Question 18488

Jan 22, 2017

The degree of dissociation $\textsf{\alpha = 0.0158}$

$\textsf{{K}_{b} = 2.51 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

Explanation:

Triethyamine is a weak base and ionises:

$\textsf{{\left(C {H}_{3}\right)}_{3} N + {H}_{2} O r i g h t \le f t h a r p \infty n s {\left(C {H}_{3}\right)}_{3} \stackrel{+}{N} H + O {H}^{-}}$

For which:

$\textsf{{K}_{b} = \frac{\left[{\left(C {H}_{3}\right)}_{3} \stackrel{+}{N} H\right] \left[O {H}^{-}\right]}{\left[{\left(C {H}_{3}\right)}_{3} N\right]}}$

Rearranging and taking -ve logs of both sides we get the expression:

$\textsf{p O H = \frac{1}{2} \left[p {K}_{b} - \log b\right]}$

$\textsf{b}$ is the concentration of the base at equilibrium. Because the dissociation is small we can assume that the initial concentration is a good enough approximation to that at equilibrium.

We can get $\textsf{p O H}$ using the expression:

$\textsf{p H + p O H = 14}$ at 298K

$\therefore$$\textsf{p O H = 14 - p H = 14 - 10.2 = 3.8}$

$\therefore$$\textsf{3.8 = \frac{1}{2} \left[p {K}_{b} - \log \left(0.01\right)\right] = \frac{1}{2} p {K}_{b} + 1}$

$\therefore$$\textsf{\frac{1}{2} p {K}_{b} = 2.8}$

$\textsf{p {K}_{b} = 5.6}$

$\therefore$$\textsf{- \log {K}_{b} = 5.6}$

$\therefore$$\textsf{{K}_{b} = {10}^{- 5.6} = 2.51 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

The degree of dissociation $\alpha$ is the fraction of the base which has dissociated.

We can find the concentration of $\textsf{O {H}^{-}}$:

$\textsf{p O H = 3.8 = - \log \left[O {H}^{-}\right]}$

$\therefore$$\textsf{\left[O {H}^{-}\right] = {10}^{- 3.8} = 1.585 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\alpha = \frac{\left[O {H}^{-}\right]}{\left[{\left(C {H}_{3}\right)}_{3} N\right]}}$

$\therefore$$\textsf{\alpha = 1.585 \times {10}^{- 4} / \left(0.01\right) = 0.0158}$

The standard, if not more long - winded method, is to set up an ICE table in concentrations (M):

$\text{ } \textsf{{\left(C {H}_{3}\right)}_{3} N + {H}_{2} O r i g h t \le f t h a r p \infty n s {\left(C {H}_{3}\right)}_{3} \stackrel{+}{N} H + O {H}^{-}}$

$\textsf{I \text{ "c" "0" } 0}$

$\textsf{C \text{ "-calpha" "+calpha" } + c \alpha}$

$\textsf{E \text{ "c(1-alpha)" "calpha" } c \alpha}$

$\therefore$sf(K_b=(c^cancel(2)alpha^(2))/(cancel(c)(1-alpha))=(calpha^2)/((1-alpha))

Because the dissociation is small we assume that $\textsf{\left(1 - \alpha\right) \Rightarrow 1}$

$\therefore$$\textsf{{K}_{b} = c {\alpha}^{2}}$

$\therefore$$\textsf{{\alpha}^{2} = {K}_{b} / c}$

$\therefore$$\textsf{\alpha = \sqrt{{K}_{b} / c}}$

sf(alpha=sqrt((2.511xx10^(-6))/0.01)#

$\textsf{\alpha = 0.0158}$

Whichever method you prefer gets you the same answer.