For use in titrations, we want to prepare 20 L of HCl(aq)...

For use in titrations, we want to prepare $20 \mathrm{L}$ of HCl(aq) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about $0.10 \mathrm{M}$ HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration. (a) How many milliliters of concentrated $\mathrm{HCl}(\mathrm{aq})$ $(d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })$ must be diluted with water to 20.0 L to prepare $0.10 \mathrm{M}$ HCl? (b) $\mathrm{A} 25.00 \mathrm{mL}$ sample of the approximately $0.10 \mathrm{M}$ HCl prepared in part (a) requires $20.93 \mathrm{mL}$ of $0.1186 \mathrm{M}$ NaOH for its titration. What is the molarity of the $\mathrm{HCl}(\mathrm{aq}) ?$ (c) Why is a titration necessary? That is, why not prepare a standard solution of $0.1000 \mathrm{M} \mathrm{HCl}$ simply by an appropriate dilution of the concentrated HCl(aq)?

For use in titrations, we want to prepare $20 \mathrm{L}$ of HCl(aq) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about $0.10 \mathrm{M}$ HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration.
(a) How many milliliters of concentrated $\mathrm{HCl}(\mathrm{aq})$ $(d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })$ must be diluted with water to 20.0 L to prepare $0.10 \mathrm{M}$ HCl?
(b) $\mathrm{A} 25.00 \mathrm{mL}$ sample of the approximately $0.10 \mathrm{M}$ HCl prepared in part (a) requires $20.93 \mathrm{mL}$ of $0.1186 \mathrm{M}$ NaOH for its titration. What is the molarity of the $\mathrm{HCl}(\mathrm{aq}) ?$
(c) Why is a titration necessary? That is, why not prepare a standard solution of $0.1000 \mathrm{M} \mathrm{HCl}$ simply by an appropriate dilution of the concentrated HCl(aq)?

Key Concepts

Dilution Principles

This concept involves reducing the concentration of a solution by mixing it with additional solvent. The dilution equation (M1V1 = M2V2) is widely used to calculate the required volume of a concentrated stock solution needed to achieve a target concentration in a given final volume.

Acid-Base Titration

Acid-base titration is an analytical technique used to determine the concentration of an acid or base in a solution. It involves gradually adding a titrant of known concentration to the analyte until neutralization is achieved, and the point of neutralization (end point) is used along with stoichiometric relationships to calculate the unknown concentration.

Stoichiometry of Neutralization Reactions

This concept deals with the quantitative relationships between reactants and products in a chemical reaction, particularly in neutralization reactions where an acid reacts with a base. Understanding the mole ratios allows for accurate calculation of the moles of acid or base present in a solution based on the titration data.

Standardization of Solutions

Standardization is a process used to accurately determine the concentration of a solution, often through titration with a primary standard. This ensures that a solution can be used confidently in subsequent quantitative analyses, which is crucial when precise concentrations are necessary for further chemical work.

Density and Percent Composition

These concepts are essential when converting the properties of concentrated solutions into molarities. Density allows conversion between volume and mass, while percent composition indicates the mass fraction of the solute in a solution. Combined, they enable the calculation of the concentration of the active chemical for use in dilution and stoichiometric computations.

Transcript

00:01 So our task is to pick out the concentration to 4 -6 -fix of 20 liters hydrochloric acid that we need to make.

00:06 In order to do this, you first need to dilute concentrated acetic acid and then neutralize it with sodium hydroxide to find the concentration that you actually made because you are not going to make 0 .10 molar because there's errors to carry forward and all the steps that you make.

00:23 So you will get closer, so you need to figure out what the molarity is.

00:27 But the idea is to make this concentration.

00:30 So the dilution formula, it's m1 v1 equals m2 v2, where m1 is the initial concentration, v1 is the initial volume, m2 is the final concentration, which is point in molar, and then v2 is the final volume, which is 20 liters.

00:48 Now, we don't know m1 and we don't know v1, and you can't have two unknowns, right? so we actually do know the concentrated hydrochloric acid.

01:00 We know this information of the concentrated hydrochloric acid, so we can find the concentration of it, and then using that we can figure out the volume.

01:08 So that way we only have one unknown.

01:10 So we have to figure out this first, so we can figure out the second unknown.

01:14 So let's do that.

01:15 So we know we have 38 % by mass concentration.

01:19 That means that there are 38 grams of hydrochloric acid in 100 grams of solution.

01:28 And the idea is to get this in molarity, which is moles per liter, right? so we have to get this moles per liter.

01:39 So we're going to get grams of hydrochloric acid to moles of hydrochloric acid first.

01:44 Whenever you're stuck, always go to moles first.

01:48 And then that will cancel grams, and that will give you moles.

01:50 And then we're going to get grams of solution to milliliters of solution using the density.

01:58 So we're going to do grams per milliliter of solution.

02:02 So grams per mole and one mole of hydrochloric acid, the molar mass is 36 .45 grams.

02:10 And then grams per millimeter is the density, 1 .19 grams and 1 millimeter.

02:15 That will cancel this.

02:17 And then grams of hydrochloric acid cancel, and you get moles per milliter.

02:21 So we need it in liter.

02:23 So we need to do one more step because there's moles per liter.

02:27 So i'm going to put a milliliters opposite to it in order to cancel it, and then liters of the bottom, because liters needs to be at the bottom.

02:33 We know in one liter there's a thousand milliliters, so then milliliters cancel.

02:38 Multiply cross, divide by the bottom, and the resulting answer is anything you didn't cancel.

02:42 So we multiplied your quirk acid per liters of solution, and then you multiply across the numbers and divide by the bottom.

02:49 And then your answer is 12 .4 molar is the concentration of the concentrated hydrochloric acid which is usually what you buy when you buy a concentrated hydrochloric acid so this is 12 .4 molar so we can use the dilution formula now so i'm gonna do m1 v1 equals m2 so m1 is my initial concentration it's 12 .4 molar v1 is the volume that i need and then m2 is .10 molar, and then the volume that i'm trying to make is 20 liters.

03:30 So then we solve it, multiply those two, divide by 12 .4, and you get 0 .16 liters, which is equal to 160 milliliters...

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