Given the pH of the solution is 12.20, we can find the pOH using the relationship:
\[
\text{pOH} = 14 - \text{pH}
\]
\[
\text{pOH} = 14 - 12.20 = 1.80
\]
Next, we calculate the concentration of OH\(^-\) ions:
\[
[\text{OH}^-] = 10^{-\text{pOH}}
\]
\[
[\text{OH}^-] = 10^{-1.80} \approx 0.0158 \, \text{M}
\]
For a weak base, the equilibrium expression is:
\[
\mathrm{C}_3\mathrm{H}_7\mathrm{NH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{C}_3\mathrm{H}_7\mathrm{NH}_3^+ + \mathrm{OH}^-
\]
Let \( x \) be the concentration of \(\mathrm{OH}^-\) ions at equilibrium. Since \( x = 0.0158 \, \text{M} \), the concentration of \(\mathrm{C}_3\mathrm{H}_7\mathrm{NH}_3^+\) is also \( 0.0158 \, \text{M} \).
The initial concentration of propylamine is \( 0.59 \, \text{M} \). At equilibrium, the concentration of propylamine is:
\[
[ \mathrm{C}_3\mathrm{H}_7\mathrm{NH}_2 ]_{\text{eq}} = 0.59 - 0.0158 \approx 0.5742 \, \text{M}
\]
The base dissociation constant \( K_b \) is given by:
\[
K_b = \frac{[\mathrm{C}_3\mathrm{H}_7\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_3\mathrm{H}_7\mathrm{NH}_2]}
\]
\[
K_b = \frac{(0.0158)(0.0158)}{0.5742}
\]
\[
K_b \approx \frac{0.0002496}{0.5742} \approx 4.347 \times 10^{-4}
\]
\(\boxed{K_b \approx 4.347 \times 10^{-4}}\)
Let \( x \) be the concentration of \(\mathrm{OH}^-\) ions at equilibrium for the 0.82 M solution. The equilibrium expression is:
\[
K_b = \frac{x^2}{0.82 - x}
\]
Assuming \( x \) is small compared to 0.82, we approximate:
\[
K_b \approx \frac{x^2}{0.82}
\]
\[
4.347 \times 10^{-4} \approx \frac{x^2}{0.82}
\]
\[
x^2 \approx (4.347 \times 10^{-4}) \times 0.82
\]
\[
x^2 \approx 3.563 \times 10^{-4}
\]
\[
x \approx \sqrt{3.563 \times 10^{-4}} \approx 0.0189 \, \text{M}
\]
The concentration of \(\mathrm{OH}^-\) ions is \( 0.0189 \, \text{M} \). The pOH is:
\[
\text{pOH} = -\log(0.0189) \approx 1.723
\]
Finally, the pH is:
\[
\text{pH} = 14 - \text{pOH} = 14 - 1.723 = 12.277
\]
\(\boxed{\text{pH} \approx 12.28}\)
\(\boxed{K_b \approx 4.347 \times 10^{-4}}\)
\(\boxed{\text{pH} \approx 12.28}\)
Answered
