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Chapter 4: Problem 81

(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH}\) ? (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) ? (c) If \(25.8 \mathrm{~mL}\) of an \(\mathrm{AgNO}_{3}\) solution is needed to precipitate all the \(\mathrm{Cl}^{-}\)ions in a \(785-\mathrm{mg}\) sample of \(\mathrm{KCl}\) (forming \(\mathrm{AgCl}\) ), what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of a \(0.108 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of KOH must be present in the solution?

Short Answer

Expert verified
In summary, (a) 38.04 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH. (b) 769 mL of 0.128 M HCl solution is needed to neutralize 2.87 g of Mg(OH)2. (c) The molarity of the AgNO3 solution needed to precipitate all the Cl- ions in a 785 mg sample of KCl is 0.408 M. (d) There are 0.2744 g of KOH present in the solution that was neutralized by 45.3 mL of 0.108 M HCl solution.

Step by step solution

01

a) Volume of HClO4 solution

First, we'll calculate the moles of NaOH in the given solution. Moles of NaOH \(= \text{Volume (L)} \times \text{Molarity}\) Moles of NaOH \(= 50.00\, \mathrm{mL} \times \frac{1\, \mathrm{L}}{1000\, \mathrm{mL}} \times 0.0875\, \mathrm{M} \) Moles of NaOH \(= 0.004375\, \mathrm{mol}\) Since the neutralization reaction between HClO4 and NaOH is a 1:1 stoichiometry, the moles of HClO4 needed will be equal to the moles of NaOH. Moles of HClO4 \(= 0.004375\, \mathrm{mol}\) Now, we can find the volume of the HClO4 solution needed using the molarity given. Volume of HClO4 solution \(= \frac{\mathrm{Moles\, of\, HClO_4}}{\mathrm{Molarity\, of\, HClO_4}}\) Volume of HClO4 solution \(= \frac{0.004375\, \mathrm{mol}}{0.115\, \mathrm{M}}\) Volume of HClO4 solution \(= 0.038043\, \mathrm{L}\) Converting to milliliters: Volume of HClO4 solution \(= 38.04\, \mathrm{mL}\) So, for part (a), 38.04 mL of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH.
02

b) Volume of HCl needed for Mg(OH)2 neutralization

First, we need to convert the mass of Mg(OH)2 to moles. Moles of Mg(OH)2 \(= \frac{2.87\, \mathrm{g}}{58.32\, \mathrm{g/mol}}\) Moles of Mg(OH)2 \(= 0.0492\, \mathrm{mol}\) The stoichiometry between Mg(OH)2 and HCl in the reaction is 1:2, so we need twice the moles of HCl. Moles of HCl needed \(= 0.0492\, \mathrm{mol} \times 2\) Moles of HCl needed \(= 0.0984\, \mathrm{mol}\) Now, we can find the volume of the HCl solution needed using the given molarity. Volume of HCl solution \(= \frac{\mathrm{Moles\, of\, HCl}}{\mathrm{Molarity\, of\, HCl}}\) Volume of HCl solution \(= \frac{0.0984\, \mathrm{mol}}{0.128\, \mathrm{M}}\) Volume of HCl solution \(= 0.769\, \mathrm{L}\) Converting to milliliters: Volume of HCl solution \(= 769\, \mathrm{mL}\) So, for part (b), 769 mL of 0.128 M HCl solution is needed to neutralize 2.87 g of Mg(OH)2.
03

c) Molarity of AgNO3 solution

First, we need to convert the mass of KCl to moles of Cl- ions. Moles of Cl- \(= \frac{785\, \mathrm{mg}}{74.55\, \mathrm{g/mol}}\) Moles of Cl- \(= 0.01053\, \mathrm{mol}\) Since the stoichiometry between AgNO3 and Cl- ions in the reaction is 1:1, the moles of AgNO3 needed will be equal to the moles of Cl- ions. Moles of AgNO3 needed \(= 0.01053\, \mathrm{mol}\) Now, we can find the molarity of the AgNO3 solution using the given volume. Molarity of AgNO3 solution \(= \frac{\mathrm{Moles\, of\, AgNO_3}}{\mathrm{Volume\, of\, AgNO_3\, solution\, in\, L}}\) Molarity of AgNO3 solution \(= \frac{0.01053\, \mathrm{mol}}{25.8\, \mathrm{mL} \times \frac{1\, \mathrm{L}}{1000\, \mathrm{mL}}}\) Molarity of AgNO3 solution \(= 0.408\, \mathrm{M}\) So, for part (c), the molarity of the AgNO3 solution needed to precipitate all the Cl- ions in a 785 mg sample of KCl is 0.408 M.
04

