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Chapter 15: Problem 89

Derive an equation analogous to the Henderson-Hasselbalch equation but relating 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 short answer based on the step-by-step solution is: To derive an equation analogous to the Henderson-Hasselbalch equation relating pOH and pKb of a buffered solution composed of a weak base (B) and its conjugate acid (HB+), we first write the expression for the base dissociation constant (Kb): \[K_{b} = \frac{[HB^{+}][OH^{-}]}{[B]}\] After assuming that the dissociation is small compared to the initial concentrations and isolating the concentration of hydroxide ions, we obtain the relationship between pOH and pKb: \[pOH = pK_{b} - \log{\frac{[B]_{0}}{[HB^{+}]_{0}}}\]

Step by step solution

01

Write the dissociation expression for the weak base

Write the expression for the dissociation of the weak base (B) in water: \[B + H_{2}O \rightleftharpoons HB^{+} + OH^{-}\]
02

Write the expression for the base dissociation constant, Kb

Write the expression for the base dissociation constant (Kb): \[K_{b} = \frac{[HB^{+}][OH^{-}]}{[B]}\]
03

Express the concentrations of the species in terms of the initial concentrations and dissociation amount

Let x be the amount dissociated, we can write the following: \[[HB^{+}] = [HB^{+}]_{0} + x\] \[[B] = [B]_{0} - x\] \[[OH^{-}] = x\]
04

Substitute the expressions in Step 3 in the Kb expression

Substituting the concentrations of the species from step 3 into the Kb expression from step 2, we get: \[K_{b} = \frac{([HB^{+}]_{0} + x)(x)}{([B]_{0} - x)}\]
05

Assume that the dissociation is small compared to the initial concentrations

For a weak base, we can assume that x (dissociation) is much smaller than the initial concentrations. Therefore, we can simplify the equation in step 4: \[K_{b} \approx \frac{([HB^{+}]_{0})(x)}{([B]_{0})}\]
06

Isolate x (concentration of OH- ions)

Isolate x in the equation from step 5 to find the concentration of hydroxide ions: \[[OH^{-}] = x = \frac{K_{b}[B]_{0}}{[HB^{+}]_{0}}\]
07

Write the expressions for pOH and pKb

Write the expressions for pOH and pKb: \[pOH = -\log{[OH^{-}]}\] \[pK_{b} = -\log{K_{b}}\]
08

Substitute the expression for the concentration of OH- ions from step 6 in the pOH expression

Substituting the expression for [OH-] from step 6 into the expression for pOH, we get: \[pOH = -\log{\left(\frac{K_{b}[B]_{0}}{[HB^{+}]_{0}}\right)}\]
09

Manipulate the expression to obtain a relationship between pOH and pKb

Use the properties of logarithms to simplify the expression from step 8: \[pOH = - \left(\log{K_{b}} + \log{\frac{[B]_{0}}{[HB^{+}]_{0}}}\right) = pK_{b} -\log{\frac{[B]_{0}}{[HB^{+}]_{0}}}\] And our final relation between pOH and pKb is: \[pOH = pK_{b} - \log{\frac{[B]_{0}}{[HB^{+}]_{0}}}\]

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Most popular questions from this chapter

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 \(\mathrm{HSO}_{3}\) .

Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 \(\mathrm{mL}\) of each of the following buffered solutions. $$ \begin{array}{l}{\text { a. } 0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{MNH}_{4} \mathrm{Cl}} \\ {\text { b. } 0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}}\end{array} $$ 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?

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5} )\) and 0.100 \(\mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the pH after 20.0\(\%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH} :\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(b\) and \(c\) agree? Explain.

A best buffer has about equal quantities of weak acid and conjugate base present as well as having a large concentration of each species present. Explain.

Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of $\mathrm{H}_{2} \mathrm{CO}_{3}\( and \)\mathrm{HCO}_{3}$ - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.
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