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Chapter 14: 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 for weak base and its conjugate acid systems, relating pOH and pKb, is given by: \[ \mathrm{pOH} = \mathrm{p}K_b + \log \frac{[BH^+]}{[B]} \]

Step by step solution

01

Write the base dissociation equilibrium expression

Write the base equilibrium dissociation expression for a weak base B, which accepts a proton from the solvent water to yield its conjugate acid BH+ and hydroxide ion OH-: \[ B + H_2O \leftrightarrows BH^+ + OH^- \]
02

Write the dissociation constant expression

Write the equilibrium constant expression, Kb, for the base dissociation equilibrium: \[ K_b = \frac{[BH^+][OH^-]}{[B]} \]
03

Rewrite equation using pOH and pKb

Convert the equation to pOH and pKb by taking the negative base-10 logarithm of both sides. This allows us to define pOH as -log10[OH-] and pKb as -log10Kb: \[ \mathrm{p}K_b = \mathrm{pOH} - \log \frac{[BH^+]}{[B]} \]
04

Rearrange the equation to solve for pOH

Rearrange the equation to calculate the pOH value, which is needed to derive an equation analogous to the Henderson-Hasselbalch equation: \[ \mathrm{pOH} = \mathrm{p}K_b + \log \frac{[BH^+]}{[B]} \] Our final equation analogous to the Henderson-Hasselbalch equation, relating pOH, pKb, and the concentrations of the weak base and its conjugate acid, is: \[ \mathrm{pOH} = \mathrm{p}K_b + \log \frac{[BH^+]}{[B]} \]

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Most popular questions from this chapter

In the titration of \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with \(0.50 M\) HCl, calculate the pH under the following conditions. a. after \(50.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl has been added b. at the stoichiometric point

Calculate the \(\mathrm{pH}\) after 0.010 mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{mL}\) of each of the following buffered solutions. a. \(0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 M \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) HONH \(_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 M\) HONH \(_{3}\) Cl c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing 0.100 \(M \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

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

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between 7.1 and 7.2 a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)
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