Answer :
To find the hydroxide ion concentration [tex]\([OH^-]\)[/tex] in a 0.255 M solution of pyridine (C₅H₅N), given that the base dissociation constant (Kb) of C₅H₅N is [tex]\(1.69 \times 10^{-9}\)[/tex], we can use the following steps:
1. Understanding the Chemical Equilibrium:
Pyridine (C₅H₅N) is a weak base and ionizes in water as follows:
[tex]\[
\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightleftharpoons \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-
\][/tex]
2. Set Up the Expression for Kb:
The base ionization constant expression for pyridine is:
[tex]\[
Kb = \frac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]}
\][/tex]
3. Assumptions and Simplification:
Assume that the concentration of [tex]\([\text{OH}^-]\)[/tex] is [tex]\(x\)[/tex]. Initially, the change in concentration for pyridine when it dissociates is negligible compared to the original concentration, so:
[tex]\[
[\text{C}_5\text{H}_5\text{N}] \approx 0.255
\][/tex]
Then, after a small amount dissociates:
[tex]\[
[\text{C}_5\text{H}_5\text{NH}^+] = [\text{OH}^-] = x
\][/tex]
4. Plug into the Kb Expression:
Substituting these into the Kb equation gives:
[tex]\[
Kb = \frac{x \times x}{0.255} = \frac{x^2}{0.255}
\][/tex]
5. Solve for [tex]\([OH^-]\)[/tex]:
Rearrange the equation to solve for [tex]\(x\)[/tex]:
[tex]\[
x^2 = (1.69 \times 10^{-9}) \times 0.255
\][/tex]
[tex]\[
x = \sqrt{(1.69 \times 10^{-9}) \times 0.255}
\][/tex]
6. Calculate x as [tex]\([OH^-]\)[/tex]:
After performing the calculation, we find that:
[tex]\[
[\text{OH}^-] \approx 2.08 \times 10^{-5} \text{ M}
\][/tex]
Thus, the hydroxide ion concentration [tex]\([OH^-]\)[/tex] in the solution is approximately [tex]\(2.08 \times 10^{-5} \, \text{M}\)[/tex].
1. Understanding the Chemical Equilibrium:
Pyridine (C₅H₅N) is a weak base and ionizes in water as follows:
[tex]\[
\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightleftharpoons \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-
\][/tex]
2. Set Up the Expression for Kb:
The base ionization constant expression for pyridine is:
[tex]\[
Kb = \frac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]}
\][/tex]
3. Assumptions and Simplification:
Assume that the concentration of [tex]\([\text{OH}^-]\)[/tex] is [tex]\(x\)[/tex]. Initially, the change in concentration for pyridine when it dissociates is negligible compared to the original concentration, so:
[tex]\[
[\text{C}_5\text{H}_5\text{N}] \approx 0.255
\][/tex]
Then, after a small amount dissociates:
[tex]\[
[\text{C}_5\text{H}_5\text{NH}^+] = [\text{OH}^-] = x
\][/tex]
4. Plug into the Kb Expression:
Substituting these into the Kb equation gives:
[tex]\[
Kb = \frac{x \times x}{0.255} = \frac{x^2}{0.255}
\][/tex]
5. Solve for [tex]\([OH^-]\)[/tex]:
Rearrange the equation to solve for [tex]\(x\)[/tex]:
[tex]\[
x^2 = (1.69 \times 10^{-9}) \times 0.255
\][/tex]
[tex]\[
x = \sqrt{(1.69 \times 10^{-9}) \times 0.255}
\][/tex]
6. Calculate x as [tex]\([OH^-]\)[/tex]:
After performing the calculation, we find that:
[tex]\[
[\text{OH}^-] \approx 2.08 \times 10^{-5} \text{ M}
\][/tex]
Thus, the hydroxide ion concentration [tex]\([OH^-]\)[/tex] in the solution is approximately [tex]\(2.08 \times 10^{-5} \, \text{M}\)[/tex].
