To find the pH of a 2.3 M aqueous solution of methylamine (CH₃NH₂), we first need to calculate the pOH. Methylamine is a weak base, and we can use its base dissociation constant (\(K_b\)) to find the concentration of hydroxide ions \([OH^-]\) in the solution.
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Set up the equilibrium expression for the dissociation of methylamine in water:
\[ CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^- \]
The equilibrium constant (\(K_b\)) expression for this reaction is:
\[ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \]
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Define the variables for the initial concentrations and the changes that occur at equilibrium:
Let \(x\) be the concentration of \(OH^-\) produced at equilibrium.
- Initial concentration of \(CH_3NH_2\) = 2.3 M
- Change in concentration of \(CH_3NH_2\) = -\(x\)
- Change in concentration of \(CH_3NH_3^+\) and \(OH^-\) = +\(x\)
At equilibrium, we have:
- \([CH_3NH_2] = 2.3 - x\)
- \([CH_3NH_3^+] = x\)
- \([OH^-] = x\)
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Substituting these values into the \(K_b\) expression:
\[ K_b = \frac{x \cdot x}{2.3 - x} = \frac{x^2}{2.3 - x} \]
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Now we can plug in the given \(K_b = 3.6 \times 10^{-4}\):
\[ 3.6 \times 10^{-4} = \frac{x^2}{2.3 - x} \]
Assuming that \(x\) is small compared to 2.3, we can simplify the expression:
\[ 3.6 \times 10^{-4} \approx \frac{x^2}{2.3} \]
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Solving for \(x\):
Rearranging the equation gives:
\[ x^2 = 3.6 \times 10^{-4} \times 2.3 \] \[ x^2 = 8.28 \times 10^{-4} \] \[ x \approx \sqrt{8.28 \times 10^{-4}} \approx 0.0288 , \text{M} \]
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Find the concentration of \(OH^-\), which is \(x \approx 0.0288\).
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Calculate the pOH:
\[ pOH = -\log[OH^-] = -\log(0.0288) \approx 1.54 \]
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Calculate the pH:
Using the relationship \(pH + pOH = 14\):
\[ pH = 14 - pOH = 14 - 1.54 \approx 12.46 \]
So, the final answer is:
\[ \boxed{12.46} \]