Precipitates and Calculations - AP Chemistry

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Question

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?

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Answer

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

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Question

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

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Answer

First, you must recognize that the chemical formula for sodium hydroxide is . 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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Question

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?

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Answer

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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Question

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?

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Answer

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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Question

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

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Answer

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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Question

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

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Answer

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 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 :

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

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

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Question

Given a pKa of 6.37 for the first deprotonation of carbonic acid (), what is the ratio of bicarbonate () to carbonic acid () at pH 5.60?

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

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Answer

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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Question

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

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Answer

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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Question

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?

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Answer

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

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Question

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

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Answer

First, you must recognize that the chemical formula for sodium hydroxide is . 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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Question

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?

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Answer

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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Question

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?

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Answer

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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Question

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

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Answer

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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Question

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

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Answer

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 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 :

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

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

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Question

Given a pKa of 6.37 for the first deprotonation of carbonic acid (), what is the ratio of bicarbonate () to carbonic acid () at pH 5.60?

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

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Answer

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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Question

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

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Answer

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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Question

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?

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Answer

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

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Question

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

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Answer

First, you must recognize that the chemical formula for sodium hydroxide is . 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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Question

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?

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Answer

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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Question

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?

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Answer

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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