Solution Equilibria

From MyMCAT

Introduction

Consider a glass of pure water to which a scientist begins to add salt. As the scientist stirs the mixture, salt continues to dissolve and disappear. Like many other reactions, a solid dissolving is a reversible process and the aqueous components can recrystallize back into solids. Thus, as the salt is added to the water and goes to solution some will go back to solid Eventually, as more and more is added, there will come a point at which if any more salt is added it will no longer dissolve as the reverse reaction of recrystallization has caught up. The solution has reached its saturation point and will not allow any more solid to dissolve. If any more salt is added to a solution that is at saturation no more salt actually dissolves, it just accumulates at the bottom.

This phenomena can be explained by the unique equilibrium between the solid salt and the salt's ions in solution. This is known as a solubility equilibrium.

Solubility

A substance is said to be a saturated solution if the dissolved substance is in equilibrium with some of the undissolved substance. The solubility of a substance is the equilibrium concentration of the substance in solution at a given temperature. The solubility is referred to as molar solubility when it is expressed in terms of concentration, or moles/liter.

A Quick Review of Ion Concentrations in Solution

Just as with any other chemical reaction, it is often important to determine how many moles are created/consumed and/or the concentrations of the species involved. Consider Ag₂CO₃ when it dissolves. At saturation, the solubility of Ag₂CO₃ is 1.2x10^-4. How would we determine what the concentration of the dissolved species are?

First, lets look at the reaction,

$Ag_{2}CO_{3}(s) \rightleftharpoons 2 Ag^{+}(aq) + CO_{3}^{2-}(aq)$

from this we can conclude that for every mole of Ag₂CO₃ which dissolves, 2 moles of Ag and 1 mole of CO₃ will be formed in solution.

Since we are given the concentration at which Ag₂CO₃ completely dissolves, we can conclude that the concentration of Ag will be twice as much as this, and the concentration of CO₃ will be equal to this at saturation.

	1.680
	0.168
	0.230
	0.023
→	As with the previous question, we must convert the solubility we are given as g/mL into molar terms of concentration, mol/L. To do this, we need the molecular weight of PbBr₂. Having found this to be ~367g/mol, we can then determine the molar solubility as (0.844g/100mL) * (1000mL/L) / (367g/mol) = 0.023. Notice that we also had to convert to terms of L as mL is not correct.

	0.046
	0.092
	1.81
	3.62
→	Using the previously determined molar concentration of PbBr₂, we can determine what the ratios of products will be. For every mole of PbBr₂ dissolved, twice as much Br^- is produced. (Writing out the reaction may be necessary to see this.) Since the concentration is 0.023, then the concentration of Br^- is 0.046.

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K_sp

Just as with any equilibrium, we can determine the equilibrium constant K_eq by considering the concentration of the products over the reactants to their coefficients (Reaction Quotient). For a solubility reaction it is merely the dissociation of a solid into ions.

$NaCl_s \rightleftharpoons Na_{aq}^+ + Cl_{aq}^-$

Thus for this equation, we can derive the K_eq like any other,

$K_{eq} = \frac{ \left[ Na^{+} (aq) \right] \left[ Cl^{-} (aq) \right] ... } {\left[ NaCl (s) \right]}$

This however is not quite correct, nor complete. The concentration of a solid (NaCl) is meaningless as a solid is pure and by convention in equilibrium chemistry, solids and pure liquids (H₂0) are never included in equilibrium equations. Thus the final expression, which we now call K_sp is,

$K_{sp} = \left[ Na^{+} (aq) \right] \left[ Cl^{-} (aq) \right]$

Notice that there is no fraction (nothing on the reactant side) and the K value depends only on the concentration of products. This product, as with any other equilibrium expression tells us where the equilibrium is, and thus in this case, tells us how much can be dissolved at equilibrium.

