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The \(\mathrm{pH}\) of an acidic buffer according to the Henderson equation is given by (a) \(\mathrm{p} K_{a}-\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\) (b) \(\mathrm{pK}_{a}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\) (c) \(\mathrm{p} \boldsymbol{K}_{a}+\log \frac{[\mathrm{Acid}]}{[\mathrm{Salt}]}\) (d) \(-p k_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

Short Answer

Expert verified
The correct formulation of the Henderson equation that represents the pH of an acidic buffer solution is (b) \(\mathrm{pK}_{a}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\).

Step by step solution

01

Understanding the Henderson Equation

The Henderson-Hasselbalch equation is a simplified expression of the law of mass action which is used to calculate the pH of buffer solutions. The equation is given as \(pH = pKa + \log \frac{[Salt]}{[Acid]}\) .
02

Analyzing the Options

We will analyze each option and compare it with the original Henderson equation. (a) \(\mathrm{p} K_{a}-\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\): In this option, minus sign appears instead of plus in the equation which is incorrect. In options (c) and (d), the expressions after the logarithm are reversed or incorrectly sign as compared to the actual equation and hence they are incorrect. However, option (b) \(\mathrm{pK}_{a}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\) perfectly matches to the actual Henderson equation and hence, it is the correct formulation.

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Key Concepts

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

pH of Buffer Solutions
Buffer solutions are special solutions that help maintain a stable pH when small amounts of acid or base are added. They are typically made from a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. To calculate the pH of these solutions, we use the Henderson-Hasselbalch equation.

The equation is given as:
  • \( \mathrm{pH} = \mathrm{pK_a} + \log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} \)
The formula relates the pH of a buffer solution to the pKa of the acid, which indicates the acid strength, and the ratio of the concentrations of salt (conjugate base) and acid.

By applying this equation, one can predict how the buffer reacts to changes. This is crucial in biochemical experiments where maintaining a specific pH is vital for enzyme function.
Acidic Buffer Calculation
An acidic buffer is a solution that has a pH less than 7 and is typically composed of a weak acid and its salt. For example, acetic acid (CH₃COOH) and its salt, sodium acetate (CH₃COONa), can make an acidic buffer.

Calculating pH using the Henderson-Hasselbalch equation involves:
  • Determining the concentration of the acid and its salt.
  • Knowing the pKa value of the acid - a constant that varies for each acid indicating its strength.
  • Using the ratio \( \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} \) in the equation:
  • Substituting into \( \mathrm{pH} = \mathrm{pK_a} + \log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} \)
This computation helps foresee the pH precisely and ensures the buffer's effectiveness. Understanding this calculation is essential for chemistry students and professionals dealing with buffer solutions in real-world applications like pharmaceuticals and environmental science.
Law of Mass Action
The law of mass action describes how the rate of a chemical reaction is related to the concentrations of the reactants. This law plays a critical role in establishing the equilibrium state, where the forward reaction rate equals the backward reaction rate, so no net changes are observed.

For buffer solutions, this law underpins the Henderson-Hasselbalch equation by providing an understanding of acid-base equilibrium. When combined with pH calculations, it helps elucidate why buffers maintain their stability.
  • It predicts how changes in concentration affect the equilibrium position and thus the buffer's pH.
  • Allows for approximation through the Henderson-Hasselbalch equation under the assumption of minimal change in reactant concentrations.
Therefore, understanding both the law and its practical application via the equation helps one grasp the core concepts of chemical reactions and buffer systems.

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Most popular questions from this chapter

Choose the correct statement from the following: (a) \(\mathrm{pOH}=-\mathrm{p} K_{b}+\log \frac{[\mathrm{Salt}]}{[\text { Base }]}\) (b) \(\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{[\mathrm{Base}]}{[\mathrm{Salt}]}\) (c) \(\mathrm{pOH}=-\mathrm{p} K_{b}-\log \frac{[\text { Salt }]}{[\text { Base }]}\) (d) \(\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}\)

An example of a basic buffer is a solution of (a) \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NH}_{4} \mathrm{Cl}\) (b) \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{HCl}\) (c) \(\mathrm{CH}_{3} \mathrm{COONH}_{4}\) (d) \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NaOH}\)

The \(\mathrm{pH}\) of a solution of salt of strong acid and weak base is given by a) \(\frac{1}{2}\left(\log K_{w^{\prime}}+\log K_{b}+\log C\right)\) (b) \(\frac{1}{2}\left(\log K_{w}-\log K_{b}-\log C\right)\) c) \(\frac{1}{2}\left(\log K_{w^{\prime}}-\log K_{b}+\log C\right)\) d) \(\frac{1}{2}\left(-\log K_{w^{\prime}}-\log K_{b}-\log C\right)\)

The \(\mathrm{pH}\) value of a solution obtained by mixing \(5 \mathrm{~g}\) of acetic acid and \(7.5 \mathrm{~g}\) of sodium acetate and making the volume to \(500 \mathrm{~mL}\left(\mathrm{~K}_{\mathrm{CH}_{3} \mathrm{COOH}}=1.8 \times 10^{-5}\right)\) is (a) \(4.78\) (b) \(3.78\) (c) \(5.78\) (d) zero

An example of an acid buffer is a solution of (a) ammonium acetate (b) a mixture of \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NH}_{4} \mathrm{Cl}\) (c) a mixture of \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{COONa}\) (d) a solution of \(\mathrm{CH}_{3} \mathrm{COONa}\)

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