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AP Chemistry : Solutions and States of Matter

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Example Question #1 : Precipitates And Calculations

A chemist has 5.2L of a 0.3M $Li_3 PO_4$ solution. If the solvent were boiled off, what would be the mass of the remaining solid?

Possible Answers:

$181g\ Li_3PO_4$

$108g\ Li_3PO_4$

$116g\ Li_3PO_4$

$53.9g\ Li_3PO_4$

$1.56g\ Li_3PO_4$

Correct answer:

$181g\ Li_3PO_4$

Explanation:

First we will figure out the number of moles of $Li_3 PO_4$ that we have. We have 5.2 L of a 0.3 M solution, so:

$0.3\hspace{1mm} \frac{mol}{L} \times 5.2\hspace{1 mm} L = 1.56 \hspace{1 mm}moles$

Now the problem is to find the mass of 1.56 moles of $Li_3 PO_4$ . First, we need to know the molecular weight of $Li_3 PO_4$ . We will go to the periodic table and add up the mass of each element present.

$(3\hspace{1 mm}\times6.94\hspace{1 mm}\frac{grams\hspace{1 mm} Li}{mol\hspace{1 mm}Li})+30.97\hspace{1 mm}\frac{grams\hspace{1 mm}P}{mol\hspace{1 mm}P}+(4\hspace{1 mm}\times 16\hspace{1 mm}\frac{grams\hspace{1 mm}O}{mol\hspace{1mm}O})= 115.79\hspace{1mm}\frac{grams\hspace{1 mm} Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}$

Now the problem is simple as we have the molar mass and the number of desired moles.

$115.79\hspace{1 mm}\frac{grams\hspace{1 mm}Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}\times1.56\hspace{1 mm}moles\hspace{1 mm}Li_3 PO_4 = 181\hspace{1 mm}grams\hspace{1 mm}Li_3 PO_4$

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Example Question #6 : Precipitates And Calculations

Given a pK_aof 6.37 for the first deprotonation of carbonic acid ( H_2CO_3 ), what is the ratio of bicarbonate ( HCO_3^- ) to carbonic acid ( H_2CO_3 ) at pH 5.60?

Assume that the effect of the deprotonation of bicarbonate is negligible in your calculations.

Possible Answers:

$1.07 * 10^{-12}$

5.89

0.170

$9.33 * 10^{11}$

Correct answer:

0.170

Explanation:

Use the Henderson-Hasselbalch equation to determine the ratio of bicarbonate to carbonic acid in solution:

$pH = pK_a + \log\frac{\left [A^- \right ]}{\left [HA \right ]}$

$pH = pK_a + \log\frac{[HCO_3^-]}{[H_2CO_3]}$

$5.60-6.37 = -0.77 = \log\frac{\left [HCO_3^- \right ]}{\left [H_2CO_3 \right ]}$

Solve for the ratio we need to answer the question:

$10^{-0.77} = \frac{\left [ HCO_3^-\right ]}{[H_2CO_3]} = 0.170$

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Example Question #7 : Precipitates And Calculations

The K_s_p of Ca(OH)_2 (at 298K) is $5.5*10^{-6}$ . What is the molar solubility of the hydroxide ion ( OH^- ) in a saturated solution of ?

Possible Answers:

0.022 M

0.0023 M

0.011 M

0.0044 M

Correct answer:

0.022 M

Explanation:

The dissociation of calcium hydroxide in aqueous solution is:

$Ca(OH)_{2}\rightleftharpoons Ca^{2+} + 2OH^{-}$

The $K_{sp}$ of calcium hydroxide is related to the dissolved concentrations of its counterions:

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

$Ca^{2+}$ and OH^- are produced in a molar ratio of 1:2; for each molecule of calcium hydroxide that dissolves:

$[OH^{-}] = 2[Ca^{2+}]$

Given a $K_{sp}$ value of $5.5 * 10^{-6}$ , the molar solubilities of each counterion may be determined by setting $[Ca^{2+}] = x$ . It follows that:

$K_{sp} = 5.5 * 10^{-6} = x (2x)^{2}$

$5.5 * 10^{-6} = 4x^{3}$

Now, we can use basic algebra to solve for :

$1.38 * 10^{-6} = x^{3}$

0.011 M = x

[OH^-] = 2(0.011M) = 0.022M

Since we set $[Ca^{2+}] = x$ , and [OH^-] = 2x , multiplying the value of by two gives the correct answer, which is 0.022M.

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Example Question #1 : Precipitates And Calculations

What type of reaction is also known as a precipitation reaction?

Possible Answers:

Decomposition

Single replacement

Combustion

Double replacement

Combination

Correct answer:

Double replacement

Explanation:

Double replacement reactions can be further categorized as precipitation reactions since it is possible to make a precipitate (solid) from mixing two liquids. Combustion reactions involve using oxygen to burn another species, and the products are carbon dioxide and water. Combination reactions involve the synthesis of one molecule from two separate ones; decomposition is the opposite.

