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Solubility Product Calculations

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Our changing oceans are a worrying problem. We must take action to prevent not only rising sea levels and temperatures, but also to limit ocean acidification. One consequence of ocean acidification is the dissolution of coral reefs. Coral reefs are made up of **calcium carbonate**, a slightly soluble ionic compound. Calcium carbonate is fairly insoluble in neutral solutions, but as the pH drops, its solubility increases dramatically. Predicting how future ocean acidification will destroy coral reefs and impact our aquatic ecosystems is just one example of **solubility product calculations.**

- This article is about
**solubility product calculations**in physical chemistry. - We'll recap the
**relationship between solubility and the solubility product**before looking at**solubility product expressions**. - We'll then focus on
**different types of solubility product calculations**. - This will involve
**calculating the solubility product from solubility**,**calculating solubility from the solubility product**,**predicting precipitation**, and**calculating solubility using the common ion effec**t. - We'll teach you the
**method**for each style of question, then let you practice your skills with the help of our**worked examples**.

In the article **Solubility Product**, you learned the definition of **solubility**.

**Solubility** is the maximum amount of a species that can dissolve in a given volume of solvent. In other words, it is the amount of a compound needed to completely saturate a solution.

We measure solubility in the **number of grams or moles of solute** that can dissolve in either **100 grams (1 decilitre) or 1 kilogram (1 litre or 1 decimetre cubed) of solvent**.

In solubility calculations, you typically work with solubility in** mol dm**^{-3}.

You should have also learned that **different ionic species have varying solubility**. Some are **fully soluble** in solution, whilst others are only **slightly soluble **and readily form **saturated solutions**. Any system containing an ionic solid in contact with its own saturated solution forms a **solubility equilibrium**. Like with all equilibria, solubility equilibria can be represented with an **equilibrium constant**. This particular equilibrium constant is known as the **solubility product**.

The **solubility product** is the equilibrium constant for the dissolution of an ionic species in a solubility equilibrium. It tells you about the **relative concentration of aqueous ions **in a dissolution reaction at equilibrium.

The solubility product gives us a clue about a species' solubility. However, solubility and the solubility product are *not* the same thing. To help you understand the differences between these two terms, we've made a handy comparison table.

Solubility | Solubility product | |

Definition | The maximum amount of a species that can dissolve in a given volume of solvent | The equilibrium constant for the dissolution of an ionic species in a solubility equilibrium. |

Units | Variable, typically mol dm^{-3} | A variable exponential involving mol dm^{-3} |

Constant/variable | Variable for a certain reaction at a specific temperature | Constant for a certain reaction at a specific temperature |

Affected by | Temperature, pH, concentration | Temperature |

Before we have a go at any solubility product calculations, we need to make sure that we are confident in finding expressions for the solubility product, no matter the dissolution reaction.

For the general solubility equilibrium \(A_aB_b(s)\rightleftharpoons aA^{b+}(aq)+bB^{a-}(aq)\) , the solubility product has the following expression

$$K_{sp}={[A^{b+}]_{eqm}}^a\space {[B^{a-}]_{eqm}}^b$$

It is easy to derive a solubility product expression yourself for any solubility equilibrium. Here's how:

- Write a balanced solubility equilibrium equation.
- Take the equilibrium concentration of each of the aqueous ion products.
- Raise each equilibrium concentration to the power of its molar coefficient in the balanced equilibrium equation.
- Multiply these terms together.

You also need to find the units of the solubility product. They vary according to the exact reaction:

- Substitute the units you used for the equilibrium concentrations of aqueous ions into each term in the solubility product expression. If not given, the typical units of concentration are mol dm
^{-3}. - Expand each term according to its exponential, then simplify the expression to find your final units.

Try your hand at the questions down below. If you get stuck, don't worry - our worked answers walk you through the solution.

