AP Chemistry : Precipitates and Calculations

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

Example Question #1 : Precipitates And Calculations

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

Possible Answers:

\(\displaystyle 233.4 g\)

No precipitate is formed

\(\displaystyle 58.5 g\)

\(\displaystyle 21.0 g\)

\(\displaystyle 18.7g\)

Correct answer:

\(\displaystyle 18.7g\)

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

By solubility rules, NaCl\(\displaystyle NaCl\) is soluble in water and BaSO_4\(\displaystyle BaSO_4\) 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)}\(\displaystyle Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl_{(aq)}\hspace{1 mm}+BaSO_{4\hspace{1 mm}(s)}\)

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

Now we perform the same calculation beginning with BaCl_2\(\displaystyle BaCl_2\):

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

The limiting reagent is BaCl_2\(\displaystyle BaCl_2\) and this reaction produces 18.7 g precipitate.

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\(\displaystyle 234\hspace{1 mm}g\hspace{1 mm}KCl\)

6.98\hspace{1 mm}g\hspace{1 mm}KCl\(\displaystyle 6.98\hspace{1 mm}g\hspace{1 mm}KCl\)

29.82\hspace{1 mm}g\hspace{1 mm}KCl\(\displaystyle 29.82\hspace{1 mm}g\hspace{1 mm}KCl\)

74.55\hspace{1 mm}g\hspace{1 mm}KCl\(\displaystyle 74.55\hspace{1 mm}g\hspace{1 mm}KCl\)

None of the available answers

Correct answer:

6.98\hspace{1 mm}g\hspace{1 mm}KCl\(\displaystyle 6.98\hspace{1 mm}g\hspace{1 mm}KCl\)

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

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:

\(\displaystyle 4.82\hspace{1 mm}g\hspace{1 mm}NaOH\)

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

\(\displaystyle 4.67\hspace{1 mm}g\hspace{1 mm}NaOH\)

None of the available answers

\(\displaystyle 13.9\hspace{1 mm}g\hspace{1 mm}NaOH\)

Correct answer:

\(\displaystyle 4.82\hspace{1 mm}g\hspace{1 mm}NaOH\)

Explanation:

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

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

Example Question #1 : Using Stoichiometry In Solutions

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

Possible Answers:

\(\displaystyle 0.04g\)

\(\displaystyle 0.0445g\)

\(\displaystyle 2.35g\)

None of the available answers

\(\displaystyle 2.34g\)

Correct answer:

\(\displaystyle 2.34g\)

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.

\(\displaystyle 0.100L*\frac{0.4mol}{1L}=0.040mol\ NaCl\)

\(\displaystyle 0.040mol\ NaCl*\frac{58.5g\ NaCl}{1mol\ NaCl}=2.34g\ NaCl\)

Example Question #2 : Precipitates And Calculations

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

Possible Answers:

\(\displaystyle 181g\ Li_3PO_4\)

\(\displaystyle 53.9g\ Li_3PO_4\)

\(\displaystyle 108g\ Li_3PO_4\)

\(\displaystyle 1.56g\ Li_3PO_4\)

\(\displaystyle 116g\ Li_3PO_4\)

Correct answer:

\(\displaystyle 181g\ Li_3PO_4\)

Explanation:

First we will figure out the number of moles of Li_3 PO_4\(\displaystyle Li_3 PO_4\) 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\(\displaystyle 0.3\hspace{1mm} \frac{mol}{L} \times 5.2\hspace{1 mm} L = 1.56 \hspace{1 mm}moles\)

Now the problem is to find the mass of 1.56 moles of Li_3 PO_4\(\displaystyle Li_3 PO_4\). First, we need to know the molecular weight of Li_3 PO_4\(\displaystyle Li_3 PO_4\). 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}\(\displaystyle (3\hspace{1 mm}\times6.94\hspace{1 mm}\frac{grams\hspace{1 mm} Li}{mol\hspace{1 mm}Li})+30.97\hspace{1 mm}\frac{grams\hspace{1 mm}P}{mol\hspace{1 mm}P}+(4\hspace{1 mm}\times 16\hspace{1 mm}\frac{grams\hspace{1 mm}O}{mol\hspace{1mm}O})= 115.79\hspace{1mm}\frac{grams\hspace{1 mm} Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}\)

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

Example Question #2 : Precipitates And Calculations

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

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

Possible Answers:

\(\displaystyle 0.170\)

\(\displaystyle 1.07 * 10^{-12}\)

\(\displaystyle 5.89\)

\(\displaystyle 9.33 * 10^{11}\)

Correct answer:

\(\displaystyle 0.170\)

Explanation:

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

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

\(\displaystyle pH = pK_a + \log\frac{[HCO_3^-]}{[H_2CO_3]}\)

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

Solve for the ratio we need to answer the question:

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

Example Question #7 : Precipitates And Calculations

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

Possible Answers:

\(\displaystyle 0.011 M\)

\(\displaystyle 0.0044 M\)

\(\displaystyle 0.022 M\)

\(\displaystyle 0.0023 M\)

Correct answer:

\(\displaystyle 0.022 M\)

Explanation:

The dissociation of calcium hydroxide in aqueous solution is:

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

The \(\displaystyle K_{sp}\) of calcium hydroxide is related to the dissolved concentrations of its counterions:

\(\displaystyle K_{sp} = [Ca^2^{+}] [OH^{-}]^{2}\)

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

\(\displaystyle [OH^{-}] = 2[Ca^{2+}]\)

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

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

\(\displaystyle 5.5 * 10^{-6} = 4x^{3}\)

Now, we can use basic algebra to solve for \(\displaystyle x\):

\(\displaystyle 1.38 * 10^{-6} = x^{3}\)

\(\displaystyle 0.011 M = x\)

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

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

 

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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