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 \(0.004 mol ÷ 0.500 M = 0.008 L (or 8.0 mL)\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The concentration of sulfate in water can be determined by adding a solution of barium chloride to precipitate the sulfate ion. Write the net ionic equation for this reaction. Treating a 145-mL sample of water with excess \(\mathrm{BaCl}_{2}(a q)\) precipitated \(0.330 \mathrm{~g}\) of \(\mathrm{BaSO}_{4} .\) Determine the concentration of sulfate in the original water sample.

You have \(505 \mathrm{~mL}\) of a \(0.125 \mathrm{M} \mathrm{HCl}\) solution and you want to dilute it to exactly \(0.100 \mathrm{M}\). How much water should you add? Assume volumes are additive.

The antibiotic gramicidin A can transport \(\mathrm{Na}^{+}\) ions into a certain cell at the rate of \(5.0 \times 10^{7} \mathrm{Na}^{+}\) ions \(\mathrm{s}^{-1}\). Calculate the time in seconds to transport enough \(\mathrm{Na}^{+}\) ions to increase its concentration by \(8.0 \times 10^{-3} M\) in a cell whose intracellular volume is \(2.0 \times 10^{-10} \mathrm{~mL}\).

Give a chemical explanation for each of the following: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. (b) In the activity series, aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. (c) Sodium and potassium lie above copper in the activity series. In your explanation, discuss why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride \(\left(\mathrm{AlCl}_{3}\right)\) with an active metal. What metals would you use to produce aluminum in that way?

Describe the steps involved in preparing a solution of known molar concentration using a volumetric flask.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free