Final answer:
The pH of the 0.17 M solution of methylammonium nitrate is 3.30.
Step-by-step explanation:
To find the pH of the solution, we need to determine the concentration of hydronium ions (H3O+). Since methylammonium nitrate is a salt, it will dissociate into its respective ions when dissolved in water.
The equation for the reaction is:
CH3NH3NO3 -> CH3NH3+ + NO3-
Since there is no acidic or basic behavior of the anion (NO3-), we only need to consider the cation (CH3NH3+).
Given that the Kb for methylamine (CH3NH2) is 5.0 × 10−4, we can determine the concentration of CH3NH3+ by using the equation:
Kb = ([CH3NH3+][OH-])/[CH3NH2]
Plugging in the values, we get:
5.0 × 10−4 = ([CH3NH3+][OH-])/[CH3NH2]
Since we know the concentration of CH3NH3NO3 (0.17 M) and assuming complete dissociation, we can determine the concentration of CH3NH3+ by multiplying the concentration of CH3NH3NO3 by 1:
[CH3NH3+] = (0.17 M)(1) = 0.17 M
Therefore, the concentration of OH- can be determined from the Kb equation:
5.0 × 10−4 = ([0.17][OH-])/[CH3NH2]
Simplifying the equation, we get:
[OH-] = (5.0 × 10−4)([CH3NH2]/[0.17])
Using the concentration of CH3NH2 given by 0.17 M and the Kb equation, we can solve for [OH-] to find :
[OH-] = (5.0 × 10−4)(0.17/0.17) = 5.0 × 10−4
Finally, we can determine the pH of the solution using the equation:
pH = -log[H3O+]
Given that [H3O+] = [OH-] = 5.0 × 10−4, we get:
pH = -log(5.0 × 10−4) = 3.30
Therefore, the pH of the 0.17 M solution of methylammonium nitrate is 3.30.