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Chapter 15: Problem 62

A student dissolves \(0.0100 \mathrm{~mol}\) of an unknown weak base in \(100.0 \mathrm{~mL}\) water and titrates the solution with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\). After \(40.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

Short Answer

Expert verified
The $K_{\mathrm{b}}$ value for the unknown weak base is approximately \(6.66 \times 10^{-7}\).

Step by step solution

01

Calculate the initial moles of the weak base and strong acid

First, let's determine the initial amount (in moles) of the weak base and HNO3. For the weak base, we are given that it is 0.0100 mol. For HNO3, we'll use the volume and concentration: moles HNO3 = 0.100 M * 0.040 L = 0.0040 mol
02

Determine the moles of the weak base and its conjugate acid after titration

The titration of the weak base with HNO3 will produce the conjugate acid of that weak base. However, not all of the weak base reacts with the strong acid, so we need to find the moles of the weak base and its conjugate acid after the titration is complete. Since each mole of HNO3 reacts with one mole of the weak base: moles weak base remaining = 0.0100 mol - 0.0040 mol = 0.0060 mol moles conjugate acid formed = 0.0040 mol
03

Calculate the volume of the resulting solution

We need the volume of the resulting solution in order to find the concentrations of the species involved. The initial volume of the weak base solution is 100 mL, and we added 40 mL of HNO3, so the final volume is: V = 100.0 mL + 40.0 mL = 140.0 mL = 0.140 L
04

Determine the concentrations of the weak base and conjugate acid

Using the moles of the weak base and conjugate acid, we can find their concentrations in the resulting solution: [weak base] = moles weak base remaining / V = 0.0060 mol / 0.140 L ≈ 0.0429 M [conjugate acid] = moles conjugate acid formed / V = 0.0040 mol / 0.140 L ≈ 0.0286 M
05

Find the concentration of OH- ions from pH

We are given that the pH of the resulting solution after titration is 8.00. To find the concentration of OH- ions, we first need to find the pOH: pOH = 14 - pH = 14 - 8.00 = 6.00 Now, we can find the OH- concentration: [OH-] = 10^(-pOH) = 10^(-6.00) = 1.00 × 10^(-6) M
06

Calculate the Kb of the weak base

Now that we have the concentrations of the weak base, conjugate acid, and OH- ions after titration, we can find the Kb value for the weak base. The Kb expression for a weak base (A-) and its conjugate acid (HA) is: Kb = ([HA][OH-])/[A-] We substitute our calculated concentrations: Kb = (0.0286 * 1.00 × 10^(-6)) / 0.0429 ≈ 6.66 × 10^(-7) Kb = 6.66 × 10^(-7) The Kb value for the unknown weak base is approximately 6.66 × 10^(-7).

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Most popular questions from this chapter

Could a buffered solution be made by mixing aqueous solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) ? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?

An aqueous solution contains dissolved \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). The concentration of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\) is \(0.50 \mathrm{M}\) and \(\mathrm{pH}\) is \(4.20\). a. Calculate the concentration of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{3}^{+}\) in this buffer solution. b. Calculate the \(\mathrm{pH}\) after \(4.0 \mathrm{~g} \mathrm{NaOH}(s)\) is added to \(1.0 \mathrm{~L}\) of this solution. (Neglect any volume change.)

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