**Find solubility product expressions and units for the solubility equilibria involving the following ionic species: **

**MgCO**_{3}**CaF**_{2}

First, let's consider part a. One mole of MgCO_{3} dissolves into one mole of Mg^{2+} ions and one mole of CO_{3}^{2-} ions:

$$MgCO_3(s)\rightleftharpoons Mg^{2+}(aq)+CO_3^{2-}(aq)$$

To find an expression for the solubility product, we take the equilibrium concentrations of each of the aqueous ion products and raise them to the power of their molar coefficients in the balanced equilibrium equation. We then multiply the two terms together. In this case, both molar coefficients are 1. Raising something to the power of 1 doesn't change its value, and so we get the following expression:

$$K_{sp}=[Mg^{2+}]_{eqm}\space [CO_3^{2-}]_{eqm}$$

To find the units of the solubility product, we substitute the units of concentration into the solubility product expression:

$$units=(mol\space dm^{-3})\space (mol\space dm^{-3})$$ $$units=mol^2\space dm^{-6}$$

That's part a done. In part b, we need to pay attention to the molar coefficients in the balanced equilibrium equation. One mole of CaF_{2} dissolves into one mole of Ca^{2+ }ions but *two* moles of F^{-} ions:

$$CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-{aq}$$

This means that in our solubility product expression, the equilibrium concentration of Ca^{2+} doesn't change but the equilibrium concentration of F^{-} is raised to the power of 2. We end up with the following expression:

$$K_{sp}=[Ca^{2+}]_{eqm}\space {[F^-]_{eqm}}^2$$

Here are the units:

$$units=(mol\space dm^{-3})\space (mol\space dm^{-3})^2$$ $$units=mol^3\space dm^{-9}$$

Both of these expressions should help you with our solubility product calculations in the rest of the article. If you're ready, we'll look at some now.

We'll spend the rest of this article looking at **different types of solubility product calculations**. For each style of question, we'll provide you with a step-by-step method and walk you through a worked example. We suggest you give the question a go by yourself first, then check out our solution if you get stuck.

We'll focus on four types of solubility product calculations:

- Calculating the solubility product using solubility.
- Calculating solubility using the solubility product.
- Predicting precipitation using the solubility product and concentration.
- Calculating solubility in systems involving the common ion effect.

The first type of calculation involving solubility equilibria that you should be able to do entails **finding the solubility product using a species' solubility**.

How does solubility help us find the solubility product? Well, remember that **solubility** is the maximum amount of solute that dissolves in a solution. In other words, tt is the concentration of dissolved solute that we'll find in a system at equilibrium. If we know the concentration of solute dissolved in the equilibrium, then we can use the balanced equilibrium equation to find the equilibrium concentrations of aqueous solute ions. We then substitute these into the solubility product expression to reach our final answer.

Here are the steps:

- Write a balanced equilibrium equation.
- Write an expression for the solubility product.
- Use solubility to work out the equilibrium concentrations of the aqueous ions in the equilibrium equation.
- Substitute the equilibrium concentrations into the expression for the solubility product and simplify your answer by expanding the brackets. This gives you a numerical value for the solubility porduct.
- Substitute the units of the equilibrium concentration of aqueous ions into the expression for the solubility product and simplify your answer by expanding the brackets. This gives you the units of the solubility product.

**Calculate the solubility product of a saturated solution of CaF _{2}, which has a solubility of 2.05 **

First, we need to write an equilibrium equation and a solubility product expression for this species. Fortunately, we already found those out earlier in the article:

$$CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-{aq}$$ $$K_{sp}=[Ca^{2+}]_{eqm}\space {[F^-]_{eqm}}^2$$

The next step is to find the concentrations of aqueous ions in solution. We do this using the solubilty given to us in the question. Here, CaF_{2} has a solubility of 2.05 × 10^{-4} mol dm^{-}^{3}, which simply means that 2.05 × 10^{-4} mol dm^{-}^{3} CaF_{2} dissolves to form a saturated solution at equilibrium. Our equilibrium equation tells us that for each mole of CaF_{2} that dissolves, we end up with *one* mole of Ca^{2+} ions and *two* moles of F^{-} ions. As a result, we can work out the equilibrium concentrations of Ca^{2+} and F^{-}:

$$[Ca^{2+}]_{eqm}=2.05\times 10^{-4}\space mol\space dm^{-3}$$ $$[F^-]_{eqm}=2(2.05\times 10^{-4})=4.10\times 10^{-4}\space mol\space dm^{-3}$$

Now we need to substitute these values into the solubility product expression:

$$K_{sp}=(2.05\times 10^{-4})\space (4.10\times 10^{-4})^2$$$$K_{sp}=3.45\times 10^{-11}$$

