Questions: A solution is prepared at 25°C that is initially 0.68 M in diethylamine ((C2H5)2NH), a weak base with Kb=1.3 x 10^-3, and 1.5 M in diethylammonium bromide ((C2H5)2NH2Br). Calculate the pH of the solution. Round your answer to 2 decimal places. pH=

Solution Steps

The solution contains diethylamine, a weak base, and diethylammonium bromide, its conjugate acid. This forms a buffer solution. The base dissociation constant \( K_b \) for diethylamine is given as \( 1.3 \times 10^{-3} \).

For a buffer solution, the pH can be calculated using the Henderson-Hasselbalch equation for bases:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

First, we need to find \(\text{pK}_a\). The relationship between \(K_a\) and \(K_b\) is:

\[ K_w = K_a \times K_b \]

where \(K_w = 1.0 \times 10^{-14}\) at \(25^\circ \text{C}\).

\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-3}} = 7.6923 \times 10^{-12} \]

\[ \text{pK}_a = -\log(K_a) = -\log(7.6923 \times 10^{-12}) \approx 11.11 \]

Given:

• \([\text{Base}] = 0.68 \, \text{M}\)
• \([\text{Acid}] = 1.5 \, \text{M}\)

\[ \text{pH} = 11.11 + \log \left( \frac{0.68}{1.5} \right) \]

\[ \text{pH} = 11.11 + \log(0.4533) \approx 11.11 - 0.3434 = 10.77 \]

Final Answer