The \(K_{b}\) of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\), at \(25^{\circ} \mathrm{C}\), is \(1.7 \times 10^{-9}\). Find " \(x\) " (CTQ 13), and enter the equilibrium concentration values in the last row of the following table. $$\begin{array}{|l|c|c|c|} \hline & \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} & \mathrm{OH}^{-} & \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \\ \hline \text { initial moles } & 0.30 & 0 & 0 \\ \hline \text { change in moles } & & & \\ \hline \text { equilibrium moles } & & & \\ \hline \begin{array}{l} \text { equilibrium conc. } \\ \text { expression } \end{array} & & & \\ \hline \begin{array}{l} \text { equilibrium conc. } \\ \text { value } \end{array} & & & \\ \hline \end{array}$$ Verify that your equilibrium concentrations are correct.

Short Answer

Expert verified
The value of 'x' is found to be 7.14 x 10^-6 M. The equilibrium concentrations are [C5H5N] ≈ 0.30 M, [OH-] = 7.14 x 10^-6 M and [C5H5NH+] = 7.14 x 10^-6 M. These values, when substituted back into the equilibrium expression, satisfy the given Kb value, therefore are correct.

Step by step solution

01

Understand the ionization

According to the problem, a nitrogen-containing compound, C5H5N ionizes into OH- and C5H5NH+ in water, and we have the equilibrium: C5H5N + H2O ↔ C5H5NH+ + OH-. Since initially, no OH- or C5H5NH+ is present, the concentration changes during the reaction can be defined by the variable 'x'. Thus, the table updates as: C5H5N/Initial moles is 0.30, Change in moles –x; OH-/Initial moles is 0, Change in moles +x; C5H5NH+/Initial moles is 0, Change in moles +x.
02

Set up Equilibrium Expression and Solve for 'x'

The equilibrium expression according to the ionization constant (Kb) is given by: Kb = [C5H5NH+] [OH-]/[C5H5N]. Using values from the table and solving for 'x', we have: \(1.7 x 10^{-9} = x^2/(0.30 -x)\). Assuming 'x' is very small compared to 0.30, the equation approximates to \(1.7 x 10^{-9} = x^2/0.30\), leading to \(x = sqrt(1.7 x 10^{-9} x 0.30) = 7.14 x 10^{-6}\). This is the value of 'x'.
03

Determine Equilibrium Concentrations

The equilibrium moles and concentration values fill in the table. Equilibrium moles are the initial moles plus the change in moles: C5H5N has 0.30-x, OH- has x, C5H5NH+ has x. Thus, the concentrations at equilibrium are: [C5H5N] = 0.30 - × 7.14 x 10^-6 ≈ 0.30 M (molar); [OH-] = 7.14 x 10^-6 M and [C5H5NH+] = 7.14 x 10^-6 M. This completes the table.
04

Verify that equilibrium concentrations are correct.

To verify the concentrations obtained, substitute these equilibrium concentrations back into the equilibrium expression, they should satisfy the given Kb value. Calculation would be like \(Kb = [C5H5NH+] [OH-]/[C5H5N] = (7.14 x 10^{-6})^2 / 0.30 ≈ 1.7 x 10^{-9}\). This matches closely with the given value, confirming the results to be accurate.

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