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Chapter 14: Problem 30

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

Short Answer

Expert verified
The relationship between the pKa and pKb values for a conjugate acid-base pair can be derived as follows: \[pK_a = \log (K_w) + pK_b\] where \(K_w\) is the ion product constant for water, approximately equal to \(1 \times 10^{-14}\) at 25°C.

Step by step solution

01

Define Ka and Kb

For a conjugate acid-base pair, Ka is the acidity constant for the acid, and Kb is the basicity constant for the base. They are related by the following equation: \[K_w = K_a \times K_b\] where Kw is the ion product constant for water, which is approximately equal to \(1 \times 10^{-14}\) at 25°C.
02

Define pKa and pKb

The pKa and pKb values are defined as the negative logarithm of the acidity and basicity constants, respectively. Mathematically, the expressions are written as: \[pK_a = -\log K_a\] \[pK_b = -\log K_b\]
03

Write the relationship between Ka and Kb using pKa and pKb

We can rewrite the equation from Step 1 using logarithm rules to express the relationship between Ka and Kb in terms of pKa and pKb. First, we can take the logarithm of both sides of the equation: \[\log (K_w) = \log(K_a \times K_b)\] Now, we can use the logarithm rule that states: \(\log(ab) = \log a + \log b\). Applying this rule, we get: \[\log (K_w) = \log(K_a) + \log(K_b)\]
04

Replace log(Ka) and log(Kb) with pKa and pKb

Using the expressions for pKa and pKb from Step 2, we can replace \(\log(K_a)\) with \(-pK_a\) and \(\log(K_b)\) with \(-pK_b\). Therefore, the equation becomes: \[\log (K_w) = -pK_a - pK_b\]
05

Rearrange the equation to express the relationship between pKa and pKb

Now, we can rearrange the equation to isolate pKa, which will give us the relationship between pKa and pKb: \[pK_a = \log (K_w) - (-pK_b)\] \[pK_a = \log (K_w) + pK_b\] Thus, we have derived the expression for the relationship between the pKa and pKb values for a conjugate acid-base pair: \[pK_a = \log (K_w) + pK_b\]

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Most popular questions from this chapter

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

What are the major species present in the following mixtures of bases? a. 0.050 M NaOH and 0.050 M LiOH b. 0.0010\(M \mathrm{Ca}(\mathrm{OH})_{2}\) and 0.020\(M \mathrm{RbOH}\) What is [OH \(^{-} ]\) and the pH of each of these solutions?

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasant tart taste. Assuming that in cola drinks the concentration of phosphoric acid is \(0.007 M,\) calculate the \(\mathrm{pH}\) of this solution.

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between $\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\( and \)\left[\mathrm{CO}_{3}^{2-}\right] ?$ c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

Calculate the \(\mathrm{pH}\) of a $0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\( solution \)\left(K_{\mathrm{b}}=\right.$ \(1.3 \times 10^{-3} )\)
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