Answer
pH = 3.24
Work Step by Step
$$K_b = 10^{-pK } = 10^{ 8.82 } = 1.5 \times 10^{-9} $$
1. $C_5H_5NH^{+}$ is the conjugate acid of a weak base; therefore, it is a weak base.
2. Find its $K_b$:
$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{ 1.5 \times 10^{-9} } = 6.7 \times 10^{-6} $$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_5H_5NH^{+} ]& [ C_5H_5NH ]& [ H_3O^+ ]\\
Initial& 0.0482 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.0482 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ C_5H_5NH ][ H_3O^+ ]}{[ C_5H_5NH^{+} ]}$$
$$K_a = \frac{(x)(x)}{[ C_5H_5NH^{+} ]_{initial} - x}$$
3. Assuming $ 0.0482 \gt\gt x:$
$$K_a = \frac{x^2}{[ C_5H_5NH^{+} ]_{initial}}$$
$$x = \sqrt{K_a \times [ C_5H_5NH^{+} ]_{initial}} = \sqrt{ 6.7 \times 10^{-6} \times 0.0482 }$$
$x = 5.7 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 5.7 \times 10^{-4} }{ 0.0482 } \times 100\% = 1.2 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 5.7 x 10^{-4} $
6. $$[H_3O^+] = x = 5.7 \times 10^{-4} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 5.7 \times 10^{-4} ) = 3.24 $$
