Question

use the references to access important values if needed for this question. the ph of an aqueous solution of 0.501 m isoquinoline (a weak base with the formula c₉h₇n) is 9 more group attempts remaining submit answer retry entire group

Explanation:

Step1: Find $K_b$ for isoquinoline

The base dissociation constant for isoquinoline is $K_b = 2.5 \times 10^{-9}$.

Step2: Define dissociation equilibrium

Isoquinoline ($\text{C}_9\text{H}_7\text{N}$) dissociates as:

$$\text{C}_9\text{H}_7\text{N} + \text{H}_2\text{O} ightleftharpoons \text{C}_9\text{H}_7\text{NH}^+ + \text{OH}^-$$

Let $x = [\text{OH}^-] = [\text{C}_9\text{H}_7\text{NH}^+]$, $[\text{C}_9\text{H}_7\text{N}] = 0.501 - x \approx 0.501$ (since $K_b$ is very small, $x \ll 0.501$).

Step3: Solve for $[\text{OH}^-]$

$$K_b = \frac{[\text{C}_9\text{H}_7\text{NH}^+][\text{OH}^-]}{[\text{C}_9\text{H}_7\text{N}]}$$
$$2.5 \times 10^{-9} = \frac{x^2}{0.501}$$
$$x = \sqrt{2.5 \times 10^{-9} \times 0.501} = \sqrt{1.2525 \times 10^{-9}} = 3.54 \times 10^{-5}\ \text{M}$$

Step4: Calculate pOH

$$\text{pOH} = -\log([\text{OH}^-]) = -\log(3.54 \times 10^{-5}) = 4.45$$

Step5: Calculate pH

$$\text{pH} = 14 - \text{pOH} = 14 - 4.45$$

Answer:

9.55