Calculate the pH of a 1.60 mol L^-1 CH₃NH₃Cl solution.K_b for methylamine,CH₃NH₂,is $3.7 \times 10 ^- 4$ A)1.61 B)5.18 C)8.82 D)12.39

Question

Quizplus · Accepted Answer

First, we need to write the chemical equation for the reaction of methylamine with water:
CH3NH2 + H2O → CH3NH3+ + OH-
The equilibrium constant expression for this reaction is:
Kb = [CH3NH3+][OH-]/[CH3NH2]
We are given the value of Kb as 3.7 × 10^-4. We can use Kb to find the concentration of hydroxide ions in the solution.
Kb = [CH3NH3+][OH-]/[CH3NH2]
3.7 × 10^-4 = x^2/1.60
x = 0.009
Therefore, [OH-] = 0.009 M
Now, we can use the fact that Kw = [H+][OH-] = 1.0 × 10^-14 to calculate the concentration of hydronium ions in the solution.
[H+] = Kw/[OH-] = (1.0 × 10^-14)/0.009
[H+] = 1.11 × 10^-12
Finally, we can calculate the pH of the solution using the definition of pH:
pH = -log[H+]
pH = -log(1.11 × 10^-12)
pH = 5.18

Calculate the PH of a 1 3.7×10−43.7 \times 10 ^- 43.7×10−4

Calculate the PH of a 1 $3.7 \times 10 ^- 4$