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Example Question #81 : Solutions And States Of Matter

Calculate the molar solubility of AgBr in 0.050 M AgNO₃ at room temperature. The K_sp of AgBr is 5.4 x 10^-13.

Possible Answers:

2.16 x 10^-11 M

2.16 M

1.08 x 10^-11 M

1.08 M

1.57 x 10^-12 M

Correct answer:

1.08 x 10^-11 M

Explanation:

$K_sp = [Ag^+ ][Br^- ] = 5.4 x 10^{-13}$

$5.4 x 10^{-13} = [0.05+x][x] \sim [0.05][x]$

$x = 1.08 x 10^{-11}$

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Example Question #2 : Solubility And Equilibrium

Would the molar solubility of Cr(OH)₃ increase or decrease as the pH is lowered (i.e. made more acidic)?

Possible Answers:

Decrease

Increase

There is no change

Can not be determined

Correct answer:

Increase

Explanation:

Since Cr(OH)₃ is a basic salt, decreasing the pH makes it more soluble.

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Example Question #82 : Solutions And States Of Matter

Calculate the molar solubility of SrF₂ in 0.023M NaF. The K_sp for SrF₂ is 4.3 x 10^-9.

Possible Answers:

3.2 x 10^-3 M

5.2 x 10^-4 M

3.2 x 10^-5 M

1.6 x 10^-3 M

8.1 x 10^-6 M

Correct answer:

8.1 x 10^-6 M

Explanation:

$K_{sp} = [Sr^{2+} ] [F^- ]^2 = 4.3 x 10^{-9}$

$4.3 x 10^{-9} = [x] [0.023+2x]^2 \sim [x][0.023]^2$

$x = 8.1 x 10^{-6}$

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Example Question #83 : Solutions And States Of Matter

Calculate the molar solubility of Mn(OH)₂ at pH 9.5. The K_sp for Mn(OH)₂ is 1.6 x 10^-13.

Possible Answers:

1.6 x 10^-4 M

1.5 x 10^-3 M

3.5 x 10^-4 M

2.1 x 10^-5M

2.4 x 10^-3 M

Correct answer:

1.6 x 10^-4 M

Explanation:

$Mn(OH)_2 \rightarrow Mn^{2+} + 2 \: OH^-$

$K_{sp} = 1.6 x 10^{-13} = [Mn^{2+} ] [OH^- ]^2$

pOH = 14 - pH = 14 - 9.5 = 4.5

$[OH^- ] = 10^{-pOH} = 10^{-4.5} = 3.16 x 10^{-5}$

$K_{sp}= 1.6 x 10^{-13} = [x] [3.16 x 10^{-5} + 2x]^2 \sim [x] [3.16 x 10^{-5} ]^2$

$x = 1.6 x 10^{-4}$

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Example Question #84 : Solutions And States Of Matter

Calculate the molar solubility of CaF₂(K_sp = 3.9 x 10^-11) in a room temperature solution of 0.010 M Ca(C₂H₃O₂)_2.

Possible Answers:

3.2 x 10^-2 M

3.1 x 10^-5 M

4.2 x 10^-4M

3.7 x 10^-4M

1.6 x 10^-2 M

Correct answer:

3.1 x 10^-5 M

Explanation:

$CaF_2 \rightarrow Ca^{2+ } + 2 F^-$

$K_{sp} = 3.9 x 10^{-11} = [Ca^{2+} ] [F^- ]^2$

$K_{sp}= 3.9 x 10^{-11} = [x+0.01] [2x]^2 \sim [0.01] [2x]^2$

$x = 3.1 x 10^{-5}$

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Example Question #1 : Other Solution Concepts

A solution on NaCl has a denisty of 1.075 g/mL. If there are 0.475 L of solution present, what is the mass?

Possible Answers:

2.263 g

0.510 g

510.6 g

2.2 g

441.9 g

Correct answer: 510.6 g

Explanation:

1.075 g / mL * 0.475 L

First, convert to mL

1.075 g / mL * 475 mL = 510.6 g

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AP Chemistry : Solutions and States of Matter

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Precipitates And Calculations

Example Question #6 : Precipitates And Calculations

Example Question #7 : Precipitates And Calculations

Example Question #1 : Precipitates And Calculations

Example Question #81 : Solutions And States Of Matter

Example Question #2 : Solubility And Equilibrium

Example Question #82 : Solutions And States Of Matter

Example Question #83 : Solutions And States Of Matter

Example Question #84 : Solutions And States Of Matter

Example Question #1 : Other Solution Concepts

All AP Chemistry Resources

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