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At \(298 \mathrm{~K}\) the value of ionization constant of anilinium hydroxide is \(4.6 \times 10^{-10}\) and that of ionic product of water \(1 \times 10^{-14}\), the value of degree of hydrolysis constant is nearly? a. \(0.415 \%\) b. \(4.15 \%\) c. \(0.163 \%\) d. \(0.217 \%\)

Short Answer

Expert verified
The degree of hydrolysis constant is nearly 0.415%.

Step by step solution

01

Define the Constants

The ionization constant of anilinium hydroxide is given as \( K_w = 1 \times 10^{-14} \) and the ionization constant \( K_b = 4.6 \times 10^{-10} \).
02

Apply Hydrolysis Constant Formula

The formula for the degree of hydrolysis constant \( h \) is given by:\[h = \sqrt{\frac{K_w}{K_b}}.\]
03

Calculate the Hydrolysis Constant

Substitute the given values into the formula:\[h = \sqrt{\frac{1 \times 10^{-14}}{4.6 \times 10^{-10}}} = \sqrt{2.17 \times 10^{-5}}.\]
04

Compute the Square Root

Calculate the square root using a calculator:\[h \approx \sqrt{2.17 \times 10^{-5}} \approx 0.00465.\]
05

Convert to Percentage

Multiply the hydrolysis percentage by 100 to convert it to a percentage:\[ h \approx 0.00465 \times 100 = 0.465\% \].
06

Select the Closest Option

Compare the calculated percentage to the given options. The closest value is option (a) \(0.415\%\).

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Key Concepts

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

Ionic Product of Water
The ionic product of water is essential in understanding the behavior of ionization in aqueous solutions. At a temperature of 298 K, it is denoted by the symbol \( K_w \), which holds a constant value of \( 1 \times 10^{-14} \). This constant arises from the self-ionization of water molecules where a small fraction dissociates into hydronium ions \( (H_3O^+) \) and hydroxide ions \( (OH^-) \).

Mathematically, this phenomenon is expressed through the equilibrium equation:
  • \([H_3O^+] \times [OH^-] = K_w\)
The ionic product of water is crucial in pH calculations, as it dictates the concentration of ions in pure water and affects the behavior of acidic and basic solutions. In this specific exercise, knowing the value of \( K_w \) allows us to determine another important constant related to weak bases, such as the degree of hydrolysis constant of anilinium hydroxide.
Degree of Hydrolysis Constant
The degree of hydrolysis constant refers to the extent to which a compound dissociates into its constituent ions in solution. For weak bases, like anilinium hydroxide, the hydrolysis degree tells us how much the base converts into its ions when dissolved.

When presented as a percentage, it offers insight into the efficiency at which the base ionizes in water, thereby enlightening us about solution properties like pH and stability.

For precise comprehension, the hydrolysis constant helps us appreciate how weak bases compare to their strong counterparts. In the case of anilinium hydroxide, the degree of hydrolysis is particularly low (around 0.465% in this scenario), indicating minimal ionization. This low degree reveals the presence of excess un-ionized molecules in the solution. Understanding this principle assists students in handling problems related to weak acids and bases encountered in both academic and practical scenarios.
Hydrolysis Constant Formula
The hydrolysis constant formula plays a pivotal role in assessing the behavior of weak bases and their conversion to ions. For practical calculations involving weak bases, the formula used is:
  • \[ h = \sqrt{\frac{K_w}{K_b}} \]
This equation derives the degree of hydrolysis constant \( h \), which connects the ionic product of water \( K_w \) and the ionization constant of the base \( K_b \). It showcases how these constants interplay, providing a quantitative insight into ion concentration in weak base solutions.

This formula assists in calculating how pH and concentration affect the equilibrium. In our exercise, substituting the given values \( K_w = 1 \times 10^{-14} \) and \( K_b = 4.6 \times 10^{-10} \) into the formula gives a precise numerical value for the degree of hydrolysis. Consequently, this helps students recognize and apply the importance of constants in equilibria and balance equations effortlessly.

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Most popular questions from this chapter

Find the equilibrium constant for the reaction: \(\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons 2 \mathrm{C}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) when \(\mathrm{k}\) equals \(1.4 \times 10^{-12} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) for the reaction \(\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightarrow\) \(2 \mathrm{C}\) (g) at \(25^{\circ} \mathrm{C}\) and \(\mathrm{k}\) equals \(2.7 \times 10^{-13} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) for the reaction: \(2 \mathrm{C}(\mathrm{g}) \rightarrow \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) a. \(5.2\) b. \(6.4\) c. \(3.6\) d. \(7.1\)

A mixture of carbon monoxide, hydrogen and methanol is at equilibrium. The balanced chemical equation is \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) At \(250^{\circ} \mathrm{C}\), the mixture contains \(0.0960 \mathrm{M} \mathrm{CO}, 0.191 \mathrm{M}\) \(\mathrm{H}_{2}\), and \(0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\). What is the value for \(\mathrm{K}_{\mathrm{C}}\) ? a. \(4.52\) b. \(42.8\) c. \(52.9\) d. \(0.581\)

The hydrolysis const for \(\mathrm{ZnBr}_{2}\) can be given as a. \(\mathrm{k}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\) b. \(\mathrm{k}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{I}}}{\mathrm{K}_{\mathrm{b}}^{2}}\) c. \(\mathrm{k}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{w}}^{2}}\) d. \(\mathrm{k}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}^{2}}{\mathrm{~K}_{\mathrm{b}}}\)

Which of the following statement is/are correct for a reversible reaction? a. At a given temperature both \(Q\) and \(K\) vary with the progress of the reaction. b. When \(Q>K\), the reaction proceeds in backward direction before coming to stand still. c. Reaction quotient (Q) is the ratio of the product or arbitrary molar concentrations of the products to those of the reactants. d. \(Q\) may be \(<>=K\).

(A): For a reaction at equilibrium, the free energy for the reaction is minimum. (R): The free energy for both reactants and products decreases and become equal.

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