM NaF at 25C?

	2.25 x 10-2
	4.44 x 10-3
	2.25 x 10-4
	4.44 x 10-5
→	First, we should write out what barium fluoride is and what its dissociation reaction looks like
→
→	At equilibrium we will have x moles of Ba and "2x + 0.15" of F.
→	Now let us write out the Ksp expression
→
→	Thus x = 4.44 × 10-5

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Solubility and pH

A special case of solubility and the common ion effect in action occurs when the ions involved are either themselves H⁺, OH^-, or can somehow interact in a acid/base reaction. Consider the solubility equilibrium of calcium hydroxide, Ca(OH)₂

What will happen if a strong base, such as NaOH, is added to this equilibrium? Well, the OH^- which is released is part of the equilibrium already and so we would expect the common ion effect to occur, reducing the solubility. This is exactly the same as the previous examples, so lets turn our attention to the other possibility.

What will happen if a strong acid, such as HCl, is added to this equilibrium? Well H⁺ can easily react with OH^- to form water, and thus we are effectively removing OH^- from the original equilibrium. According to Le Chatelier's principle then, the equilibrium will shift towards the side of the loss, which is the products side, and so more of the solid must dissolve! Notice what has occurred here, decreasing the pH (making it more acidic) has actually increased the solubility.

1. What is the solubility of Ca(OH)₂ (Ksp = 8.0 x 10^-6) in a solution that buffered at a pH of 12?

	0.004 M
	0.008 M
	0.00004 M
	0.00008 M
→	First, we should write the dissociation reaction for Ca(OH)2
→
→	Now let us write out the Ksp expression
→
→	Now we must consider the significances of a pH of 12. Knowing the pH implies we know what the concentration of OH is. pOH = 14 - 12 = 2. antilog( -2 ) = 1.0 x 10-2. So we know the concentration of OH is 1.0 x 10-2 M. Furthermore, if we let the molar solubility of Ca(OH)2 equal x, what we are interested in finding, than we know that we will get x moles of Ca2+ when it dissociates (plus x moles of OH, BUT this is already taken into account by our pOH calculations!). Using the Ksp value given then, we have enough values to solve the problem,
→
→	Thus x = 0.08 M. Notice that once all the elements are in place in the Ksp expression solving this is actually quite straight forward and does not require making any assumptions/simplifications to remove quadratics. This is vary common to these problems because we are generally given the pH/pOH and thus we already know the concentration of one of the terms.

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In general, this pH effect will occur not only with H⁺ and OH^-, but also with ions which come from weak acids/bases. These conjugate acids/bases released through the dissociation will go on to react with water and shift the original equilibrium futher. Salts which contain anions that do not hydrolyze, such as Cl^-, Br^-, I^-, NO₃^- (in other words anions from strong acids) are not affected by pH however any salt which contains ions that are themselves weak acids or bases will result in a pH dependent solubility effect.