Finally, we need units for the solubility product. We do this by substituting the units used for our equilibrium concentrations of aqueous ions into the solubility product expression. Fortunately, we worked out the units for this particular solubility product earlier in the article:

$$units=(mol\space dm^{-3})\space (mol\space dm^{-3})^2$$ $$units=mol^3\space dm^{-9}$$

Our final answer is therefore **3.45 × 10 ^{-11} mol^{3} dm**

We only carry out solubility product calculations, no matter the type or style, using **slightly soluble species**. First of all, fully soluble species have a high solubility and so they rarely reach equilibrium - the idea of the solubilty product simply doesn't apply to them. Secondly, even if soluble species *do* reach equilibrium, the concentration of aqueous ions becomes so high that other ion-ion interactions interfere with our calculations.

Working out the solubility product using solubilty is just one example of a solubility product calculation you can do. You can also work in the other direction and **find a species' solubility using the solubility product itself**. Here's the process:

- Write a balanced equilibrium equation.
- Write an expression for the solubility product.
- Assume that your ionic species has a solubility of x mol dm
^{-3}. - Find the equilibrium concentrations of aqueous ions in terms of x and substitute these into the solubilty product expression.
- Simplify and solve for x to find a value for solubility.

Sound complicated? We promise it isn't that bad. Here's an example to show you how these calculations are done.

**Work out the solubility of MgCO _{3} in mol dm^{-}^{3}. Use the following equilibrium equation to help you.**

$$MgCO_3(s)\rightleftharpoons Mg^{2+}(aq)+CO_3^{2-}(aq)\qquad K_{sp}=6.82\times 10^{-6}\space mol^2\space dm^{-6}$$

First, we need to write an equilibrium equation for this particular solubility equilibrium. Fortunately, this has been done for us. We can instead move on to writing a solubility product expression. Once again, we did this earlier on in the article, so take a look back to the first worked example if you can't quite remember how we got to our answer.

$$K_{sp}=[Mg^{2+}]_{eqm}\space [CO_3^{2-}]_{eqm}$$

Now, assume that the solubility of MgCO_{3} is x mol dm^{-3}. This means that x mol dm^{-3} MgCO_{3} dissolves in solution at equilibrium. The molar coefficients in the balanced equilibrium equation tell us that for each mole of MgCO_{3} that dissolves, we end up with one mole of Mg^{2+} ions and one mole of CO_{3}^{2- }ions. Therefore:

$$[Mg^{2+}]_{eqm}=x\space mol\space dm^{-3}$$ $$[{CO_3}^{2-}]_{eqm}=x\space mol\space dm^{-3}$$

We know what the solubility product is for this particular reaction - we were given it in the question. We can now substitute the equilibrium concentrations in terms of x into the solubility product expression and set them equal to the solubility product itself. When we expand the brackets, we end up with a value for x:

$$K_{sp}=(x)\space (x)$$ $$6.82\times 10^{-6}=(x)^2$$ $$x=2.61\times 10^{-3}\space mol\space dm^{-3}$$

How does this help us find solubility? Well, remember that we assumed the solubility of MgCO_{3} was x mol dm^{-3}. We know x - we just worked it out! Therefore, the solubility of MgCO_{3} equals **2.61 ****× 10 ^{-3} mol dm**

Sometimes, mixing two unsaturated solutions of different salts causes a new salt to form and precipitate out of solution. **P****redicting precipitation** is easy when we know the new salt's solubility product. If we substitute the concentrations of aqueous ions into the new salt's solubility product expression and we get an answer *greater* than its solubility product, we know that the salt will precipitate. This is because the solubility product must always remain **constant for a certain reaction at a specific temperature**. The system forces the concentrations of aqueous ions to decrease by turning them into a solid salt in order to keep the solubility product the same.

We've summarised the method for you below:

- Write a balanced equilibrium equation for the new salt that could form.
- Write an expression for the new salt's solubility product.
- Use the information in the question to find the concentrations of the new salt's aqueous ions.
- Substitute these concentrations into the solubility product expression.
- Compare your answer to the new salt's actual slubility product and deduce whether the salt precipitates or not.

Try the following problem and see how your answer matches up to our worked solution.

