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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conjugate Acid-Base Pair
When an acid donates a proton, it forms a base known as its conjugate base. Similarly, when a base accepts a proton, it becomes a conjugate acid. Together, these create a conjugate acid-base pair. This relationship is important to understand the interplay between acidity and basicity in chemical reactions.
  • A common example is the pair of HCl (acid) and Cl- (conjugate base).
  • In water, acids increase the concentration of H3O+, while bases increase OH-.
Recognizing these pairs helps in predicting the direction of equilibrium in reactions involving weak acids and bases. Understanding conjugate acid-base pairs also assists in linking pKa and pKb values, which offer insights into the strength of acids and bases.
Ion Product Constant for Water
The ion product constant for water, denoted as Kw, is a vital concept in chemistry. It describes the equilibrium constant for the self-ionization of water:\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]At 25°C, Kw is approximately equal to \(1 \times 10^{-14}\). This value is crucial as it sets the stage for understanding how water partially ionized to establish a basis for the pH scale.
  • Kw is temperature dependent, changing at varying temperatures.
  • It shows the concentration product of H+ and OH- ions.
Understanding Kw ties directly into the exploration of pKa and pKb, highlighting how the concentration of H+ and OH- are interrelated, specifically when relating to acidic and basic substances dissolved in water.
Acidity Constant
The acidity constant, often symbolized as Ka, is a measure of the strength of an acid in solution. It represents the equilibrium constant for the dissociation reaction of an acid:\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]A larger Ka value indicates a stronger acid, which dissociates more in solution. Consequently, we derive the pKa, calculated as:\[ pK_a = -\log K_a \]This logarithmic scale simplifies representation, making it easier to compare strengths across different acids.
  • Stronger acids have lower pKa values, indicating greater degree of dissociation.
  • Knowledge of pKa helps in identifying the capacity of an acid to donate a proton.
The concept of Ka is essential when discussing the relationship between acids and bases and their respective conjugates in chemical equilibrium systems.
Basicity Constant
The basicity constant, or Kb, quantifies the strength of a base. It is the equilibrium constant for its reaction with water to produce hydroxide ions:\\[ \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \]A higher Kb value signifies a stronger base.pSimilar to the relationship of pKa for acids, the pKb, which is the negative logarithm of Kb, is given by:\[ pK_b = -\log K_b \]By understanding pKb, one can gain insight into how readily a base accepts a proton.
  • Lower pKb values represent stronger bases.
  • It is crucial in predicting the behavior and properties of base solutions.
Combining both pKa and pKb discussions provides a complete picture of acid-base balance, aiding in comprehending their roles in biochemical, industrial, and environmental processes.

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Most popular questions from this chapter

A solution is prepared by dissolving 0.56 g benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make 1.0 \(\mathrm{L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right],\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right],\) and the pH of this solution.

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Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(C_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) ii.. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}+\) b. Calculate the \(\mathrm{pH}\) of this solution.

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