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AP Chemistry : Solubility

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Example Questions

Example Question #71 : Solutions

What is the formula for the dissociation of iron (II) phosphate?

Possible Answers:

$FeP\rightleftharpoons Fe^{2+}+P^{2-}$

$Fe_2(PO_4)_3\rightleftharpoons 2Fe^{3+}+3PO_4^{2-}$

$FePO_4\rightleftharpoons Fe^{2+}+PO_4^{2-}$

$Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$

$Ir_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$

Correct answer:

$Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$

Explanation:

Iron (II) has a positive two charge: $Fe^{2+}$ .

Phosphate has a negative three charge: $PO_4^{3-}$ .

The initial compound must be constructed to cancel these charges. The dissociation is: $Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}$ .

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Example Question #2 : Solubility

Pb(OH)₂ has a K_sp of 1.43 * 10^-20.

What is its solubility?

Possible Answers:

2.43 * 10^-7M

1.20 * 10^-10M

8.42 * 10^-12M

1.53 * 10^-7M

Correct answer:

1.53 * 10^-7M

Explanation:

Lead (II) hydroxide dissociates according to this reaction: $Pb(OH)_2\rightleftharpoons Pb^{2+}+2OH^-$ . Solubility is equal to the number of moles that will dissociate at equilibrium, and can be found using this reaction and the value of K_sp.

K_sp = [Pb²⁺][OH^-]²

If [Pb²⁺] = x, then [OH^-] = 2x. They exist in a 1:2 ratio.

K_sp = 1.43 * 10^-20 = x(2x)² = 4x³

x = 1.53 x 10^-7M

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Example Question #3 : Solubility

Which of the following factors will generally lead a solute to dissolve more easily into a solution (i.e. increase solubility)?

Possible Answers:

Increasing the temperature of the solution

The common ion effect

Decreasing the temperature of the solution

None of these choices will increase solubility

Correct answer:

Increasing the temperature of the solution

Explanation:

Increasing the temperature of a solution will generally increase its solubility, while decreasing temperature will have the opposite effect. The common ion effect refers to the phenomenon when two compunds in a solution dissociate to produce at least one similar ion, and will make it harder for additional amounts of that ion in the solution to dissociate, decreasing solubility.

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Example Question #4 : Solubility

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

If the K_sp of the above dissociation is found to be 2.70 * 10^-4, what is the solubility of NaCl?

Possible Answers:

1.44 * 10^-2M

1.89 * 10^-2M

1.64 * 10^-2M

7.29 * 10^-8M

Correct answer:

1.64 * 10^-2M

Explanation:

First, set up a balanced reaction and enter X for the amount of each solute that is going to be dissolved. As NaCl dissociates, each ion amount will increase by some amount, X.

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

$\wr$

Then set up the equation for the value of K_sp and enter in the value of X for each solute. Use this equation to solve for X, the solubility (in mol/L). Remember that NaCl si not included in the equation, as it is a pure solid.

$K_{sp}= [Na^{+}][Cl^{-}] = (X)(X) = X^{2}$

$2.70 * 10^{-4} = X^{2}$

$X= 1.64 * 10^{-2}M$

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Example Question #5 : Solubility

$CaF_{2}(s) \rightarrow Ca^{2+}(aq) + 2F^{-}(aq)$

If the solubility of CaF₂ in the dissociation reaction above was determined to be 1.54 * 10^-3M, what is the K_sp value for CaF₂?

Possible Answers:

2.24 * 10^-6

1.98 * 10^-8

1.42 * 10^-6

1.46 * 10^-8

Correct answer:

1.46 * 10^-8

Explanation:

Use the balanced reaction of the dissociation to assign values of X for the solubility of each ion. As CaF₂ dissociates, each ion amount will increase by some amount; calcium will increase by X and flouride will increase by 2X.

$CaF_{2}(s) \rightarrow Ca^{2+}(aq) + 2F^{-}(aq)$

$\wr$ +2X

$\wr$

Then plug in the equation to solve K_sp, using the X values assigned for each solute. Use the given value for solubility to find K_sp. Remember that NaCl si not included in the equation, as it is a pure solid.

$K_{sp}= [Ca^{2+}][F^-]^2=(X)(2X)^{2} = 4X^{3}$

$K_{sp}= 4(1.9 * 10^{-2})^{3} = 1.46 * 10^{-8}$

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Example Question #1 : Solubility

Consider the following balanced equation for the solubility of barium hydroxide in an aqueous solution.

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

What is the solubility of barium hydroxide?

Possible Answers:

$\small 0.05M$

$\small 0.108M$

$\small 0.071M$

$\small 0.035M$

Correct answer:

$\small 0.108M$

Explanation:

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated $\small +x$ and $\small +2x$ , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

$\begin{matrix} Ba(OH)_2 & Ba^{2+} &OH^- \\ /& 0 & 0\\ / & x & 2x \end{matrix}$

$K_{sp}=[Ba^{2+}][OH^-]^2$

$\small 5*10^{-3} = [x][2x]^{2}$

$\small x = 0.108M$

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Example Question #7 : Solubility

Four different insoluble salts are added to an aqueous solution in equal amounts. Assume that the salts do not interact with one another in any way. The following list shows the salts and their solubility product constants.