**A 4.00 ****× 10**^{-2}** ****mol dm**^{-3}** ****solution of ****KCl is mixed with a 2.50 ****× 10**^{-3}** ****mol dm**^{-3}** solution of AgNO _{3}_{ }in a 1:1 ratio. Predict whether the slightly soluble salt AgCl precipitates out of solution or not. The solubility product for AgCl equals 1.77 **

First, we write an equilibrium equation and solubility product expression for the dissolution of the new salt, AgCl:

$$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$$ $$K_{sp}=[Ag^+]_{eqm}\space [Cl^-]_{eqm}$$

Next, we consider the concentrations of the new salt's aqueous ions, which are Ag^{+ }and Cl^{-}. Ag+ ions come from the 2.50 × 10^{-3} mol dm^{-3} AgNO3 solution; Cl- ions come from the 4.00 × 10^{-2} mol dm^{-3} KCl solution. The two solutions are combined in a 1:1 ratio, which means that the concentrations of all their ions are halved. Hence:

$$[Ag^+]=\frac{2.50\times 10^{-3}}{2} =1.25\times 10^{-3}$$ $$[Cl^-]=\frac{4.00\times 10^{-2}}{2} =2.00\times 10^{-2}$$

Let's now substitute these concentrations into the solubility product expression:

$$(1.25\times 10^{-3})\space (2.00\times 10^{-2})=2.50\times 10^{-5}\space mol^2\space dm^{-6}$$

Compare this value to the actual solubility product given in the question. We find that it is larger than the solubilty product, and so the salt precipitates.

$$2.50\times 10^{-5}>1.77 × 10^{-10}$$

You're almost done with solubility product calculations. We have one final application of the solubility product to look at: **Calculating solubility in solutions involving common ions**. In these equilibrium systems, we need to consider the **common ion effect**.

A **common ion** is an ion found in two different compounds. The **common ion effect** states that ionic species are less soluble in solutions containing a common ion.

In **Solubility Product**, you learned that the solubility of an ionic species *decreases* if the solvent contains a common ion. For example, MgCO_{3} is less soluble in a solution of the fully soluble salt MgSO_{4} than in pure water. This phenomenon can be explained by **Le Chatelier's Principle** and the solubility product.

- The solubility product is
**constant for a certain reaction at a specific temperature**. - Adding an aqueous common ion to a saturated solution of a slightly soluble salt increases the concentration of dissolved salt and disturbs the position of the equilibrium.
- This causes the position of the equilibrium to shift to the left in order to counteract the disturbance, meaning that more of the dissolved salt precipitates.
- Thus, equilibrium is resotred and the salt's solubility decreases.
- Overall, the value of the solubility product remains constant.

Calculating solubility in solutions involving common ions uses a similar method to calculating solubility in pure water, which we looked at just a second ago. However, there are a few changes. Like before, we assume that the solubility of the salt in question is x mol dm^{-3}. This allows us to find the concentration of the non-common ion in terms of x. But this time, we need to consider the concentration of the common ion at equilibrium using the information about the solvent given in the question. Here's how:

- Write a balanced equilibrium equation.
- Write an expression for the solubility product.
- Assume that your ionic species has a solubility of x mol dm
^{-3}. - Find the equilibrium concentration of the
*non-common*aqueous ion in terms of x. - Find the equilibrium concentration of the
*common*aqueous ion using the concentration of the solvent given in the question. - Substitute both equilibrium concentrations into the solubility product expression, then simplify and solve for x to find a value for solubility.

Remember that we are finding the solubility of a **slightly soluble salt**, which has a low solubility. As a result, its concentration of dissolved ions at equilibrium is very low. But this is helpful because it means we can pretend that the concentration of the *common ion *is entirely due to the solvent. We simply ignore any common ions that come from the dissolved salt, because their number is insignificantly small compared to the number from the solvent. This makes our calculation a lot simpler!

Have a go at the following problem.

