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Chapter 7: Problem 51

The dissociation constants of two acids \(/ / A_{1}\), and \(/ I A_{2}\) are \(3.6 \times 10^{-4}\) and \(1.8 \times 10^{-5}\), respectively. The relative strengths of the acids will be (1) \(1: 4\) (2) \(4: 1\) (3) \(1: 16\) (4) \(16: 1\)

Short Answer

Expert verified
The relative strengths of the acids are closest to \(16:1\) (Option 4).

Step by step solution

01

- Understand the Dissociation Constants

The dissociation constant of an acid, often denoted as \(K_a\), measures the strength of the acid. The larger the value of \(K_a\), the stronger the acid because it implies a greater degree of dissociation in water.
02

- Compare the Given Dissociation Constants

For acid \(A_1\), the dissociation constant \(K_{a1}\) is \(3.6 \times 10^{-4}\). For acid \(A_2\), the dissociation constant \(K_{a2}\) is \(1.8 \times 10^{-5}\).
03

- Calculate the Relative Strengths

To determine the relative strengths of the acids, compare their dissociation constants: \[ \text{Relative strength} = \frac{K_{a1}}{K_{a2}} = \frac{3.6 \times 10^{-4}}{1.8 \times 10^{-5}} \] Simplify the fraction: \[ \frac{3.6 \times 10^{-4}}{1.8 \times 10^{-5}} = \frac{3.6}{1.8} \times \frac{10^{-4}}{10^{-5}} = 2 \times 10^1 = 20 \] Thus, acid \(A_1\) is 20 times stronger than acid \(A_2\).
04

- Determine the Closest Option

The calculated relative strength (20) is closest to the option 16:1. Therefore, the relative strengths of the acids are approximately \(16:1\).

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Key Concepts

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

Acid Strength
Acid strength refers to the ability of an acid to donate protons (H鈦� ions) to a base. The stronger the acid, the more completely it dissociates in water. This dissociation process is represented by the equation: HA 鈫� H鈦� + A鈦� For strong acids, nearly all the molecules dissociate into ions. In contrast, weak acids dissociate only partially. Their inability to fully dissociate means they release fewer H鈦� ions in solution, making them less acidic compared to strong acids. Understanding acid strength helps predict the behavior of acids in chemical reactions.
Relative Strength of Acids
To compare the strengths of different acids, we use their dissociation constants. This comparison can help determine which acid is stronger or weaker. In the given exercise, the relative strength is found by comparing the dissociation constants (Ka) of two acids.For instance, if acid A鈧� has a Ka of 3.6 x 10鈦烩伌 and acid A鈧� has a Ka of 1.8 x 10鈦烩伒, we find the relative strength by dividing their Ka values:Relative strength = Ka鈧� / Ka鈧俇sing the equation above, we get:Relative strength = (3.6 x 10鈦烩伌) / (1.8 x 10鈦烩伒) = 20This means acid A鈧� is 20 times stronger than acid A鈧�. The closest option to our calculated value (20) is 16:1, so acid A鈧� is approximately 16 times stronger. This method allows for easy comparison and ranking of acid strength.
Dissociation Constant (Ka)
The dissociation constant, represented as Ka, is a key indicator of acid strength. It quantifies the tendency of an acid to lose its proton in solution. A higher Ka value signifies a stronger acid, as more of the acid dissociates into ions.The dissociation constant can be mathematically expressed as:Ka = [H鈦篯[A鈦籡 / [HA]Where:- [H鈦篯 is the concentration of hydrogen ions,- [A鈦籡 is the concentration of conjugate base ions,- [HA] is the concentration of the undissociated acid.For example, in the exercise, acid A鈧� has a Ka of 3.6 x 10鈦烩伌 while acid A鈧� has a Ka of 1.8 x 10鈦烩伒. The higher Ka value of acid A鈧� indicates it is a stronger acid compared to acid A鈧�. By understanding and comparing Ka values, we can gain insights into the reactivity and behavior of different acids in solution.

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Most popular questions from this chapter

\Lambdaqucous solution of acctic acid contains (1) \(\mathrm{CII}_{3} \mathrm{COOII}, \mathrm{II}^{+}\) (2) \(\mathrm{CII}_{3} \mathrm{COO}^{-}, \mathrm{II}_{3} \mathrm{O}^{-}, \mathrm{CH}_{3} \mathrm{COOH}\) (3) \(\mathrm{CII}_{3} \mathrm{COO}^{-}, \mathrm{II}_{3} \mathrm{O}^{-}, \mathrm{II}^{+}\) (4) \(\mathrm{CII}_{3} \mathrm{COOII}, \mathrm{CII}_{3} \mathrm{COO}^{-}, \mathrm{II}^{+}\)

The \(\mathrm{p} K_{\mathrm{a}}\) of certain weak acid is \(4.0 .\) What should be the salt to acid ratio if we have to prepare a buffer with \(\mathrm{pH}=5\) using the acid and of its salts? (1) \(4: 5\) (2) \(5: 4\) (3) \(10: 1\) (4) \(1: 10\)

An aqucous solution of hydrogen sulphide shows the cquilibrium \(\mathrm{II}_{2} \mathrm{~S} \rightleftharpoons \mathrm{II}^{-} \mathrm{I} \mathrm{IIS}^{-}\) If dilute hydrochloric acid is added to an aqucous solution of hydrogen sulphide without any change in temperature, then (1) the equilibrium constant will change (2) the concentration of HS will increase (3) the concentration of nondissociated hydrogen sulphide will decrease (4) the concentration of HS will decrease

The \(10^{4} \mathrm{Ka}\) values for the acids acetic, hydrofluoric, formic and nitrous are \(6.7,4.5,1.8\) and \(0.18\) but not in the correct order. The correct acid strengths arc (1) \(\mathrm{HF}=0.18, \mathrm{HNO}_{2}=1.8, \mathrm{HCOOH}=4.5\), \(\mathrm{CH}_{3} \mathrm{COOH}=6.7\) (2) \(\mathrm{HF}=6.7, \mathrm{HNO}_{2}=4.5, \mathrm{HCOOH}=1.8, \mathrm{CH}_{3} \mathrm{COOH}\) \(=0.18\) (3) \(\mathrm{HF}=1.8, \mathrm{HNO}_{2}=0.18, \mathrm{HCOOH}=4.5\), \(\mathrm{CH}_{3} \mathrm{COOH}=6.7\) (4) \(\mathrm{HF}=6.7, \mathrm{HNO}_{2}=0.18, \mathrm{HCOOH}=4.5\) \(\mathrm{CH}_{3} \mathrm{COOH}=1.8\)

When acid is added to a buffer solution composed of a weak base (B) and its salt with strong acid, then the reaction which occur to maintain the \(\mathrm{pH}\) is (1) \(\mathrm{B}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{BH}^{+}+\mathrm{H}_{2} \mathrm{O}\) (2) \(\mathrm{OH}^{-}+\mathrm{BH}^{+} \longrightarrow \mathrm{B}+\mathrm{H}_{2} \mathrm{O}\) (3) \(\mathrm{B}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{BH}^{-}+\mathrm{OH}^{-}\) (4) \(\mathrm{BH}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{B}+\mathrm{H}_{3} \mathrm{O}^{+}\)
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