$Ag_{2}CrO_{4}(s) \rightleftharpoons 2 Ag^+(aq) + CrO_{4}^-(aq)$

2.9 x 10^-13

1.2 x 10^-12

4.4 x 10^-9

6.6 x 10^-5

→

From the reaction given, we can write out the equilibrium expression, or Ksp as $K_{sp} = (\left[ Ag^{+} \right])^2 \left[ CrO_{4}^{-2} \right]$ . This tells us that we must determine what the concentrations of Ag and CrO₄ are first. We are given what the solubility is in g/L, this if we can convert this to moles/L, we can then determine the concentration of the dissolved species. The molecular weight is 332g/mol, therefore 0.022g/L is (0.022g/L) / (322 g/mol) = 6.6 × 10^-5. One mole of reactant turns into TWO moles of Ag and ONE mole of CrO₄. We can conclude then that the concentration of Ag will be 2 x 6.6 × 10^-5, and the concentration of CrO₄ will be 6.6 × 10^-5. If we then plug these values into the Ksp expression above, $K_{sp} = (\left[ Ag^{+} \right])^2 \left[ CrO_{4}^{-2} \right] = (\left[ 2\left( 6.6 \times 10^{-5}\right) \right])^2 \left[ 6.6 \times 10^{-5} \right] = 1.2 \times 10^{-12}$ .

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Comparing Solubilities with Ksp

Consider the following two solubility reactions,

$AgI(s) \rightleftharpoons Ag^+(aq) + I^-(aq) \left( Ksp = 8.5 \times 10^{-17} \right)$ $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) \left( Ksp = 1.6 \times 10^{-10} \right)$

because the two follow the same reaction, XY turning into X and Y, it is easy to tell which one has the higher solubility. Their Ksp expressions will both be Ksp=[X][Y] and thus the larger Ksp value must have had the larger concentrations of ions, and so the higher solubility.

This unfortunately is NOT always the case, when comparing Ksp values to determine solubility one must be aware of the dissociation reaction occurring. Consider the dissociation of AgCrO₄. Its dissociation reaction is

$Ag_{2}CrO_{4}(s) \rightleftharpoons 2 Ag^+(aq) + CrO_{4}^{2-}(aq)$

and so one must recognize that its Ksp expression is now

$K_{sp} = (\left[ Ag^{+} \right])^2 \left[ CrO_{4}^{-2} \right]$

This is nothing like the previous Ksp expressions and so comparing them would lead to meaningless results of solubility.

The table belows provides examples of common dissociation reactions and how the Ksp expressions are related to the molar solubility. Notice that if two reactions are going to be compared based on solubility the Ksp expressions must be worked out to determine how they actually relate to molar solubility. Only when the have the same expression can a simple comparison of Ksp values be made, otherwise, the molar solubility of each will have to be worked out (solving for X) and then compared.

Reaction	Ksp Expression
AB -> A⁺ + B^-	Ksp = [A⁺][B^-] = X²
AB₂ -> A²⁺ + 2B^-	Ksp = [A²⁺][B^-]² = (X)(2X)² = 4X³
AB₃ -> A³⁺ + 3B^-	Ksp = [A³⁺][B^-]³ = (X)(3X)³ = 27X⁴

Where X represents what the molar solubility of the solid was.

Other possibilities are generally not encountered on the MCAT due to the complexity of solving these problems without a calculator, however questions in which one must set up the equation (without solving) may still be present. In such a case, the strategy is the same, write out the dissociation reaction, for this reaction write out a Ksp expression. Then consider what happens when one mole of reactant dissolves into the product species, what ratios are present now? Make sure to include these constants in the final equation you have constructed.

X²

4X³

16X⁴

27X⁴

→

To answer this question, one must be able to determine how Ag₂C₂O₄ dissociates. The products will be 2 Ag⁺ and 1 C₂O₄^2-, and thus the Ksp expression will be [Ag⁺]²[C₂O₄^2-]. In terms of molar solubility, this will be [2X]²[X], or 4X³. Notice that whenever a term is squared (or cubed) the concentration is also doubled (or tripled) this is due to the dissociation creating a doubled (or tripled) amount and consequently when you write a Keq (Ksp) equation, you must include the powers in which correlate to the same thing.

