To determine the hydroxide ion concentration [tex]\([\text{OH}^-]\)[/tex] in a 1.40 M solution of pyridine (C[tex]\(_5\)[/tex]H[tex]\(_5\)[/tex]N), which has a base dissociation constant (K[tex]\(_\text{b}\)[/tex]) of 1.70 [tex]\(\times\)[/tex] 10[tex]\(^{-9}\)[/tex], we can follow these step-by-step calculations:
1. Write the balanced equation for the dissociation of pyridine in water:
[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 the base dissociation constant (K[tex]\(_\text{b}\)[/tex]):
[tex]\[
K_\text{b} = \frac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]}
\][/tex]
3. Define the initial concentrations and changes at equilibrium:
- Initially: [tex]\([\text{C}_5\text{H}_5\text{N}] = 1.40\)[/tex] M, [tex]\([\text{C}_5\text{H}_5\text{NH}^+] = [\text{OH}^-] = 0\)[/tex] M.
- At equilibrium: [tex]\([\text{C}_5\text{H}_5\text{NH}^+] = x\)[/tex] M, [tex]\([\text{OH}^-] = x\)[/tex] M, and [tex]\([\text{C}_5\text{H}_5\text{N}] = 1.40 - x\)[/tex] M.
4. Substitute these values into the K[tex]\(_\text{b}\)[/tex] expression:
[tex]\[
K_\text{b} = \frac{(x)(x)}{1.40 - x}
\][/tex]
5. Since [tex]\(K_\text{b}\)[/tex] is very small, we can assume that [tex]\(x\)[/tex] (the amount of dissociation) is small compared to the initial concentration, so [tex]\(1.40 - x \approx 1.40\)[/tex]:
[tex]\[
K_\text{b} \approx \frac{x^2}{1.40}
\][/tex]
6. Rearrange the equation to solve for [tex]\(x\)[/tex]:
[tex]\[
x^2 = K_\text{b} \times 1.40
\][/tex]
7. Insert the value of [tex]\(K_\text{b}\)[/tex]:
[tex]\[
x^2 = (1.70 \times 10^{-9}) \times 1.40
\][/tex]
8. Calculate [tex]\(x\)[/tex] (which represents the [tex]\([\text{OH}^-]\)[/tex] concentration):
[tex]\[
x = \sqrt{(1.70 \times 10^{-9}) \times 1.40}
\][/tex]
9. Thus, the concentration of hydroxide ions [tex]\([\text{OH}^-]\)[/tex] is found to be approximately:
[tex]\[
[\text{OH}^-] \approx 4.88 \times 10^{-5} \text{ M}
\][/tex]
Therefore, the [tex]\([\text{OH}^-]\)[/tex] concentration in a 1.40 M solution of pyridine is approximately [tex]\(4.88 \times 10^{-5}\)[/tex] M.
