Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 78

Piperazine is a diprotic weak base used as a corrosion inhibitor and an insecticide. Its ionization is described by the following equations. \(\mathrm{HN}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\left[\mathrm{HN}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}_{2}\right]^{+}+\mathrm{OH}^{-} \quad \mathrm{p} K_{\mathrm{b}_{1}}=4.22\) \(\left[\mathrm{HN}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}_{2}\right]^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\left[\mathrm{H}_{2} \mathrm{N}\left(\mathrm{C}_{4} \mathrm{H}_{8}\right) \mathrm{NH}_{2}\right]^{2+}+\mathrm{OH}^{-} \quad \mathrm{p} K_{\mathrm{b}_{2}}=8.67\) . The piperazine used commercially is a hexahydrate, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{N}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} .\) A \(1.00-\mathrm{g}\) sample of this hexahydrate is dissolved in \(100.0 \mathrm{mL}\) of water and titrated with 0.500 M HCl. Sketch a titration curve for this titration, indicating (a) the initial \(\mathrm{pH} ;\) (b) the pH at the halfneutralization point of the first neutralization; (c) the volume of \(\mathrm{HCl}(\text { aq })\) required to reach the first equivalence point; (d) the pH at the first equivalence point; (e) the \(\mathrm{pH}\) at the point at which the second step of the neutralization is half-completed; (f) the volume of \(0.500 \mathrm{M} \mathrm{HCl}(\) aq) required to reach the second equivalence point of the titration; (g) the pH at the second equivalence point.

Short Answer

Expert verified
Sketching the titration curve involves a nine-step process. The curve represents the titration of a diprotic base with a strong acid. The initial pH can be calculated using the first pKb and solving for [OH-] concentration. The first half-neutralization point occurs when [HB+] = [B], from which the pH can be calculated using the first pKb. The volume of HCl to reach the first equivalence point can be found using stoichiometry and knowledge about the initial concentration of the base. At the first equivalence point, the pH can be found using the second pKb and calculating the [OH-] concentration. The second half-neutralization point occurs when [A^2+] = [HB^+]. At this point, the pH can be calculated using the second pKb. The volume of HCl for the second equivalence point can be calculated using stoichiometry. The final pH (second equivalence point) can be computed considering the remaining H+ from the strong acid.

Step by step solution

01

Calculate the initial concentration

First, figure out the molar mass of the hexahydrate \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{N}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \). Since 1.00g of the hexahydrate is dissolved in 100mL of water, you can calculate the initial concentration of the base, piperazine, in moles per liter.
02

Calculate the initial pH

For finding the initial pH, apply that the Kb of a base is given by \(Kb = [OH^-][HB^+]/[B] \). From this and given \(pKb_1 \) you will be able to estimate the [OH^−] concentration, and further the pOH and initial pH.
03

Calculate the pH at the first half-neutralization point

At the first half-neutralization point, [HB^+] = [B]. Solve for the pH using the expression for \(pKb_1 \).
04

Calculate the HCl volume for the first equivalence point

The equivalence point is reached when all of the base has been reacted with the acid. Use the initial concentration and the stoichiometry of the reaction to calculate the volume of 0.500 M HCl needed to reach this point.
05

Calculate the pH at the first equivalence point

As all the base is reacted, use \(pKb_2 \) and the expression \(Kb = [OH^-][A^2+]/[HB^+] \) to estimate the [OH−] concentration and further the pOH and pH at this point.
06

Calculate the pH at the second half-neutralization point

At the second half-neutralization point, [A^2+] = [HB^+]. Use \(pKb_2 \) and the equation for Kb to find the pH at this point.
07

Calculate the volume of HCl to reach the second equivalence point

As before, use stoichiometry of the reaction to calculate the volume of 0.500 M HCl to reach the second equivalence point.
08

Calculate the pH at the second equivalence point

For finding the pH at the second equivalence point, you need to remember that all the base is reacted at this point and estimate the H+ concentration remaining from the strong acid. Use this to calculate the pH.
09

Sketch the titration curve

For sketching the curve: On the x-axis plot the volume of HCl added, on the y-axis plot the pH. Mark the points from steps 2 to 8 on the curve.

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.

Start your free trial

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

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

Using appropriate equilibrium constants but without doing detailed calculations, determine whether a solution can be simultaneously: (a) \(0.10 \mathrm{M} \mathrm{NH}_{3}\) and \(0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl},\) with \(\mathrm{pH}=6.07\) (b) \(0.10 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.058 \mathrm{M} \mathrm{HI}\) (c) \(0.10 \mathrm{M} \mathrm{KNO}_{2}\) and \(0.25 \mathrm{M} \mathrm{KNO}_{3}\) (d) \(0.050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.65 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (e) \(0.018 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(0.018 \mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) with \(\mathrm{pH}=4.20\) (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)

Is a solution that is \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}(\mathrm{aq})\) likely to be acidic, basic, or pH neutral? Explain.

The effect of adding \(0.001 \mathrm{mol} \mathrm{KOH}\) to 1.00 Lof a solution that is \(0.10 \mathrm{M} \mathrm{NH}_{3}-0.10 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) is to (a) raise the pH very slightly; (b) lower the pH very slightly; (c) raise the pH by several units; (d) lower the pH by several units.

In 1922 Donald D. van Slyke ( J. Biol. Chem., 52, 525) defined a quantity known as the buffer index: \(\beta=\mathrm{d} C_{\mathrm{b}} / \mathrm{d}(\mathrm{pH}),\) where \(\mathrm{d} C_{\mathrm{b}}\) represents the increment of moles of strong base to one liter of the buffer. For the addition of a strong acid, he wrote \(\beta=-\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH})\) By applying this idea to a monoprotic acid and its conjugate base, we can derive the following expression: \(\beta=2.303\left(\frac{K_{w}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\frac{\mathrm{CK}_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left(\mathrm{K}_{\mathrm{a}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\right)^{2}}\right)\) where \(C\) is the total concentration of monoprotic acid and conjugate base. (a) Use the above expression to calculate the buffer index for the acetic acid buffer with a total acetic acid and acetate ion concentration of \(2.0 \times 10^{-2}\) and a \(\mathrm{pH}=5.0\) (b) Use the buffer index from part (a) and calculate the \(\mathrm{pH}\) of the buffer after the addition of of a strong acid. (Hint: Let \(\left.\mathrm{d} C_{\mathrm{a}} / \mathrm{d}(\mathrm{pH}) \approx \Delta C_{\mathrm{a}} / \Delta \mathrm{pH} .\right)\) (c) Make a plot of \(\beta\) versus \(\mathrm{pH}\) for a \(0.1 \mathrm{M}\) acetic acid buffer system. Locate the maximum buffer index as well as the minimum buffer indices.

A handbook lists various procedures for preparing buffer solutions. To obtain a \(\mathrm{pH}=9.00,\) the handbook says to mix \(36.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NH}_{3}\) with \(64.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (a) Show by calculation that the pH of this solution is 9.00. (b) Would you expect the \(\mathrm{pH}\) of this solution to remain at \(\mathrm{pH}=9.00\) if the \(100.00 \mathrm{mL}\) of buffer solution were diluted to 1.00 L? To 1000 L? Explain. (c) What will be the pH of the original \(100.00 \mathrm{mL}\) of buffer solution if \(0.20 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) is added to it? (d) What is the maximum volume of \(1.00 \mathrm{M} \mathrm{HCl}\) that can be added to \(100.00 \mathrm{mL}\) of the original buffer solution so that the pH does not drop below \(8.90 ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation
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