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Chapter 5: Problem 38

The dissociation constant of two acids \(\mathrm{HA}_{1}\) and \(\mathrm{HA}_{1}\) are \(3.0 \times 10^{-4}\) and \(1.8 \times 10^{-5}\) respectively. The relative strengths of the acids is a. \(1: 16\) b. \(1: 4\) c. \(4: 1\) d. \(16: 1\)

Short Answer

Expert verified
The relative strengths of the acids are in a ratio of 16:1.

Step by step solution

01

Understand Dissociation Constant

The dissociation constant, often denoted as \(K_a\), measures how easily an acid dissociates into protons and its conjugate base. A higher \(K_a\) value indicates a stronger acid.
02

Compare Dissociation Constants

We have two dissociation constants: \(K_{a1} = 3.0 \times 10^{-4}\) for \(\mathrm{HA}_{1}\) and \(K_{a2} = 1.8 \times 10^{-5}\) for \(\mathrm{HA}_{2}\). To find their relative strengths, calculate the ratio of their dissociation constants.
03

Calculate the Ratio

The relative strength of the acids is given by the ratio \(\frac{K_{a1}}{K_{a2}}\). Compute the ratio: \[\frac{3.0 \times 10^{-4}}{1.8 \times 10^{-5}} = \frac{3.0}{1.8} \times \frac{10^{-4}}{10^{-5}} = \frac{3.0}{1.8} \times 10^1\]
04

Simplify the Ratio

Simplify \(\frac{3.0}{1.8}\):\[\frac{3.0}{1.8} = \frac{30}{18} = \frac{5}{3}\]So, the ratio becomes:\[\frac{5}{3} \times 10^1 = \frac{50}{3}\]
05

Express as Integer Ratio

Express \(\frac{50}{3}\) as a simple integer ratio. This approximates to about \(16.67\), so the nearest integer ratio is \(16:1\).

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Key Concepts

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

Relative Acid Strength
When discussing acids, one common question is about the relative strength of different acids. Acid strength refers to the ability of an acid to donate protons (H鈦�) to a solution. A stronger acid donates protons more readily. A helpful way to compare acid strength is by looking at their dissociation constants, represented as the acid dissociation constant (\( K_a \)). This constant provides a numerical value that helps indicate the strength of an acid.**Key Points to Remember:**
  • The higher the \( K_a \) value, the stronger the acid, as it dissociates more completely in solution.
  • A lower \( K_a \) value indicates a weaker acid, as it does not donate protons as readily.
  • Comparing the \( K_a \) values allows us to determine relative acid strength easily.
Understanding this principle lays the foundation for more in-depth studies in chemistry, such as analyzing weak versus strong acids in chemical reactions.
Equilibrium Constant Calculation
Equilibrium constant calculations are essential for understanding how reactions progress in solution. The acid dissociation constant, \( K_a \), specifically relates to the equilibrium of an acid dissociating into its ions:**HA** 鈬� **H鈦�** + **A鈦�**To calculate \( K_a \):
  • Identify the concentrations of each species at equilibrium. The concentrations of the ions determine how far the reaction has proceeded.
  • Use the expression \( K_a = \frac{[H^+][A^-]}{[HA]} \) to compute the constant. This ratio signifies the relationship between the dissociated and undissociated forms of the acid.
  • Remember, the magnitude of \( K_a \) will inform you on how efficiently the acid provides protons to the solution. A larger \( K_a \) means a more complete equilibrium shift toward dissociation.
These calculations are foundational as they reveal the dynamic balance of chemical reactions, especially in aqueous solutions.
Comparison of Ka Values
When two acids are compared, their \( K_a \) values are usually contrasted to evaluate their relative strengths. To perform this comparison accurately:**Steps to Compare \( K_a \) Values:**
  • Locate each acid's \( K_a \) value. For example, \( K_{a1} = 3.0 \times 10^{-4} \) and \( K_{a2} = 1.8 \times 10^{-5} \)
  • Calculate the ratio \( \frac{K_{a1}}{K_{a2}} \) to find the relative strength of one acid compared to another. In this case, the computation reflects how many times stronger one acid is than the other.
  • Simplify the ratio to get a clearer understanding of the relative strength difference. For our values, \( \frac{3.0}{1.8} \times 10^1 = \frac{50}{3} \), which approximates to about \( 16:1 \).
This numeric ratio provides clarity on which acid more effectively donates protons, thus determining their potency in reactions. This analytical approach is vital for making informed decisions in both academic and industrial chemical applications.

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Most popular questions from this chapter

(A): The value of \(\mathrm{K}\) gives us a relative idea about the extent to which a reaction proceeds. \((\mathbf{R}):\) The value of \(\mathrm{K}\) is independent of the stoichiometry of reactants and products at the point of equilibrium.

(A): Ice melts slowly at higher altitudes. (R): The melting of ice is favoured at high pressure because ice \(\longrightarrow\) water shows decrease in volume.

Which of the following condition is not favourable for formation of more ammonia in this processI. Higher temperature II. More pressure III. More concentration of \(\mathrm{NH}_{3}\) IV. More concentration of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) a. I, II, III b. I, III C. II, III, IV d. I, III, IV

The equilibrium constant \((\mathrm{Kp})\) equals \(3.40\) for the isomerization reaction: cis-2-butene \(\rightleftharpoons\) trans-2-butene If a flask initially contains \(0.250 \mathrm{~atm}\) of cis-2-butene and \(0.125\) atm of trans-2-butene, what is the equilibrium pressure of each gas? a. \(\mathrm{P}(\mathrm{cis}-2\)-butene \()=0.085 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.290 \mathrm{~atm}\) b. \(\mathrm{P}(\) cis-2-butene \()=0.058 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.230 \mathrm{~atm}\) c. \(\mathrm{P}(\mathrm{cis}-2\)-butene \()=0.028 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.156 \mathrm{~atm}\) d. \(\mathrm{P}\) (cis-2-butene) \(=0.034 \mathrm{~atm}\), \(\mathrm{P}(\) trans-2-butene \()=0.128 \mathrm{~atm}\)

The solubility of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(\mathrm{y} \mathrm{mol} \mathrm{dm}^{-3} .\) Its solubility product is a. \(6 y^{4}\) b. \(64 \mathrm{y}^{4}\) c. \(36 \mathrm{y}^{5}\) d. \(108 \mathrm{y}^{5}\)
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