Answer:
Explanation:
To find the hydroxide ion concentration (\([OH^-]\)) of a solution of pyridine (\(C_5H_5N\)), we need to use the given information about the base dissociation constant (\(K_b\)).
The equilibrium expression for the reaction of pyridine with water is:
\[ C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^- \]
The equilibrium constant expression (\(K_b\)) is given by:
\[ K_b = \frac{[C_5H_5NH^+][OH^-]}{[C_5H_5N]} \]
The concentration of \(OH^-\) at equilibrium is represented by \([OH^-]\). Since pyridine is a weak base, we can assume that the change in concentration of \(OH^-\) is small compared to the initial concentration of pyridine, and we can neglect the \(x\) term (change in concentration).
Now, we can set up the equation using the given values:
\[ K_b = \frac{x^2}{[C_5H_5N]} \]
Given that \(K_b = 1.70 \times 10^{-9}\) and the initial concentration of pyridine \([C_5H_5N] = 2.20 \, \text{M}\), we can solve for \(x\) (the concentration of \(OH^-\)).
\[ 1.70 \times 10^{-9} = \frac{x^2}{2.20} \]
Solving for \(x\):
\[ x^2 = 1.70 \times 10^{-9} \times 2.20 \]
\[ x^2 = 3.74 \times 10^{-9} \]
\[ x \approx 6.12 \times 10^{-5} \, \text{M} \]
So, the concentration of hydroxide ions (\([OH^-]\)) in the solution is approximately \(6.12 \times 10^{-5} \, \text{M}\).