**1.23 × 10 ^{-4} moles of AgOH dissolves in 1.00 dm of pure water, giving it a solubility of 1.23 × 10^{-4} mol dm^{-3 }and a solubility product of 1.52 × 10^{-8} mol dm^{-3}. Calculate its solubility in a 2.00 mol dm^{-3} solution of NaOH.**

First, we write an equilibrium equation and a solubility product expression for AgOH:

$$AgOH(s)\rightleftharpoons Ag^+(aq)+OH^-(aq)$$ $$K_{sp}=[Ag^+]_{eqm}\space [OH^-]_{eqm}$$

Next, we assume that AgOH has a solubility of x mol dm^{-3}. This means that x mol dm^{-3} AgOH dissolves at equilibrium. The equilibrium equation tells us that for each mole of AgOH that dissolves, we end up with one mole of Ag^{+} ions. Therefore:

$$[Ag^+]_{eqm}=x\space mol\space dm^{-3}$$

We also need to know the concentration of OH^{-} ions. OH^{-} is our common ion - in this equilibrium, the OH^{-} ions come from not only the dissolved AgOH, but also from our solvent, NaOH. But because AgOH is only slightly soluble, it doesn't produce many OH^{-} ions when it dissolves at equilibrium and so the concentration of OH^{-} ions from this dissolution is very low. Therefore, we can ignore it, and simply consider the OH^{-}^{ }ions from the solvent. The question tells us that the solution of NaOH used as a solvent has a concentration of 2.00 mol dm^{-3}. Therefore, the concentration of OH^{-} ions in the solvent is also 2.00 mol dm^{-3}. We use this as our overall equilibrium concentration of Cl^{-} ions:

$$[OH^-]_{eqm}=2.00\space mol\space dm^{-3}$$

Now let's substitute both of our equilibrium concentrations into the solubility product expression and compare it to the value of the solubility product given to us in the question:

$$1.52\times 10^{-8}=(x)\space (2.00)$$ $$x=7.60\times 10^{-9}\space mol\space dm^{-3}$$

Remember that we assumed the solubility of AgOH in this solution was x mol dm^{-3}. Hence, its solubility is **7.60 × 10 ^{-9} mol dm**

It is always helpful to compare your answer to the solubility of the salt in pure water, if you know it. Helpfully, the question gave us the solubility of AgOH in pure water. Thanks to the common ion effect, we'd expect this value to be** much larger** than the solubility of AgOH in 2.00 mol dm^{-3} NaOH, as this second solution has a **relatively high concentration of the common ion** **(OH ^{-})**. Because both values take the same units, we can compare them directly. We find that the solubility of AgOH does indeed decrease when we dissolve it in a solution of NaOH:

$$1.23\times 10^{-4}>7.60\times 10^{-9}$$

- The
**solubility product**is the equilibrium constant for the dissolution of an ionic species in a solubility equilibrium. It tells you about the relative concentration of aqueous ions in a dissolution reaction at equilibrium. - For the solubility equilibrium \(A_aB_b(s)\rightleftharpoons aA^{b+}(aq)+bB^{a-}(aq)\) , the solubility product has the expression \(K_{sp}={[A^{b+}]_{eqm}}^a\space {[B^{a-}]_{eqm}}^b\)
**Solubility product calculations**include:- Calculating the solubility product using solubility.
- Calculating solubility using the solubility product.
- Using the solubility product to predict whether a salt will precipitate or not.
- Calculating solubility in solutions involving a common ion.

- Solubility product calculations can only be carried out for
**slightly soluble species**.

To calculate the solubility product (K_{sp}) from solubility, first write a solubility product expression. Then, use solubility and the balanced solubility equilibrium equation to work out the equilibrium concentrations of dissolved solute ions. Substitute these equilibrium concentrations into the solubility product expression and expand and simplify to find a numerical value. To finish, work out the units of the solubility product by substituting the units of your equilibrium concentrations into the solubility product expression, expanding and simplifying once again.

Confused? Check out the worked examples we walk you through in this article.

We typically consider a species soluble if its solubility product (K_{sp}) is greater than 1.0.

_{sp} and K_{c} are both types of equilibrium constant. They measure the relative amounts of products and reactants in a system at equilibrium. K_{c} is used for any equilibria involving gaseous or aqueous species and measures the ratio of products and reactants in terms of their concentration. K_{sp}, also known as the solubility product, is used specifically for solubility equilibria. It measures the relative concentration of dissolved aqueous solute ions in a system at equilibrium and indicates the solubility of an ionic species.

_{sp}, also known as the solubility product, is a specific type of equilibrium constant used for solubility equilibria. It measures the extent of dissolution of an ionic species and tells you the concentration of aqueous solute ions in a system at equilibrium.

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