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A weak monobasic acid is half neutralized by a strong base. If the \(\mathrm{pH}\) of the solution is \(5.4\), its pKa is: (a) \(6.8\) (b) \(2.7\) (c) \(5.4\) (d) \(10.8\)

Short Answer

Expert verified
The pKa of the acid is 5.4, corresponding to option (c).

Step by step solution

01

Understand the Concept of a Half Neutralized Solution

A weak monobasic acid that is half neutralized by a strong base forms a buffer solution where the concentration of the acid is equal to the concentration of its conjugate base. This is the definition of a half-neutralization point in an acid-base titration.
02

Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for a buffer solution is \( \mathrm{pH} = \mathrm{pKa} + \log \frac{[A^-]}{[HA]} \). At half-neutralization, \([A^-] = [HA]\), so \( \log \frac{[A^-]}{[HA]} = \log 1 = 0 \). Thus, \( \mathrm{pH} = \mathrm{pKa} \).
03

Substitute the Given pH to Find pKa

Given that the \( \mathrm{pH} \) of the solution is 5.4, and since \( \mathrm{pH} = \mathrm{pKa} \) at the half-neutralization point, we substitute \( 5.4 \) for \( \mathrm{pH} \). Thus, \( \mathrm{pKa} = 5.4 \).
04

Choose the Correct Answer

Now that we calculated \( \mathrm{pKa} = 5.4 \), match this value with the options provided. The correct choice is option (c): 5.4.

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Key Concepts

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

Weak Monobasic Acid
A weak monobasic acid is an acid that only has one hydrogen atom available to donate in its structure. It partially dissociates in a solution, meaning it does not completely release its hydrogen ions like a strong acid. This partial dissociation characterizes weak acids and influences their reactions with bases.

When a weak monobasic acid reacts with a strong base, it undergoes neutralization. At the point of half neutralization, the number of unreacted acid molecules equals the number of its conjugate base molecules formed. This balance leads to the creation of a buffer solution, which resists changes in pH.

Understanding weak monobasic acids is crucial for performing calculations related to acid-base reactions and determining buffer capacities in different solutions.
pKa Calculation
Calculating the pKa value of an acid is an essential step in understanding its strength and behavior in a solution. The pKa is the negative logarithm of the acid dissociation constant, denoted as Ka. In simple terms, pKa indicates an acid's tendency to donate protons; the lower the pKa, the stronger the acid.

During half-neutralization of a weak monobasic acid, the relationship between the concentrations of the acid and its conjugate base allows for a straightforward calculation using the pH value. Since pH is equal to pKa at half-neutralization, no additional computation is required when given the pH value of the solution.

This relationship simplifies many acid-base calculations and is integral to understanding how acids and bases will interact in different environments.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a remarkable tool for understanding the pH of buffer solutions. It allows you to calculate the pH by using the concentrations of the acid and its conjugate base. The formula is given as:
  • \( \mathrm{pH} = \mathrm{pKa} + \log \left( \frac{[A^-]}{[HA]} \right) \)
This equation is especially beneficial when dealing with buffer systems. For a half-neutralized weak acid, where the acid \([HA]\) and its conjugate base \([A^-]\) are in equal concentrations, the logarithmic term becomes zero:
  • \( \log \left( \frac{[A^-]}{[HA]} \right) = \log(1) = 0 \)
Therefore, the pH is equal to the pKa at this point, illustrating an essential characteristic of buffers that resist drastic pH changes.

The Henderson-Hasselbalch equation is a cornerstone in chemistry, offering insight into the balance of acidity and basicity in a mixture.
Buffer Solution
A buffer solution is a special type of solution that resists changes to its pH when small amounts of acid or base are added. This characteristic is vital in many chemical and biological systems where maintaining a stable pH is necessary for proper function.

Buffers are typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid. When a weak monobasic acid is half-neutralized by a strong base, a buffer solution forms. At this stage, the concentrations of the acid and its conjugate base are equal, making the solution effective in maintaining pH balance.

Buffers are used in applications ranging from laboratory experiments to industrial processes and even within our bodies. For instance, the blood maintains a stable pH range due to buffer systems that neutralize excess acids or bases to minimize pH fluctuation. Understanding buffer solutions helps in designing systems that can adequately resist unwanted pH changes.

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Most popular questions from this chapter

The value of \(\mathrm{K}_{\mathrm{p}}\) in the reaction: \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{P}\left(\mathrm{CO}_{2}\right)\) (b) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{MgCO}_{3}\right)}{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}\)

For the hypothetical reactions, the equilibrium constant \((\mathrm{K})\) values are given: \(\mathrm{A} \rightleftharpoons \mathrm{B} \mathrm{K}_{1}=2\) \(\mathrm{B} \rightleftharpoons \mathrm{C} \mathrm{K}_{2}^{1}=4\) \(\mathrm{C} \rightleftharpoons \mathrm{D} \mathrm{K}_{3}^{2}=3\) The equilibrium constant \((\mathrm{K})\) for the reaction: \(\mathrm{A} \rightleftharpoons \mathrm{D}\) is (a) 3 (b) 6 (c) 12 (d) 24

The equilibrium constant for the reaction: \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}\). The value of \(\mathrm{K}_{c}\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

At \(100^{\circ} \mathrm{C}\) the vapour density of nitrogen peroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) is \(26.8\). The percentage dissociation into \(\mathrm{NO}_{2}\) molecules is: (a) \(71.64 \%\) (b) \(61.57 \%\) (c) \(83.56 \%\) (d) \(67.39 \%\)

For the reaction \(\begin{aligned}&\mathrm{PQ}_{2} \rightleftharpoons \mathrm{PQ}+\mathrm{Q} \text { the degree of } \\\&' \mathrm{~g} & \mathrm{~g}\end{aligned} \mathrm{~g}\). dissociation \((\alpha)\) of \(\mathrm{PQ}_{2}\) can be related with \(\mathrm{P}\) or \(\mathrm{V}\) as: (a) \(\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}\) (b) \(\alpha \propto \frac{1}{\mathrm{P}}\) (c) \(\alpha \propto \frac{1}{\sqrt{\mathrm{V}}}\) (d) \(\alpha \propto \mathrm{V}\)

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