$\small BaSO_{4}$ $\small K_{sp} = 1.1*10^{-10}$

$\small CaCrO_{4}$ $\small K_{sp} = \small 7.1*10^{-4}$

$\small MgF_{2}$ $\small K_{sp } = 3.7*10^{-8}$

$\small NiCO_{3}$ $\small K_{sp} = 6.6*10^{-9}$

Which of the following ions will have the highest concentration in the solution?

Possible Answers:

$\small SO_{4}^{2-}$

$\small CO_{3}^{2-}$

$\small \small CrO_{4}^{2-}$

$\small F-$

Correct answer:

$\small \small CrO_{4}^{2-}$

Explanation:

The solubility product constants help us envision which ions will have the largest concentrations after dissolving to equilibrium. $\small CaCrO_{4}$ has the largest solubility product constant, meaning it will dissolve the most out of the four options.

Upon the dissolving of $\small MgF_{2}$ , twice the number of fluoride ions will be released as magnesium ions; however, the solubility constant for magnesium fluoride is so much smaller, compared to the solubility constant of $\small CaCrO_{4}$ , that it will not become the ion with the largest concentration.

$2(3.7*10^{-8})<7.1*10^{-4}$

$[F^-]<[CrO_4^{2-}]$

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Example Question #2 : Solubility

Consider the following balanced equation for the dissociation of barium hydroxide in an aqueous solution.

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

A 1M solution of barium nitrate is added to the solution. What is the solubility of barium hydroxide after this addition?

Possible Answers:

$\small 0.003M$

$\small 0.050M$

$\small 0.035M$

$\small 0.108M$

0.118M

Correct answer:

$\small 0.035M$

Explanation:

Before we solve for the solubility, remember Le Chatelier's principle. Barium nitrate is a soluble salt that will completely dissolve in solution. This means that there will be a one molar concentration of barium ions before the salt dissolves. Since we have added to the products side of the reaction, we can predict that the equilibrium will be shifted to the left, and the barium hydroxide will be less soluble. This change in equilibrium is referred to as the common ion effect.

We will use an ICE table to determine the solubility of the salt.

I. Before the salt begins to dissolve, there is a one molar concentration of barium ions (from the barium nitrate) with no hydroxide ions in the solution.

C. Since every dissolved molecule of barium hydroxide will yield one barium ion and two hydroxide ions, the unknown increases in each ion can be designated $\small 1+x$ and $\small 2x$ respectively. Notice how the barium ion increase will be added to the previously added one molar concentration in the solution.

E. Now, we set the solubility expression equal to the solubility product constant.

$\begin{matrix} Ba(OH)_2 & Ba^{2+} & OH^-\\ / & 1M & 0M\\ /& 1+x & x \end{matrix}$

$K_{sp}=[Ba^{2+}][OH^-]^2$

$\small 5*10^{-3} = [1+x][2x]^{2}$

Since 1 is such a larger number than the value we will get for $\small x$ , we can consider the value in the barium ion expression to be negligible. This allows us to simplify the equation to the following setup.

$\small 5*10^{-3} = [1][2x]^{2}$

$\small x = 0.035M$

The solubility of barium hydroxide without the common ion effect is 0.108M. Notice how the solubility for the salt has decreased, as was predicted in the beginning by Le Chatelier's principle.

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Example Question #9 : Solubility

A chemist has a fully saturated solution of chemical X, with solid X resting on the bottom. If she adds a chemical that will react with the aqueous X, what will happen to the solid?

Possible Answers:

A layer of sediment will cover the solid

More solid will precipitate out of solution

It will start to dissolve into the solution

The solid will change color

Nothing will change

Correct answer:

It will start to dissolve into the solution

Explanation:

A fully saturated solution is at equilibrium between the solid and aqueous states, but it is a dynamic equilibrium. As the reacting chemical reacts with aqueous X, more and more of the solid will dissolve into solution.

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Example Question #3 : Solubility

The partial pressure of an arbitrary gas, $X_{2\hspace{1 mm}(g)}$ , in an enclosed chanber with a pool of water is 7.2atm at 298K.

The Henry's Law constant for $X_{2\hspace{1 mm}(g)}$ is $5.6*10^{-2}\frac{mol}{L\cdot atm}$ .

What is the solubility of $X_{2\hspace{1 mm}(g)}$ ?

Possible Answers:

$4.0\frac{mol}{L}$

$4.0*10^{-1}\frac{mol}{L\cdot atm}$

$4.0*10^{-1}\frac{mol}{L}$

$129\frac{mol}{L}$

None of the available answers

Correct answer:

$4.0*10^{-1}\frac{mol}{L}$

Explanation:

Henry's law can be expressed:

$S_{gas}=k_H\times P_{gas}$

Where $k_H$ is the Henry's Law constant, and $P_{gas}$ is the partial pressure of the gas above the solution. We can substitute our givens into this equation:

$S_{gas}=\frac{5.6\times 10^{-2}\hspace{1 mm}mol}{L\cdot atm}\times 7.2\hspace{1 mm}atm=4.0\times 10^{-1}\hspace{1 mm}\frac{mol}{L}$

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AP Chemistry : Solubility

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #71 : Solutions

Example Question #2 : Solubility

Example Question #3 : Solubility

Example Question #4 : Solubility

Example Question #5 : Solubility

Example Question #1 : Solubility

Example Question #7 : Solubility

Example Question #2 : Solubility

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Example Question #3 : Solubility

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