Answer
(a) Ratio = 0.19
(b) Ratio = 0.59
(c) Ratio = 1.0
(d) Ratio = 1.9
Work Step by Step
(a)
1. Since $C_5H_5NH^+$ is the conjugate acid of $C_5H_5N$ , we can calculate its ka by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.7\times 10^{- 9} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.7\times 10^{- 9}}$
$K_a = 5.882\times 10^{- 6}$
2. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.5}$
$[H_3O^+] = 3.162 \times 10^{- 5}$
3. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][C_5H_5N]}{[C_5H_5NH^+]}$
$5.881 \times 10^{-6} = \frac{3.162 \times 10^{-5}*[C_5H_5N]}{[C_5H_5NH^+]}$
$\frac{5.881 \times 10^{-6}}{3.162 \times 10^{-5}} = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
$0.185 = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
(b)
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 5}$
$[H_3O^+] = 1 \times 10^{- 5}$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][C_5H_5N]}{[C_5H_5NH^+]}$
$5.881 \times 10^{-6} = \frac{1 \times 10^{-5}*[C_5H_5N]}{[C_5H_5NH^+]}$
$\frac{5.881 \times 10^{-6}}{1 \times 10^{-5}} = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
$0.5881 = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
(c)
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 5.23}$
$[H_3O^+] = 5.888 \times 10^{- 6}$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][C_5H_5N]}{[C_5H_5NH^+]}$
$5.881 \times 10^{-6} = \frac{5.888 \times 10^{-6}*[C_5H_5N]}{[C_5H_5NH^+]}$
$\frac{5.881 \times 10^{-6}}{5.888 \times 10^{-6}} = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
$0.999 = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
(d)
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 5.5}$
$[H_3O^+] = 3.162 \times 10^{- 6}$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][C_5H_5N]}{[C_5H_5NH^+]}$
$5.881 \times 10^{-6} = \frac{3.162 \times 10^{-6}*[C_5H_5N]}{[C_5H_5NH^+]}$
$\frac{5.881 \times 10^{-6}}{3.162 \times 10^{-6}} = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
$1.85 = \frac{[C_5H_5N]}{[C_5H_5NH^+]}$