d) Mass of KOH in solution

First, we'll calculate the moles of HCl in the given solution. Moles of HCl \(= \text{Volume (L)} \times \text{Molarity}\) Moles of HCl \(= 0.0453 \, \mathrm{L} \) 脳 \(0.108\, \mathrm{M}\) Moles of HCl \(= 0.0048924\, \mathrm{mol}\) Since the neutralization reaction between HCl and KOH is a 1:1 stoichiometry, the moles of KOH in the solution will be equal to the moles of HCl. Moles of KOH \(= 0.0048924\, \mathrm{mol}\) Now, we can find the mass of KOH present in the solution. Mass of KOH \(= \mathrm{Moles\, of\, KOH} \times \mathrm{Molar\, mass\, of\, KOH}\) Mass of KOH \(= 0.0048924\, \mathrm{mol} \times 56.11\, \mathrm{g/mol}\) Mass of KOH \(= 0.2744\, \mathrm{g}\) So, for part (d), there are 0.2744 g of KOH present in the solution that was neutralized by 45.3 mL of 0.108 M HCl solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Understanding stoichiometry is like having a recipe for the perfect chemical reaction. Imagine you're baking cookies, and you need to know exactly how many teaspoons of sugar balance one cup of flour to make them just right. In chemistry, stoichiometry is that recipe. It tells us the exact amounts of reactants needed to create products in a chemical reaction.

For instance, when we look at a neutralization reaction where hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), the stoichiometry tells us that one molecule of HCl reacts with one molecule of NaOH, which is a 1:1 ratio. This is paramount when calculating how much of one reactant is needed to completely react with a given amount of another. However, not all reactions are as straightforward as the 1:1 ratio. Some reactions, such as the reaction between magnesium hydroxide (Mg(OH)2) and HCl, exhibit a 1:2 stoichiometry, meaning that for every one mole of Mg(OH)2, two moles of HCl are required for complete neutralization.
Molarity
If stoichiometry is the recipe, think of molarity as the concentration of your ingredients. It measures how concentrated a solution is, specifically how many moles of solute are in one liter of solution. You'll often see it written as 'M,' such as a 0.115 M HClO4 solution.

In our kitchen, if we increase the concentration of sugar in our cookies by adding more teaspoons per cup of flour, the cookies become sweeter. Similarly, by knowing the molarity of a solution in a chemical reaction, we can predict how it will behave and calculate the volumes needed to react completely with another reactant. For example, to neutralize a certain volume of NaOH, we measure how many liters of HClO4 we need based on its molarity.
Acid-Base Neutralization
An acid-base neutralization is essentially a chemical handshake; an acid offers a proton (H+), and a base is ready to accept it, resulting in water and a salt. For instance, hydrochloric acid (HCl) meeting potassium hydroxide (KOH) will turn into water (H2O) and potassium chloride (KCl).

By using stoichiometry and molarity, we calculate the precise volume of an acid like HCl to fully neutralize a given amount of a base like KOH. In practice, when the correct volumes of the two solutions mix, they react completely, leaving no excess acid or base鈥攚hich is ideal in many applications, from titrations in a laboratory to antacids neutralizing stomach acid.
Precipitation Reactions
Imagine if you could make a solid magically appear in a liquid solution. No wizardry here鈥攖his is what happens in a precipitation reaction. It occurs when two solutions combine, and an insoluble solid, or precipitate, forms and falls out of the solution. An everyday example is adding lemon juice to tea, which can cause some of the tea's components to solidify and become visible.

In our textbook problem, when silver nitrate (AgNO3) is mixed with potassium chloride (KCl), solid silver chloride (AgCl) precipitates out of the solution. We use the knowledge of stoichiometry to understand that for every chlorine ion (Cl-) there needs to be one silver ion (Ag+) to form the solid AgCl, and we use molarity to calculate how much AgNO3 we need to precipitate all the Cl- ions present in the sample.

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Most popular questions from this chapter

The average concentration of gold in seawater is \(100 \mathrm{fM}\) (femtomolar). Given that the price of gold is \(\$ 1764.20\) per troy ounce ( 1 troy ounce \(=31.103 \mathrm{~g}\) ), how many liters of seawater would you need to process to collect \(\$ 5000\) worth of gold, assuming your processing technique captures only \(50 \%\) of the gold present in the samples?

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with \(\mathrm{HCl}\) according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min}\), the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+}\), and \(\mathrm{Mn}^{2+}\). Addition of \(\mathrm{HCl}\) solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid according to: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow & \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\\\ & 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

State whether each of the following statements is true or false. Justify your answer in each case. (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\)ions, and yet its aqueous solutions are basic. (b) HF is a strong acid. (c) Although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\)ions than \(\mathrm{SO}_{4}{ }^{2-}\) ions.
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