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Chapter 17: Problem 83

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

Short Answer

Expert verified
The derived equation relating the pOH of a buffer solution to the \(pK_{b}\) of its base component is: \(pOH = pK_{b} + \log \frac{[B]}{[BH^{+}]}\)

Step by step solution

01

Write the base ionization expression

To derive the desired equation, we first need to write the base ionization expression. Let B be a weak base, and let BH+ be its conjugate acid. The ionization of the base in water can be represented as follows: \(B + H_{2}O \rightleftharpoons BH^{+} + OH^{-}\)
02

Write the base ionization constant expression

We can now write the base ionization constant expression, as follows: \(K_{b} = \frac{ [BH^{+}][OH^{-}]}{[B]}\) Here, \(K_{b}\) is the base ionization constant, and [BH+], [OH-], and [B] represent the equilibrium concentrations of the conjugate acid, hydroxide ions, and the base, respectively.
03

Express pOH and \(pK_{b}\)

Next, we need to express the pOH and the \(pK_{b}\) in terms of their respective logarithms. By definition, pOH is the negative logarithm of the hydroxide ion concentration, and \(pK_{b}\) is the negative logarithm of the base ionization constant. Therefore, we can write: \(pOH = - \log [OH^{-}]\) and \(pK_{b} = - \log K_{b}\)
04

Take the logarithm of the base ionization constant expression

Now, we will take the logarithm of both sides of the base ionization constant (Step 2) expression: \(\log K_{b} = \log \frac{ [BH^{+}][OH^{-}]}{[B]}\)
05

Simplify using logarithmic properties

Using logarithmic properties, we can simplify the equation as follows: \(\log K_{b} = \log [BH^{+}] + \log [OH^{-}] - \log [B] \)
06

Insert pOH and \(pK_{b}\) expressions

Now, replace the \(\log K_{b}\) with \(-pK_{b}\) and \(\log [OH-]\) with \(-pOH\): \(-pK_{b} = \log [BH^{+}] - pOH - \log [B] \)
07

Rearrange the equation

Finally, let's rearrange the equation to relate pOH to the \(pK_{b}\) of the base component: \(pOH = pK_{b} + \log \frac{[B]}{[BH^{+}]}\) This is the derived equation that relates the pOH of a buffer solution to the \(pK_{b}\) of its base component, similar to the Henderson-Hasselbalch equation.

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Most popular questions from this chapter

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{p} \mathrm{H}\) of a 0.050 M solution of \(\mathrm{CaCl}_{2}\) is adjusted to 8.0? (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when 100 mL of 0.050 M \(\mathrm{AgNO}_{3}\) is mixed with 10 mL of \(5.0 \times 10^{-2} \mathrm{MNa}_{2} \mathrm{SO}_{4}\) solution?

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