$$ \begin{aligned} &\text { The } \mathrm{pH} \text { of } 0.05 \mathrm{M} \text { aqueous solution of diethyl amine is } 12.0 . \text { Calculate }\\\ &K_{\mathrm{b}} \end{aligned} $$

Short Answer

Expert verified
The Kb for diethylamine when its 0.05 M aqueous solution has a pH of 12.0 is 2 x 10^{-3}.

Step by step solution

01

Understanding pH and pOH

When dealing with pH and pOH, remember that pH + pOH = 14 for aqueous solutions at 25 degrees Celsius. Since the pH is given as 12.0, we can calculate pOH as pOH = 14 - pH = 14 - 12.0 = 2.0.
02

Converting pOH to OH- concentration

To find the hydroxide ion concentration ([OH^-] ), we use the formula pOH = - og([OH^-]) . Rewriting this formula to solve for [OH^-] gives [OH^-] = 10^{-pOH} = 10^{-2.0} = 0.01 M.
03

Writing the Equilibrium Expression for Kb

The base dissociation constant, Kb, is expressed as Kb = [RNH2][OH^-] / [RNH3^+]. Given that RNH2 is diethylamine, [OH^-] is 0.01 M, and assuming RNH3^+ concentration equals the original concentration of diethylamine minus [OH^-] (although this is a very small difference and often negligible), we can write Kb.
04

Approximate Concentrations for Weak Bases

For weak bases, the concentration of the base remains close to the initial concentration because of the small amount dissociated. Therefore, [RNH3^+] is approximately equal to the initial concentration, which is 0.05 M.
05

Calculate Kb

Using the approximation that [RNH3^+] is close to the initial concentration, we can find Kb = [OH^-]^2 / [RNH3^+] = (0.01)^2 / 0.05 = 0.0001 / 0.05 = 2 x 10^{-3}.

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Key Concepts

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

Weak Base Equilibrium
Understanding weak base equilibrium is crucial in acid-base chemistry, particularly when dealing with substances like diethylamine which only partially dissociate in solution. Unlike strong bases that fully dissociate, a weak base reaches a state of equilibrium where both the base and its conjugate acid coexist in balance. This balance can be expressed with the equilibrium constant for bases, or Kb, which provides a quantitative measure of base dissociation.

Weak base equilibrium involves the transfer of a proton from water to the base, creating hydroxide ions (OH⁻) and the conjugate acid of the base. The reaction for diethylamine in water can be written as: RNH2 + H2O ↔ RNH3⁺ + OH⁻. This equilibrium is dynamic, meaning the reaction can move in either direction, and the position of the equilibrium will determine the pH of the solution.
Hydroxide Ion Concentration
The concentration of hydroxide ions, [OH⁻], in solution is a direct reflection of a base's strength and its ability to accept a proton. In the case of weak bases, such as diethylamine, the hydroxide ion concentration is not as high as it would be with a strong base. The hydroxide ion concentration affects the pH and the pOH of the solution — for every increase in concentration, the pOH decreases and vice versa. This is because pOH is the negative logarithm of the hydroxide ion concentration (pOH = -log[OH⁻]).

Quantifying [OH⁻] allows chemists to calculate both pOH and pH, which are important for understanding the acidity or basicity of a solution. In our example, the hydroxide ion concentration was found to be 0.01 M, and this value is instrumental in calculating the base's dissociation constant.
Base Dissociation Constant
The base dissociation constant, Kb, is a fixed value at a given temperature that represents the strength of a base. It is determined by the equilibrium concentrations of the products and reactants in the base's dissociation reaction in water. For a weak base like diethylamine, the value of Kb reflects how readily the base accepts a proton and forms hydroxide ions. The larger the value of Kb, the stronger the base.

The equation Kb = [RNH3⁺][OH⁻] / [RNH2] is used to calculate the dissociation constant for a weak base, where [RNH3⁺] is the concentration of the conjugate acid formed, [OH⁻] is the hydroxide ion concentration, and [RNH2] is the un-dissociated base. From the given hydroxide ion concentration and assuming the concentration of RNH2 remains approximately the initial concentration due to negligible dissociation, Kb can be computed–a crucial step in quantifying base strength.
Acid-Base Chemistry
Acid-base chemistry involves the study of proton transfer reactions and the resultant equilibrium in aqueous solutions. Acids and bases are thus classified by their ability to donate or accept protons (H⁺ ions), respectively. In water, this chemistry is defined by the self-ionization where water acts as both an acid and a base, H2O + H2O ↔ H3O⁺ + OH⁻.

The pH scale, which ranges from 0 to 14, serves as a measure of how acidic or basic a solution is, with 7 being neutral. For bases, a pH above 7 indicates basicity, with a corresponding pOH below 7. Understanding the interplay between pH, pOH, and the dissociation constants (Kb for bases and Ka for acids) is pivotal in grasping the complexity of acid-base reactions. In our example, calculating the pH from the given pKb of diethylamine reveals how these concepts are interconnected to accurately describe the nature of the solution.

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Most popular questions from this chapter

Cilculate the \(\mathrm{pH}\) of the resultant mixtures : (ii) \(10 \mathrm{ml}\). of \(0.2 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}+25 \mathrm{~mL}\) of \(0.1 \mathrm{MHCl}\), (b) \(10 \mathrm{ml}\). of \(0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\), (c) \(10 \mathrm{ml}\) of \(0.1 \mathrm{MH}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KOH}\).

The \(p K_{a}\) of acetic acid and \(p K_{b}\) of ammonium hydroxide are \(4.70\) and \(4.75\) respectively. Calculate the hydrolysis constant of ammonium acetate at \(298 \mathrm{~K}\) and also the degree of hydrolysis and \(\mathrm{pH}\) of its (a) \(0.01 M\) and (b) \(0.04 \mathrm{M}\) solutions.

Calculate the \(\mathrm{pH}\) of \(0.05 \mathrm{M}\) sodium acetate solution if the \(p K_{a}\) of acetic acid is \(4.74\)

Calculate the \(\mathrm{pH}\) of a solution of given mixtures; (a) \(\left(4 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}+6 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COONa}\right)\) in \(100 \mathrm{~mL}\) of mixture; \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}\) (b) \(5 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MOH}+250 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MCl}\); \(K_{\mathrm{a}}\) for \(M \mathrm{OH}=1.8 \times 10^{-5}\) (c) \(\left(0.25\right.\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}+0.35\) mole of \(\mathrm{CH}_{3} \mathrm{COONa}\) ) in \(500 \mathrm{ml}\) mixture; \(K_{\text {a }}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=3.6 \times 10^{-4}\)

Calculate the \(\mathrm{pH}\) of a solution made by mixing \(50 \mathrm{~mL}\) of \(0.01 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) with \(50 \mathrm{~mL}\) water. Assume complete ionisation.

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