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What volume of a \(0.500 M\) HCl solution is needed to neutralize each of the following? (a) \(10.0 \mathrm{~mL}\) of a \(0.300 \mathrm{M} \mathrm{NaOH}\) solution (b) \(10.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution

Short Answer

Expert verified
The volume of a 0.500 M HCl solution needed to neutralize (a) 10.0 mL of a 0.300 M NaOH solution is 6.0 mL and (b) 10.0 mL of a 0.200 M Ba(OH)2 solution is 8.0 mL.

Step by step solution

01

Calculate the number of moles of the base in the first solution

For part (a), the number of moles of NaOH in the solution can be calculated by multiplying the volume of the solution by its molarity. In this case, the volume of the solution is 10.0 mL (or 0.01 L when converted to liters) and the molarity is 0.300 M, so the number of moles of NaOH is \(0.01 L * 0.300 M = 0.003 mol (moles)\)
02

Calculate the volume of HCl needed to neutralize the NaOH solution

Since one mole of an acid neutralizes one mole of a base, the number of moles of HCl needed to neutralize the NaOH solution is equal to the number of moles of NaOH. Using the formula volume = moles ÷ molarity, the volume of the 0.500 M HCl solution needed is \(0.003 mol ÷ 0.500 M = 0.006 L (or 6.0 mL)\)
03

Calculate the number of moles of the base in the second solution

For part (b), note that one molecule of Ba(OH)2 has two OH- ions. So the molar concentration of Ba(OH)2 needs to be doubled to get the molar concentration of hydroxides (OH-). Therefore, the number of moles of hydroxides is \(0.01 L * 0.200 M * 2 = 0.004 mol\)
04

Calculate the volume of HCl needed to neutralize the Ba(OH)2 solution

Again, the number of moles of HCl needed is equal to the number of moles of hydroxides. The volume of the 0.500 M HCl solution needed to neutralize the solution is \(0.004 mol ÷ 0.500 M = 0.008 L (or 8.0 mL)\)

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Key Concepts

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

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. In the context of neutralization reactions, such as those between acids and bases, stoichiometry involves using the balanced equation for the reaction to determine the exact amounts of reactants needed to completely react with each other.

For example, when an acid neutralizes a base, the stoichiometric relationship is often a one-to-one ratio, meaning one mole of acid will react with one mole of base. This knowledge allows us to calculate the required volume of an acid solution to neutralize a known amount of a base, as shown in the exercise. Understanding stoichiometry is essential in chemical analysis and industry, where precise amounts of substances must be mixed to yield the desired product.
Molarity
Molarity represents the concentration of a solution and is defined as the number of moles of solute dissolved in one liter of solution, commonly expressed in moles per liter (M). It is an integral part of the stoichiometry in acid-base titrations because it directly relates the volume of solution to the number of moles of solute.

When calculating volume or molarity in a neutralization reaction, you can rearrange the relationship molarity (M) = moles of solute (n) / volume of solution (V) depending on what information you have and what you need to find. In the given exercise, molarity helps in calculating the number of moles of base initially present and the volume of acid solution needed for neutralization. Molarity is very practical, as it allows chemists to easily scale reactions up or down by simply adjusting the volumes of solutions they use.
Acid-Base Titration
Acid-base titration is a laboratory method used to determine the concentration of an acid or base in a solution by neutralization. A measured volume of an acid or base of known concentration (the titrant) is gradually added to a volume of the other with unknown concentration (the analyte), until the reaction reaches an end-point, typically indicated by a color change in an added indicator or by reaching a certain pH value measured by a pH meter.

During the titration, the stoichiometry of the neutralization reaction guides the amount of titrant needed to reach the end-point. As demonstrated in the original exercise, we use stoichiometry and molarity to calculate the exact volumes of an HCl solution required to neutralize given amounts of NaOH and Ba(OH)2 solutions. Through acid-base titration, students learn to execute precise techniques to measure and analyze solutions, an essential skill in many branches of chemistry.

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Most popular questions from this chapter

What are the similarities and differences between acid-base titrations and redox titrations?

A 325-mL sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\). (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

Determine how many grams of each of the following solutes would be needed to make \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.100 M\) solution: (a) cesium iodide (CsI), (b) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right),\) (c) sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right),\) (d) potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)\) (e) potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\).

The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of a stock solution. Instead, it is done by a series of dilutions. A sample of \(0.8214 \mathrm{~g}\) of \(\mathrm{KMnO}_{4}\) was dissolved in water and made up to the volume in a 500-mL volumetric flask. A \(2.000-\mathrm{mL}\) sample of this solution was transferred to a \(1000-\mathrm{mL}\) volumetric flask and diluted to the mark with water. Next, \(10.00 \mathrm{~mL}\) of the diluted solution were transferred to a \(250-\mathrm{mL}\) flask and diluted to the mark with water. (a) Calculate the concentration (in molarity) of the final solution. (b) Calculate the mass of \(\mathrm{KMnO}_{4}\) needed to directly prepare the final solution.

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

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