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Chapter 4: Problem 91

Calculate the volume in milliliters of a \(1.420 \mathrm{M}\) \(\mathrm{NaOH}\) solution required to titrate the following solutions. (a) \(25.00 \mathrm{~mL}\) of a \(2.430 \mathrm{M} \mathrm{HCl}\) solution (b) \(25.00 \mathrm{~mL}\) of a \(4.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution (c) \(25.00 \mathrm{~mL}\) of a \(1.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution

Short Answer

Expert verified
(a) 42.82 mL of NaOH (b) 158.45 mL of NaOH (c) 79.22 mL of NaOH

Step by step solution

01

Stoichiometry and Neutralization Reactions

Consider the neutralization reactions: \n(a) \( \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{H_2O} + \mathrm{NaCl} \) \n(b) \( \mathrm{H_2SO4} + 2\mathrm{NaOH} \rightarrow \mathrm{H_2O} + \mathrm{Na}_2\mathrm{SO_4} \) \n(c) \( \mathrm{H_3PO_4} + 3\mathrm{NaOH} \rightarrow \mathrm{H_2O} + \mathrm{Na}_3\mathrm{PO_4} \) \nThese reactions show the amount of \(\mathrm{NaOH}\) needed to neutralize the corresponding acid.
02

Calculation (a)

For every mole of \(\mathrm{HCl}\), we need one mole of \(\mathrm{NaOH}\). The moles of \(\mathrm{HCl}\) are calculated by multiplying the volume by its molarity: \n\n Moles of \(\mathrm{HCl} = 25.00 \mathrm{mL} * 2.430 \mathrm{M} = 0.0608 \mathrm{moles} \). \n\n Then, using the molarity of the \(\mathrm{NaOH}\), we can calculate the volume of \(\mathrm{NaOH}\) needed to neutralize the \(\mathrm{HCl}\): \n\n Volume of \(\mathrm{NaOH} = \frac{0.0608 \mathrm{moles}}{ 1.420 \mathrm{M}} = 42.82 \mathrm{mL}\)
03

Calculation (b)

For each mole of \(\mathrm{H_{2}SO_{4}}\), we need two moles of \(\mathrm{NaOH}\). The moles of \(\mathrm{H_{2}SO_{4}}\) are calculated by multiplying the volume by its molarity: \n\n Moles of \(\mathrm{H_{2}SO_{4}} = 25.00 \mathrm{mL} * 4.500 \mathrm{M} = 0.1125 \mathrm{moles} \). \n\n Here, we will need two times this amount of \(\mathrm{NaOH}\): \n\n Moles of \(\mathrm{NaOH} = 2 * 0.1125 \mathrm{moles} = 0.225 \mathrm{moles} \). \n\n Then, using the molarity of the \(\mathrm{NaOH}\), we can calculate the volume of \(\mathrm{NaOH}\) needed to neutralize the \(\mathrm{H_{2}SO_{4}}\) : \n\n Volume of \(\mathrm{NaOH} = \frac{0.225 \mathrm{moles}}{ 1.420 \mathrm{M}} = 158.45 \mathrm{mL}\)
04

Calculation (c)

For each mole of \(\mathrm{H_{3}PO_{4}}\), we need three moles of \(\mathrm{NaOH}\). The moles of \(\mathrm{H_{3}PO_{4}}\) are calculated by multiplying the volume by its molarity: \n\n Moles of \(\mathrm{H_{3}PO_{4}} = 25.00 \mathrm{mL} * 1.500 \mathrm{M} = 0.0375 \mathrm{moles} \). \n\n Here, we will need three times this amount of \(\mathrm{NaOH}\): \n\n Moles of \(\mathrm{NaOH} = 3 * 0.0375 \mathrm{moles} = 0.1125 \mathrm{moles} \). \n\n Then, using the molarity of the \(\mathrm{NaOH}\), we calculate the volume of \(\mathrm{NaOH}\) needed to neutralize the \(\mathrm{H_{3}PO_{4}}\): \n\n Volume of \(\mathrm{NaOH} = \frac{0.1125 \mathrm{moles}}{ 1.420 \mathrm{M}} = 79.22 \mathrm{mL}\)

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Key Concepts

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

Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the context of titration calculations, stoichiometry helps us understand how much of a reactant (in this case, a base like NaOH) is needed to completely react with a given amount of another reactant (an acid). Understanding the balanced chemical equation is crucial because it indicates the molar ratios of the substances involved in the reaction.

Applying Stoichiometry to Titration

For instance, in a reaction where one mole of HCl reacts with one mole of NaOH to produce water and salt, stoichiometry tells us that the molar ratio is 1:1. If the ratio were 1:2, as is the case with H2SO4 and NaOH, then two moles of NaOH would be required to neutralize one mole of H2SO4. These stoichiometric ratios are applied to the volume and molarity of the solutions involved to calculate the volumes for titration. Remember to first balance the chemical equation before performing any titration calculation!
Neutralization Reactions
Neutralization reactions are chemical reactions in which an acid and a base react to form water and a salt. This type of reaction is central to titration because the objective is to determine the point at which the reactants have completely reacted with each other, known as the equivalence point.

Understanding the Reaction Process

In neutralization titrations, a known concentration of base (or acid) is added to a solution containing an unknown concentration of acid (or base). By knowing the stoichiometry of the neutralization reaction, we can calculate the unknown concentration once the equivalence point is reached. The reactions provided in the exercise (a, b, and c) are all examples of neutralization. They vary in complexity, with H3PO4 requiring three moles of NaOH per mole of acid to reach equivalence. It's important to always consider the products formed, as they indicate the completion of these reactions.
Molarity
Molarity, represented as M, is a measure of the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. Knowing the molarity of a solution is fundamental in titration calculations because it allows us to relate volumes and moles of reactants.

Using Molarity in Titration Calculations

To calculate how much of an acid or base is needed for a titration, we can use the formula Molarity (M) = moles of solute/volume of solution in liters. This formula plays a vital role in finding out the exact volume of, for example, a NaOH solution to add to an acid to reach the end point of a titration. In the provided exercises, the volume of a 1.420 M NaOH solution required to titrate various acid solutions is calculated using the concept of molarity in conjunction with the stoichiometry of the corresponding neutralization reactions.

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Most popular questions from this chapter

Classify the following redox reactions. (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (b) \(\mathrm{Mg}+2 \mathrm{AgNO}_{3} \longrightarrow \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{HBr}\)

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?

The \(\mathrm{SO}_{2}\) present in air is mainly responsible for the acid rain phenomenon. Its concentration can be determined by titrating against a standard permanganate solution as follows: \(5 \mathrm{SO}_{2}+2 \mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}\) Calculate the number of grams of \(\mathrm{SO}_{2}\) in a sample of air if \(7.37 \mathrm{~mL}\) of \(0.00800 \mathrm{M} \mathrm{KMnO}_{4}\) solution are required for the titration.

A \(35.2-\mathrm{mL}, 1.66 M \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

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?
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