Problem 79 - Chemistry: Principles and Reactions

Question

Starting with the relation $$\left[\mathrm{H}^{+}\right]=K_{\mathrm{a}} \frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}$$ derive the Henderson-Hasselbalch equation $$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{B}^{-}\right]}{[\mathrm{HB}]}$$

Step-by-Step Solution

Answer

Question: Derive the Henderson-Hasselbalch equation from the given relation between the concentration of protons, the ionization constant, and the concentrations of the conjugate acid and base. Answer: The Henderson-Hasselbalch equation is derived as follows: $$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{B}^{-}\right]}{[\mathrm{HB}]}$$

Step 1: Write down the given relation

We are given the following relation: $$\left[\mathrm{H}^{+}\right]=K_{\mathrm{a}} \frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}$$

Step 2: Take the logarithm of both sides

To derive the Henderson-Hasselbalch equation, we need to take the logarithm base 10 of both sides of the equation: $$\log_{10}\left(\left[\mathrm{H}^{+}\right]\right) = \log_{10}\left(K_{\mathrm{a}}\frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}\right)$$

Step 3: Apply logarithm properties

Next, we apply the logarithm properties to simplify the equation. We can break down the logarithm on the right side into the difference of the logarithms of the numerator and denominator, and then we can separate the logarithm of the product into a sum of the logarithms: $$\log_{10}\left(\left[\mathrm{H}^{+}\right]\right) = \log_{10} K_{\mathrm{a}}+\log_{10} \left(\frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}\right)$$

Step 4: Use definitions of pH and pKa

Now, we'll use the definitions of pH and pKa. Recall that pH = \(-\log_{10}\left(\left[\mathrm{H}^{+}\right]\right)\) and pKa = \(-\log_{10} K_{\mathrm{a}}\). We can rewrite the equation in terms of pH and pKa: $$- \mathrm{pH} = - \mathrm{p} K_{\mathrm{a}}+\log_{10} \left(\frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}\right)$$

Step 5: Derive the Henderson-Hasselbalch equation

Finally, we move the pKa term to the left side of the equation to obtain the Henderson-Hasselbalch equation: $$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{B}^{-}\right]}{[\mathrm{HB}]}$$

Key Concepts

Acid-Base Chemistry

Acid-base chemistry is a central topic in chemistry. It helps us understand how acids and bases behave and interact in various environments.
Acids are substances that can donate a proton ( H^+ ), while bases are substances that can accept a proton. This interaction is crucial because it affects the pH of a solution, a measure of its acidity or basicity.

• Acids increase the concentration of hydrogen ions ( [H^+] ) in a solution.
• Bases decrease the concentration of hydrogen ions, often by providing hydroxide ions ( OH^- ) that combine with H^+ to form water.

Understanding how acids and bases react allows chemists to predict changes in pH and the chemical equilibrium of reactions.
Acid-base reactions are common in everyday life, from digestion in the human body to the mixing of household cleaning products.

pH Calculation

The pH calculation is an important tool in chemistry, as it measures how acidic or basic a solution is. The pH scale runs from 0 to 14, with 7 being neutral.
Calculating pH involves determining the concentration of hydrogen ions in a solution using the equation:
\[\text{pH} = -\log_{10}([H^+])\]
This equation shows that pH is a logarithmic measure based on the concentration of hydrogen ions. As the hydrogen ion concentration increases, the pH decreases, indicating a more acidic solution.

• A pH less than 7 is acidic.
• A pH of 7 is neutral.
• A pH greater than 7 is basic or alkaline.

By knowing the pH, chemists can understand the properties of the solution and predict how it will react with other substances.
The concept also ties in with pKa, which is the negative logarithm of the acid dissociation constant, providing a valuable understanding of acid strength.

Equilibrium Constant

The equilibrium constant (K_a for acids and K_b for bases) is a crucial concept in understanding how reactions reach a state of balance. It indicates the extent to which a chemical reaction will progress before reaching equilibrium.
For weak acids, K_a helps determine how much an acid dissociates in solution. The equation for an acid dissociation reaction is:
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]
The equilibrium constant expression is:
\[K_a = \frac{[H^+][A^-]}{[HA]}\]
This indicates the ratio of the concentration of products to reactants at equilibrium. A higher K_a value means a stronger acid that dissociates more in the solution.

• Strong acids have larger K_a values.
• Weak acids have smaller K_a values.

This understanding aids in deriving equations like the Henderson-Hasselbalch equation, which elegantly links pH, pKa, and the concentration of acid and base components, thus providing insight into buffer solutions in biochemistry and medicine.