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Example Questions

Example Question #31 : Solutions

In order to prepare a needed solution for an experiment, 0.082 grams of NaCl was dissolved in water to give a 35mL solution. What is the molarity of this solution?

Possible Answers:

0.40M

0.040M

0.34M

0.409M

0.45M

Correct answer:

0.040M

Explanation:

In order to calculate the concentration, we must use molarity formula:

$Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

We must use the molecular weight of sodium chloride to calculate the moles of solute:

$MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)$

$MW_{NaCl}=(23 \frac{g}{mole})+(35 \frac{g}{mole})= 58 \frac{g}{mole}$

$0.082g*\frac{mole}{58g}=0.0014\ moles\ NaCl$

$\frac{0.0014\ moles\ CH_{3}CH_{2}OH}{0.035\L} = 0.040\ M$

Therefore, the concentration in molarity of this solution is 0.040M.

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Example Question #34 : Solutions

How many moles are in 1000mL solution with a concentration that is $3*10^{-5}M$ ?

Possible Answers:

$6.5*10^{-5}mol$

$3*10^{-8}mol$

$3*10^{-5}mol$

$9*10^{-5}mol$

$8*10^{-5}mol$

Correct answer:

$3*10^{-5}mol$

Explanation:

By simply using the concentration as a conversion factor, the number of moles can be calculated by multiplying the concentration by the number of liters. Before calculating the number of moles, the number of milliliters must be converted to liters using the fact that $1\ liter = 1000\ mL$ .

$\frac{3*10^{-5}\ moles}{L} * 1\ L = 3*10^{-5}\ moles$

Therefore, there are $3*10^{-5}\ moles$ in $0.010\ L$ of a $5*10^{-3}M$ solution.

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Example Question #1 : Precipitates And Calculations

A chemist combines 300 mL of a 0.3 M $Na_2SO_4$ solution with 200 mL of 0.4 M $BaCl_2$ solution. How many grams of precipitate form?

Possible Answers:

21.0 g

18.7g

233.4 g

58.5 g

No precipitate is formed

Correct answer:

18.7g

Explanation:

First, let us write out an ion exchance reaction for the reactants:

$Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl\hspace{1 mm}+BaSO_4$

By solubility rules, $NaCl$ is soluble in water and $BaSO_4$ is not. Our new reaction is:

$Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl_{(aq)}\hspace{1 mm}+BaSO_{4\hspace{1 mm}(s)}$

Now we will calculate the theoretical yield of each reactant.

$300\hspace{1 mm}mL\times\frac{1 L}{1000 mL}\times \frac{0.3\hspace{1 mm}moles\hspace{1 mm}Na_2SO_4}{1 L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}Na_2SO_4}\times \frac{233.4\hspace{1 mm}g\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}=21.0\hspace{1 mm}g\hspace{1 mm}BaSO_4$

Now we perform the same calculation beginning with $BaCl_2$ :

$200\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.4\hspace{1 mm}moles\hspace{1 mm}BaCl_2}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaCl_2}\times\frac{233.4\hspace{1 mm}g\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}=18.7\hspace{1 mm}g\hspace{1 mm}BaSO_4$

The limiting reagent is $BaCl_2$ and this reaction produces 18.7 g precipitate.

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Example Question #2 : Precipitates And Calculations

A chemist boils off the water from 234 mL of a 0.4 M solution of KCl. What is the mass of the remaining solid?

Possible Answers:

$234\hspace{1 mm}g\hspace{1 mm}KCl$

$6.98\hspace{1 mm}g\hspace{1 mm}KCl$

$29.82\hspace{1 mm}g\hspace{1 mm}KCl$

$74.55\hspace{1 mm}g\hspace{1 mm}KCl$

None of the available answers

Correct answer:

$6.98\hspace{1 mm}g\hspace{1 mm}KCl$

Explanation:

$234\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.4\hspace{1 mm}moles\hspace{1 mm}KCl}{1\hspace{1 mm}L}\times\frac{74.55\hspace{1 mm}g\hspace{1 mm}KCl}{1\hspace{1 mm}mole\hspace{1 mm}KCl}=6.98\hspace{1 mm}g\hspace{1 mm}KCl$

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Example Question #2 : Precipitates And Calculations

A chemist boils off 344 mL of a 0.35 M solution of sodium hydroxide. How much solid remains?

Possible Answers:

$4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

No solid will remain as it will boil off with the water

$4.67\hspace{1 mm}g\hspace{1 mm}NaOH$

None of the available answers

$13.9\hspace{1 mm}g\hspace{1 mm}NaOH$

Correct answer:

$4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

Explanation:

First, you must recognize that the chemical formula for sodium hydroxide is NaOH . The mass of the boiled solution is

$344\hspace{1 mm}mL\times\frac{0.35\hspace{1 mm}mol\hspace{1 mm}NaOH}{1000\hspace{1 mm}mL}\times\frac{39.997\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mol\hspace{1 mm}NaOH}= 4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

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Example Question #2 : Precipitates And Calculations

What is the mass of the solid left over after boiling off 100mL of 0.4M NaCl solution?

Possible Answers:

0.0445g

2.35g

None of the available answers

2.34g

0.04g

Correct answer:

2.34g

Explanation:

The remaining mass with be equal to the mass of the sodium chloride in the solution. Once the solvent (water) evaporates, the solute will remain.

Atomic mass of sodium is ~23. Atomic mass of chlorine is ~35.5. Molecular mass of NaCl is ~58.5.

$0.100L*\frac{0.4mol}{1L}=0.040mol\ NaCl$

$0.040mol\ NaCl*\frac{58.5g\ NaCl}{1mol\ NaCl}=2.34g\ NaCl$

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Example Question #3 : 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:

$53.9g\ Li_3PO_4$

$181g\ Li_3PO_4$

$116g\ Li_3PO_4$

$1.56g\ Li_3PO_4$

$108g\ Li_3PO_4$

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 #2 : 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:

0.170

$1.07 * 10^{-12}$

5.89

$9.33 * 10^{11}$

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:

0.011 M

0.0044 M

0.022 M

0.0023 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:

Double replacement

Single replacement

Combustion

Combination

Decomposition

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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AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #31 : Solutions

Example Question #34 : Solutions

Example Question #1 : Precipitates And Calculations

Example Question #2 : Precipitates And Calculations

Example Question #2 : Precipitates And Calculations

Example Question #2 : Precipitates And Calculations

Example Question #3 : Precipitates And Calculations

Example Question #2 : Precipitates And Calculations

Example Question #7 : Precipitates And Calculations

Example Question #1 : Precipitates And Calculations

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