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Answer :
To find the concentration of hydroxide ions, [tex]\([OH^-]\)[/tex], in a 0.255 M solution of pyridine ([tex]\(C_5H_5N\)[/tex]), we use the base dissociation constant, [tex]\(K_b\)[/tex], provided as [tex]\(1.69 \times 10^{-9}\)[/tex].
### Step-by-step solution:
1. Write the equilibrium expression:
[tex]\[
C_5H_5N + H_2O \rightleftharpoons C_5H_6N^+ + OH^-
\][/tex]
2. Set up the equilibrium expression using [tex]\(K_b\)[/tex]:
[tex]\[
K_b = \frac{[C_5H_6N^+][OH^-]}{[C_5H_5N]}
\][/tex]
3. Assuming initial concentration of [tex]\(C_5H_5N\)[/tex]:
- Initial concentration: [tex]\( [C_5H_5N] = 0.255 \, \text{M} \)[/tex].
- At equilibrium, [tex]\( [C_5H_5N] = 0.255 - x \)[/tex], where [tex]\(x\)[/tex] is the concentration of [tex]\(OH^-\)[/tex].
4. Make an approximation:
Given [tex]\(K_b\)[/tex] is small, assume [tex]\( x \ll 0.255 \)[/tex]. Thus, [tex]\(0.255 - x \approx 0.255\)[/tex].
5. Substitute into the equilibrium expression:
[tex]\[
K_b = \frac{x \cdot x}{0.255} = \frac{x^2}{0.255}
\][/tex]
6. Solve for [tex]\(x^2\)[/tex]:
[tex]\[
x^2 = K_b \times 0.255
\][/tex]
[tex]\[
x^2 = 1.69 \times 10^{-9} \times 0.255
\][/tex]
[tex]\[
x^2 = 4.3095 \times 10^{-10}
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \sqrt{4.3095 \times 10^{-10}}
\][/tex]
[tex]\[
x = 2.076 \times 10^{-5}
\][/tex]
8. Conclusion:
The concentration of hydroxide ions, [tex]\([OH^-]\)[/tex], is approximately [tex]\(2.076 \times 10^{-5} \, \text{M}\)[/tex].
This method accurately calculates the [tex]\([OH^-]\)[/tex] concentration in a solution with the given conditions.
### Step-by-step solution:
1. Write the equilibrium expression:
[tex]\[
C_5H_5N + H_2O \rightleftharpoons C_5H_6N^+ + OH^-
\][/tex]
2. Set up the equilibrium expression using [tex]\(K_b\)[/tex]:
[tex]\[
K_b = \frac{[C_5H_6N^+][OH^-]}{[C_5H_5N]}
\][/tex]
3. Assuming initial concentration of [tex]\(C_5H_5N\)[/tex]:
- Initial concentration: [tex]\( [C_5H_5N] = 0.255 \, \text{M} \)[/tex].
- At equilibrium, [tex]\( [C_5H_5N] = 0.255 - x \)[/tex], where [tex]\(x\)[/tex] is the concentration of [tex]\(OH^-\)[/tex].
4. Make an approximation:
Given [tex]\(K_b\)[/tex] is small, assume [tex]\( x \ll 0.255 \)[/tex]. Thus, [tex]\(0.255 - x \approx 0.255\)[/tex].
5. Substitute into the equilibrium expression:
[tex]\[
K_b = \frac{x \cdot x}{0.255} = \frac{x^2}{0.255}
\][/tex]
6. Solve for [tex]\(x^2\)[/tex]:
[tex]\[
x^2 = K_b \times 0.255
\][/tex]
[tex]\[
x^2 = 1.69 \times 10^{-9} \times 0.255
\][/tex]
[tex]\[
x^2 = 4.3095 \times 10^{-10}
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \sqrt{4.3095 \times 10^{-10}}
\][/tex]
[tex]\[
x = 2.076 \times 10^{-5}
\][/tex]
8. Conclusion:
The concentration of hydroxide ions, [tex]\([OH^-]\)[/tex], is approximately [tex]\(2.076 \times 10^{-5} \, \text{M}\)[/tex].
This method accurately calculates the [tex]\([OH^-]\)[/tex] concentration in a solution with the given conditions.
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