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Chapter 15: Problem 74

A student dissolves 0.0100 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 Kb value for the unknown weak base is approximately 4.29 x 10⁻⁵.

Step by step solution

01

Determine moles of weak base and HNO3 added

Initially, we have 0.0100 mol of the weak base dissolved in 100.0 mL of water. During titration, we add 40.0 mL of 0.100 M HNO3. To determine the moles of HNO3 added, multiply its volume and concentration: Moles of HNO3 = volume x concentration Moles of HNO3 = 40.0 mL x 0.100 M = 0.0040 mol
02

Calculate the moles of unreacted base

Since the weak base and HNO3 react in a 1:1 ratio during titration, the moles of unreacted base can be found by subtracting the moles of HNO3 added from initial moles of the weak base: Moles of unreacted base = initial moles of base - moles of HNO3 Moles of unreacted base = 0.0100 mol - 0.0040 mol = 0.0060 mol
03

Calculate moles of OH- present after titration

We are given the pH of the solution after adding 40 mL of HNO3, which is 8.00. Using the relation between pH and pOH: pOH = 14 - pH pOH = 14 - 8.00 = 6.00 Now, using the pOH value, we can calculate the concentration of OH- ions present in the solution: [OH-] = \(10^{-pOH}\) = \(10^{-6.00}\) = 1.0 x 10⁻⁶ M To find the moles of OH-, we will multiply this concentration by the total volume of the solution, which is the initial 100 mL of water plus the 40 mL of HNO3 added: Total volume = 100.0 mL + 40.0 mL = 140.0 mL Moles of OH- = [OH-] x total volume Moles of OH- = 1.0 x 10⁻⁶ M x 140.0 mL = 1.4 x 10⁻⁴ mol
04

Determine moles of reacted base

Since each mole of reacted base produces one mole of OH-, we can determine the moles of reacted base: Moles of reacted base = moles of OH- Moles of reacted base = 1.4 x 10⁻⁴ mol
05

Calculate the Kb value for the weak base

Now, we can calculate the Kb value for the weak base using the relation: Kb = \(\frac{[OH-][unreacted\ base]}{[reacted\ base]}\) First, divide the moles by the total volume to get the concentrations: [OH-] = \(\frac{1.4 \times 10^{-4}\ mol}{140.0\ mL}\) = 1.0 x 10⁻⁶ M [unreacted base] = \(\frac{0.0060\ mol}{140.0\ mL}\) = 4.29 x 10⁻⁵ M [reacted base] = \(\frac{1.4 \times 10^{-4}\ mol}{140.0\ mL}\) = 1.0 x 10⁻⁶ M Now, substitute these concentrations into the Kb expression: Kb = \(\frac{(1.0 \times 10^{-6})(4.29 \times 10^{-5})}{(1.0 \times 10^{-6})}\) = 4.29 x 10⁻⁵ The Kb value for the weak base is approximately 4.29 x 10⁻⁵.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a solution that is 1.00 \(\mathrm{M}\) HNO, and 1.00 \(\mathrm{M} \mathrm{NaNO}_{2}\)

Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the pH of the solution related to the \(\mathrm{p} K_{\mathrm{a}}\) value of the weak acid? If [conjugate base \(]>[\text { weak acid }]\) , how is pH related to \(\mathrm{p} K_{\mathrm{a}} ?\)

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HONH}_{2}$ by 0.100 $\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)$ a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

Some \(\mathrm{K}_{2} \mathrm{SO}_{3}\) and \(\mathrm{KHSO}_{3}\) are dissolved in 250.0 \(\mathrm{mL}\) of solution and the resulting \(\mathrm{pH}\) is \(7.25 .\) Which is greater in this buffer solution, the concentration of \(\mathrm{SO}_{3}^{2-}\) or the concentration of \(\mathrm{HSO}_{3}-7\) If \(\left[\mathrm{SO}_{3}^{2-}\right]=1.0 \mathrm{M}\) in this solution, calculate the concentration of \(\mathrm{HSO}_{3}\) .

Two drops of indicator HIn \(\left(K_{2}=1.0 \times 10^{-9}\right),\) where HIn is yellow and In - is blue, are placed in 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{MCl}\) . a. What color is the solution initially? b. The solution is titrated with 0.10\(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after 200.0 \(\mathrm{mL}\) NaOH has been added?
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