For a 1.85 ×10^-3 M aqueous solution of aniline, C6 H5 NH2, do the following: (a) Identify major

For a 1.85 ×10^-3 M aqueous solution of aniline, C6 H5 NH2,...

For a $1.85 \times 10^{-3} \mathrm{M}$ aqueous solution of aniline, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$, do the following: (a) Identify major and minor species. (b) Compute concentrations of all species. (c) Find the $\mathrm{pH}$. (d) Draw a molecular picture illustrating the equilibrium reaction that determines the $\mathrm{pH}$.

Key Concepts

Molecular Representation of Equilibrium Reactions

Drawing a molecular picture that shows the equilibrium reaction helps to visualize the process at the molecular level. This involves illustrating the base molecule, water molecules, and the resulting conjugate acid and hydroxide ion. Such diagrams aid in understanding the dynamic nature of acid-base equilibria and the role of each species in determining the pH of the solution.

Weak Base Equilibrium

In aqueous solution, weak bases such as aniline do not completely react with water but establish an equilibrium. This equilibrium involves the base accepting a proton from water to form its conjugate acid and hydroxide ion. Understanding this behavior is crucial for determining the concentrations of all species present in solution and calculating the resulting pH.

Equilibrium Constant and ICE Table Method

For weak base equilibria, the equilibrium constant (Kb) quantifies the extent of protonation of the base. The ICE table (Initial, Change, Equilibrium) is a systematic approach to calculate the concentrations of reactants and products at equilibrium by relating initial concentrations to changes and final equilibrium values. This method is essential for solving for unknown concentrations in weak acid-base systems.

pH and pOH Calculations in Basic Solutions

The production of hydroxide ions in the equilibrium of a weak base with water leads to a basic solution. Calculating the pOH from the concentration of hydroxide ions and then converting it to pH is a fundamental concept in acid-base chemistry. This step involves using the relationships pOH = -log[OH?] and pH + pOH = 14 at standard temperature.

Transcript

00:01 Knowing the volume and molarity of hcl required to reach the equivalence point, and knowing the stoichiometry with anilin is one -to -one, the moles of aniline will be equal to the moles of hcl we added.

00:13 The moles of hcl added will be the volume of hcl added, converted to liters, multiplied by the concentration of hcl.

00:20 That is the moles of aniline we had at the very beginning.

00:23 We then divide that by the volume of analin, which is 25 milliliters, or 0 .025 liters, in order to get a concentration of 0 .1797 molar of anolin at the beginning before any of the hcl was added.

00:40 Then to determine the concentration of anilin at the equivalence point, we'll recognize that, i'm sorry, not aniline, but the conjugate acid to analin, which is c6h5nh3 plus, will recognize that the moles of this compound will be equal to the moles of aniline we started with, which is the volume of the anelin multiplied by the molarity of the anelin.

01:10 Well, let me back up, which will be equal to, so the moles of anilin is equal to the moles of hcl added, and the moles of the anelin conjugate acid, c6h5nh3 +, is also equal to the moles of hcl added because looking at the balanced chemical reaction, for every mole of hcl that we added, we made a mole of the conjugate acid of aniline.

01:38 So this then is the moles of the c6h5 nh3 plus that was formed at the equivalence point.

01:48 We then divide that by the new volume, the 25 milliliters we started with, plus the 25 .67 milliliters we just added, and we get 0 .0866 .6.

01:58 Molar is the concentration of that conjugate acid at the equivalence point...

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