CuCO₃; Ksp = 2.0 x 10^-10

AgCl; Ksp = 1.8 x 10^-10

AgCN; 1.2 x 10^-16

Zn(OH)₂; Ksp = 4.4 x 10^-18

→

From the answers we are given a list of Ksp's for selected compounds. The Ksp is directly related to the molar solubility, which is what we are after, however the exact formula changes depending on how the compound dissolves. A simple AB compound (the first three cases) will yield A + B. The Ksp then would be [A][B] or [X]², where X is the molar solubility we need to compare between the compounds (this involves rearranging to see that X=sqrroot(Ksp) ). The last choice unfortunately, does not dissolve the same way, instead it follows an AB₂ -> A²⁺ + 2B^- where then the Ksp = [A²⁺][B^-]² or in terms of molar solubility (X)(2X)² = 4X³ where X again is the molar solubility we need to find.

→

The long way to answer this question would be to determine the solubility for all four compounds, however from the previous explanation it should be clear that we only really need to compare two. We need to determine the molar solubility for the best of the first three, and then the molar solubility for the last one and then compare those two. Of the first three, CuCO₃ has the largest Ksp, and so it will have the largest value for X when we solve. This value will be X = sqrroot( Ksp ) = sqrroot( 2.0 x 10^-10 ) = 1.4 x 10^-5. For the last one, Zn(OH)₂, X = cuberoot( Ksp / 4 ) = cuberoot( 4.4 x 10^-18 / 4 ) = cuberoot( 1.1 x 10^-18 ) = 1.0 x 10^-6. Thus, between the two, CuCO₃ is an order of magnitude biggest and so will be the most soluble.

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Ion Product

Just like any other system where we can calculate the reaction quotient, Q, to determine the system is going to behave with respect to the equilibrium, the same can be done for solubility expressions. In this case, Q is called the Ion Product, and will tell us if a solution is presently understaturated (IP < K_sp), at equilibrium (IP = K_sp), or supersaturated (IP > K_sp).

An important thing to remember is that when a solution is supersaturated, it does NOT mean that there is too much solid dissolved. Any extra present exists as solid and the more supersaturated the system gets the more solid at the bottom of the beaker as it cannot dissolve this!

Another important consideration is that of temperature. Saturation equilibrium, K_eq, values are always dependent on temperature, thus the solubility of a solid in solution also depends on temperature. Specifically, as temperature increases so does solubility. K_sp values are always considered to be measured at 25 Celsius unless otherwise stated.

In order to create a supersaturated solution, one typically raises the temperature of the solution to dissolve more solute, then lower the temperature back to normal. If more solute stays dissolved in the solution when returned to normal temperature than would normally be dissolved the solution is termed supersaturated.

The Common Ion Effect

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 compares the solubility expressions they are the same, thus one would assume they have the same value and so they should be able to dissolve the same amount of salt. But if we actually consider what is in the solution, 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, we have a LOT of Cl^-, which as we can see, is also in the Ksp equilibrium expression. Thus this additional acid actually changes the value of the Ion Product! If we compare IPs at any given point, the IP will be larger in the acidic solution than the simple case and thus our IP will reach the true equilibrium value K_sp much quicker. As a result, we wont be able to dissolve as much solid NaCl in the acidic solution.

This experiment can be extended to any ion. If there are any ions already present in the solution always check to see if they are the same as the ions in the K_sp, if they are, then reduced solubility will occur.

	0.0016
	0.00679
	6.79
	1.6
→	Asking for the molar solubility, in essences, is just asking us to convert to solubility into a concentration, or into the moles/L form. We are given what the solubility is as grams/L and so we must convert the grams to moles leaving us with moles/L. The molecular weight of AgBrO₃ is 235.8g/mol, and so if we divide the mass we are given, 1.6g, by 235.g/mol we are left with 0.00679 moles. Because the question stated that this was done in exactly one liter, we do not need to adjust for the volume as the answer is complete, 0.00679moles/L.

Solution Equilibria

From MyMCAT

Contents

Introduction

Solubility

A Quick Review of Ion Concentrations in Solution

K_sp

Comparing Solubilities with Ksp

Ion Product

The Common Ion Effect

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