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Chapter 15: Problem 81

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+} .\)

Short Answer

Expert verified
The analogous equation to the Henderson-Hasselbalch equation relating pOH and pKb for a buffered solution composed of a weak base and its conjugate acid is given by: \[pOH = pK_b - \log_{10}\frac{[Conjugate\: Acid]}{[Base]}\]

Step by step solution

01

Write the dissociation reaction of a weak base and its conjugate acid

For a weak base and its conjugate acid, we can use NH₃ as a base and NH₄⁺ as the conjugate acid. The reaction will be: \[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\]
02

Write the equilibrium expression for the reaction

The equilibrium expression for the above reaction can be written using the equilibrium constant, Kb. \[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\]
03

Isolate the Kb for the reaction

We can rewrite the equilibrium expression to isolate Kb. \[K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\]
04

Convert the expression to involve pOH, pKb, and the base-to-acid concentration ratio

Now, let's introduce the following logarithmic expressions: \[pOH = -\log_{10}([OH^-])\] \[pK_b = -\log_{10}(K_b)\] \[\frac{[NH_4^+]}{[NH_3]} = \frac{[Conjugate\: Acid]}{[Base]}\] Taking the logarithmic form of the Kb expression, we get: \[\log_{10}(K_b) = \log_{10}\frac{[NH_4^+][OH^-]}{[NH_3]}\] \[-pK_b = \log_{10}\frac{[NH_4^+][OH^-]}{[NH_3]}\]
05

Rearrange the expression to form the desired equation

Now, let's rearrange the previous equation to isolate the pOH. \[-pK_b = \log_{10}\frac{[NH_4^+][OH^-]}{[NH_3]}\] \[pOH = pK_b - \log_{10}\frac{[NH_4^+]}{[NH_3]}\] This equation relates pOH, pKb, and the concentration ratio of the conjugate acid to the weak base. It is analogous to the Henderson-Hasselbalch equation for pOH and pKb in a buffered solution.

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Most popular questions from this chapter

A buffer is made using \(45.0 \mathrm{~mL}\) of \(0.750 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=\right.\) \(1.3 \times 10^{-5}\) ) and \(55.0 \mathrm{~mL}\) of \(0.700 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\). What volume of \(0.10 M\) NaOH must be added to change the \(\mathrm{pH}\) of the original buffer solution by \(2.5 \% ?\)

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 few drops of each of the indicators shown in the accompanying table were placed in separate portions of a \(1.0-M\) solution of a weak acid, HX. The results are shown in the last column of the table. What is the approximate \(\mathrm{pH}\) of the solution containing HX? Calculate the approximate value of \(K_{\mathrm{a}}\) for HX. $$ \begin{array}{|lcccc|} \hline \text { Indicator } & \begin{array}{c} \text { Color } \\ \text { of Hin } \end{array} & \begin{array}{c} \text { Color } \\ \text { of In }^{-} \end{array} & \begin{array}{c} \mathrm{pK}_{\mathrm{a}} \\ \text { of Hin } \end{array} & \begin{array}{c} \text { Color of } \\ 1.0 \mathrm{MHX} \end{array} \\ \text { Bromphenol blue } & \text { Yellow } & \text { Blue } & 4.0 & \text { Blue } \\ \text { Bromcresol purple } & \text { Yellow } & \text { Purple } & 6.0 & \text { Yellow } \\ \text { Bromcresol green } & \text { Yellow } & \text { Blue } & 4.8 & \text { Green } \\ \text { Alizarin } & \text { Yellow } & \text { Red } & 6.5 & \text { Yellow } \\\ \hline \end{array} $$

Which of the following can be classified as buffer solutions? a. \(0.25 \mathrm{M} \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}\) b. \(0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}\) c. \(0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{M} \mathrm{KOCl}\) d. \(0.70 \mathrm{MKOH}+0.70 \mathrm{M} \mathrm{HONH}_{2}\) e. \(0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}\)

Consider a buffered solution containing \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{NH}_{2}\). Which of the following statements concerning this solution is(are) true? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}=2.3 \times 10^{-11}\).) a. A solution consisting of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) would have a higher buffering capacity than one containing \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\). b. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then the \(\mathrm{pH}\) is larger than the \(\mathrm{p} K_{\mathrm{a}}\) value. c. Adding more \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\right]\) to the initial buffer solution will decrease the \(\mathrm{pH}\). d. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]<\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}<3.36\). e. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}=10.64\).
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