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Chapter 15: Problem 68

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 \(K_b\) value for the unknown weak base is approximately \(6.67 \times 10^{-7}\).

Step by step solution

01

Calculate the initial concentration of the weak base and HNO3

First, we find the initial concentration of weak base and HNO3. The weak base is dissolved in 100.0 mL of water, so we can calculate its initial concentration by dividing the given moles by the volume in liters: \[Initial \ concentration \ of \ weak \ base = \frac{0.0100 \ mol}{0.100 \ L} = 0.100 \ M\] The concentration of HNO3 is given as 0.100 M.
02

Calculate the moles of HNO3 added during titration

Next, we will find the moles of HNO3 added during titration. We are given the volume of HNO3 used (40.0 mL) and its concentration (0.100 M). To find the moles of HNO3 added, multiply the volume in liters by the concentration: \[Moles \ of \ HNO_3 = 0.100 \ M \times 0.0400 \ L = 0.00400 \ mol\]
03

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

During the titration, the weak base reacts with HNO3 to form its conjugate acid. The moles of weak base and HNO3 reacted are equal. Therefore, since 0.00400 moles of HNO3 were added, 0.00400 moles of the weak base have reacted: \[Moles \ of \ weak \ base \ after \ titration = 0.0100 \ mol - 0.00400 \ mol = 0.00600 \ mol\] Since the weak base reacts 1:1 with HNO3, the moles of the conjugate acid will be the same as the moles of reacted HNO3: \[Moles \ of \ conjugate \ acid = 0.00400 \ mol\]
04

Calculate the concentrations of the weak base and its conjugate acid after titration

After the titration, the total volume of the solution is 100.0 mL + 40.0 mL = 140.0 mL or 0.140 L. Now, we can find the concentrations of the weak base and its conjugate acid after titration by dividing their moles by 0.140 L: \[Concentration \ of \ weak \ base = \frac{0.00600 \ mol}{0.140 \ L} = 0.0429 \ M\] \[Concentration \ of \ conjugate \ acid = \frac{0.00400 \ mol}{0.140 \ L} = 0.0286 \ M\]
05

Calculate the Kb value for the weak base

First, we will find the pOH of the solution from the given pH value. The sum of pH and pOH is equal to 14, so: \[pOH = 14 - pH = 14 - 8.00 = 6.00\] Now we can find the hydroxide ion concentration, [OH-], from the pOH: \[[OH^-] = 10^{-pOH} = 10^{-6.00} = 1.00 \times 10^{-6} \ M\] We will use the Kb expression for the weak base dissociation equilibrium: \[K_b = \frac{[OH^-][conjugate \ acid]}{[weak \ base]}\] Plugging in the concentrations we have calculated: \[K_b = \frac{(1.00 \times 10^{-6})(0.0286)}{0.0429} = 6.67 \times 10^{-7}\] So, the Kb value for the unknown weak base is approximately \(6.67 \times 10^{-7}\).

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Most popular questions from this chapter

Consider the titration of \(150.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) HI by \(0.250 \mathrm{M}\) \(\mathrm{NaOH} .\) a. Calculate the \(\mathrm{pH}\) after \(20.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) has been added. b. What volume of \(\mathrm{NaOH}\) must be added so that the \(\mathrm{pH}=\) \(7.00 ?\)

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}}\), would you add \(\mathrm{HCl}\) or \(\mathrm{NaOH}\) ? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{~L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) ?

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like \(\mathrm{NaOH}\) is added, the HA reacts with the \(\mathrm{OH}^{-}\) to form \(\mathrm{A}^{-}\). Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding \(\mathrm{HCl}\) to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right) .\) Thus how can we claim that a buffered solution resists changes in the \(\mathrm{pH}\) of the solution?" How would you explain buffering to this friend?

Calculate the \(\mathrm{pH}\) of a solution prepared by mixing \(250 . \mathrm{mL}\) of \(0.174 m\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{~g} / \mathrm{mL}\) ) with \(38.7 \mathrm{~g}\) of an aqueous solution that is \(1.50 \% \mathrm{NaOH}\) by mass (density \(=\) \(1.02 \mathrm{~g} / \mathrm{mL}) \cdot\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HF}=7.2 \times 10^{-4} .\right)\)

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \begin{array}{l} \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \\ \text { Adipic acid } \quad \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{2}}=5.28 \end{array} $$ In two separate experiments the pH was measured during the titration of \(5.00\) mmol of each acid with \(0.200 M \mathrm{NaOH}\). Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at \(25.00 \mathrm{~mL}\) added \(\mathrm{NaOH}\), and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise \(113 .\) )
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