A 20.00 -mL sample of 0.220 M triethylamine, (CH3 CH2)3 N, is titrated with 0.544 M HCl .(Kb(CH

A 20.00 -mL sample of 0.220 M triethylamine, (CH3 CH2)3 N,...

A 20.00 -mL sample of $0.220 \mathrm{M}$ triethylamine, $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N},$ is titrated with $0.544 \mathrm{M} \mathrm{HCl} .\left(\mathrm{K}_{\mathrm{b}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}=5.2 \times 10^{-4}\right)$ (a) Write a balanced net ionic equation for the titration. (b) How many milliliters of $\mathrm{HCl}$ are required to reach the equivalence point? (c) Calculate $\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right],\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right],$ and $\left[\mathrm{Cl}^{-}\right]$ at the equivalence point. (Assume that volumes are additive.) (d) What is the $\mathrm{pH}$ at the equivalence point?

A 20.00 -mL sample of $0.220 \mathrm{M}$ triethylamine, $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N},$ is titrated with $0.544 \mathrm{M} \mathrm{HCl} .\left(\mathrm{K}_{\mathrm{b}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}=5.2 \times 10^{-4}\right)$
(a) Write a balanced net ionic equation for the titration.
(b) How many milliliters of $\mathrm{HCl}$ are required to reach the equivalence
point?
(c) Calculate $\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right],\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right],$ and $\left[\mathrm{Cl}^{-}\right]$ at the
equivalence point. (Assume that volumes are additive.)
(d) What is the $\mathrm{pH}$ at the equivalence point?

Key Concepts

pH Calculation at the Equivalence Point

At the equivalence point in a titration of a weak base with a strong acid, the solution contains the conjugate acid, which slightly dissociates in water. Calculating the pH in such cases requires setting up an equilibrium expression for the conjugate acid hydrolysis, taking into account its Ka value, and solving for the hydrogen ion concentration.

Relationship Between Kb and Ka

The equilibrium constant for a weak base (Kb) and its conjugate acid (Ka) are related by the water ionization constant (Kw), as expressed by the relationship Kw = Ka × Kb. This relationship is essential in determining the acidity of the conjugate acid formed in the titration of a weak base with a strong acid.

Weak Base and Conjugate Acid Chemistry

In titrations involving a weak base and a strong acid, the weak base reacts with the acid to form its conjugate acid. This conjugate acid can undergo hydrolysis in water, influencing the pH of the solution at the equivalence point. Understanding this transformation is vital for accurately calculating the pH after titration.

Titration Stoichiometry

Titration stoichiometry refers to the quantitative relationship between the reactants in a titration reaction. It is based on the mole ratios dictated by the balanced chemical equation and is used to calculate how much titrant is required to react completely with the analyte, especially at the equivalence point.

Net Ionic Equation

A net ionic equation shows only the species that are directly involved in the chemical reaction, omitting the spectator ions that do not participate. In acid-base titrations, this helps to focus on the actual proton transfer or electron donation processes, clarifying the chemical change occurring during the titration.

Equivalence Point

The equivalence point is reached in a titration when the moles of titrant added exactly equal the moles of analyte present in the solution, based on the stoichiometry of the reaction. At this precise moment, the reaction between the acid and the base is complete, although the pH may not necessarily be neutral if one of the reactants is a weak acid or base.

Acid?Base Titration

This concept involves the gradual addition of a titrant (an acid or a base of known concentration) to a solution containing an analyte until the reaction is complete. It is commonly used to determine the concentration of an unknown solution. The working principles include mixing, reaction completion, and precise measurement of the volume of titrant delivered.

Transcript

00:01 Okay, in this problem, we're going to react an amine, right, which is this triethylamine.

00:08 The ch3, ch2 group is called ethel, so there's three of them, and that's where you get your triethylamine.

00:15 And we're going to react it with a strong acid, so we'll just write h -plus, and these will combine to make the conjugate acid, right? so that's your net ionic equation.

00:34 And let's start by figuring out how many moles of our mean we start with.

00:41 So we were given the molarity, so 0 .220 molar.

00:47 We'll multiply by its volume and liters.

00:53 So we've got 0 .044 moles of our triethylamine.

01:01 So of our mean right here.

01:05 I'm just going to write amine to shorten things up.

01:08 Little bit.

01:10 So that means that we would also need 0 .0044 moles of h plus to react with that.

01:20 Okay? so if we divide that by the molarity, we'll get the volume that we needed to add.

01:30 So 0 .00809.

01:34 That's our liters.

01:36 So 8 .09 milliliters would have been added.

01:43 I'm going to figure out our total volume right now, because we're going to need that in a little bit, i imagine.

01:50 Okay, so that's the 20 milliliters we started with of our mean, and then we'll add the 8 .09.

02:01 So 28 .09 milliliters is our total volume.

02:11 Okay, so we're being asked to calculate a whole bunch of concentrations.

02:15 So at the equivalence point, we make this weak acid.

02:21 So in order to figure out what's going to happen, and we'll go ahead and write its reaction where it ionizes, so that we can figure out how much of this amine we make, plus some h -plus.

02:41 So i need the initial concentration of this substance right here, and i know that i'm going to make the same number of moles, right? so 0 .0044 moles, and if i divide that by the total volume, 0 .2809, i'll see that i'm starting with 0 .157 molar of this.

03:09 So let's see what else is happening here with an icebox...

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