what is the pH of a solution that is 0.200 M in methylamine, CH3 NH2
CH3NH2+ H20<======> CH3NH3+ + OH-
Kb= 4.2 x 10 ^-4
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Set up an ICE chart, substitute, and solve for the unknown (OH^-), then convert to pH.
if i have 1.00 grams of N2 in 0.180 grams of H2 how many grams of NH3 is produced
11.97
To determine the pH of a solution that is 0.200 M in methylamine (CH3NH2), you can use the equilibrium constant for the reaction between methylamine and water.
The reaction is as follows:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
The equilibrium constant for this reaction is given as Kb = [CH3NH3+][OH-]/[CH3NH2].
Given the Kb value of 4.2 x 10^-4, we can use this equation to find the concentration of hydroxide ion (OH-) in the solution and then calculate the pH.
First, let's define the initial concentration of CH3NH2 as 0.200 M. If x is the concentration of OH-, then the concentration of CH3NH3+ is also x.
Using the equilibrium constant expression, we can write:
Kb = (x)(x) / (0.200 - x)
Since Kb is a small number (4.2 x 10^-4), we can make the approximation that x is very small compared to 0.200. This allows us to simplify the equation to:
Kb = x^2 / 0.200
Now, we can rearrange the equation to solve for x:
x^2 = Kb * 0.200
x = √(Kb * 0.200)
Plugging in the given Kb value:
x = √(4.2 x 10^-4 * 0.200)
x = √(8.4 x 10^-5)
x ≈ 2.90 x 10^-3
This gives us the concentration of hydroxide ion (OH-) in the solution. To find the pOH, we can use the equation: pOH = -log10(OH- concentration).
pOH = -log10(2.90 x 10^-3)
pOH ≈ 2.54
The pH of a solution is related to the pOH by the equation: pH + pOH = 14. Therefore, to find the pH, we can subtract the pOH from 14:
pH = 14 - 2.54
pH ≈ 11.46
So, the pH of the solution that is 0.200 M in methylamine (CH3NH2) is approximately 11.46.