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Chapter 14: Problem 28

Derive an expression for the relationship between \(\mathrm{p} K_{\mathrm{a}}\) and \(\mathrm{p} K_{\mathrm{b}}\) for a conjugate acid-base pair. \((\mathrm{p} K=-\log K .)\)

Short Answer

Expert verified
The relationship between pKa and pKb for a conjugate acid-base pair can be derived using the relationship between Ka and Kb. Using the equation \(Ka \times Kb = Kw\), taking the negative logarithm of both sides, applying logarithm properties, and substituting the value of \(-\log(Kw) = 14\), we obtain the final expression: \(pKa + pKb = 14\).

Step by step solution

01

Write down the relationship between Ka and Kb

We have the relationship between the acid dissociation constant (Ka) and the base dissociation constant (Kb) as follows: \[Ka \times Kb = Kw\] where Kw is the ion-product constant for water, which is equal to \(1 \times 10^{-14}\) at 25°C.
02

Take the negative logarithm of both sides of the equation

Now, we will take the negative logarithm (base 10) of both sides of the equation, which gives us the relationship between pKa and pKb: \[-\log(Ka \times Kb) = -\log(Kw)\]
03

Apply logarithm properties

Recall that \(-\log(a \times b) = -\log(a) - \log(b)\). We will use this property to separate the terms on the left side of the equation: \[-\log(Ka) - \log(Kb) = -\log(Kw)\]
04

Simplify the equation

Now, rewrite the equation in terms of pKa and pKb. We know that pKa = -log(Ka) and pKb = -log(Kb): \(pKa + pKb = -\log(Kw)\)
05

Find the value of -log(Kw)

We know that Kw is equal to \(1 \times 10^{-14}\) at 25°C. We can calculate -log(Kw) as follows: \[-\log(Kw) = -\log(1 \times 10^{-14}) = -(-14) = 14\]
06

Write the final expression for the relationship between pKa and pKb

Finally, substitute the value of -log(Kw) into the equation: \(pKa + pKb = 14\) This is the expression for the relationship between pKa and pKb for a conjugate acid-base pair.

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Most popular questions from this chapter

A solution of formic acid \(\left(\mathrm{HCOOH}, K_{\mathrm{a}}=1.8 \times 10^{-4}\right)\) has a \(\mathrm{pH}\) of \(2.70 .\) Calculate the initial concentration of formic acid in this solution.

A typical aspirin tablet contains \(325 \mathrm{mg}\) acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right) .\) Calculate the \(\mathrm{pH}\) of a solution that is prepared by dissolving two aspirin tablets in enough water to make one cup \((237 \mathrm{~mL})\) of solution. Assume the aspirin tablets are pure acetylsalicylic acid, \(K_{\mathrm{a}}=3.3 \times 10^{-4}\).

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

An acid HX is \(25 \%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M}\), calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\).

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.10 \mathrm{MCH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.050 M \mathrm{NaCN}\)
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