NEWW-
0.10 M methylamine (CH3NH2)
kb= 4.38e-4
calc the percent ionization
so i did the ICE table to find that x=6.62e-3 (that was a typo the first time)
and then i divided by .1 and then multiplied by 100. but i got the wrong answer whats did i do wrong?
my answer was 6.62% but that was wrong... -__-
I solved the quadratic and obtained 0.0064 which is appreciably different than 6.62 x 10^-3 so %ion = 6.40 Why don't you solve it by the quadratic and see if you get the same thing?
To calculate the percent ionization of a weak base, such as methylamine (CH3NH2), you need to use the equilibrium constant expression and the given equilibrium concentration.
Here's the correct procedure:
1. Write the balanced equation for the reaction:
CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)
2. Set up the equilibrium expression using the equilibrium constant (Kb) for the reaction:
Kb = [CH3NH3+][OH-] / [CH3NH2]
3. Since methylamine is a weak base, you can assume that the reaction reaches equilibrium, and the initial concentration of CH3NH3+ and OH- will be negligible compared to the initial concentration of CH3NH2. Therefore, you can simplify the expression to:
Kb ≈ [CH3NH3+][OH-] / [CH3NH2]
4. Since this is a 0.10 M solution of methylamine, the initial concentration of CH3NH2 is also 0.10 M.
5. Let's assume x is the concentration of CH3NH3+, OH-, and the percent ionization. So, the equilibrium concentrations will be:
[CH3NH3+] = [OH-] = x
[CH3NH2] = 0.10 M - x
6. Substitute these values into the simplified equilibrium expression:
Kb ≈ x * x / (0.10 M - x)
7. Now, solve this equation for x. Note that Kb equals 4.38e-4:
4.38e-4 ≈ x^2 / (0.10 - x)
Rearrange and restructure:
4.38e-4 * (0.10 - x) ≈ x^2
4.38e-4 * 0.10 - 4.38e-4 * x ≈ x^2
4.38e-5 - 4.38e-4 * x ≈ x^2
x^2 + 4.38e-4 * x - 4.38e-5 ≈ 0
Use the quadratic formula to solve for x:
x = [-4.38e-4 ± √((4.38e-4)^2 - 4(1)(-4.38e-5)) ] / (2)
x ≈ -4.38e-4 ± √((4.38e-4)^2 + 1.752e-4) / (2)
x ≈ -4.38e-4 ± √(1.914e-7 + 1.752e-4) / (2)
x ≈ -4.38e-4 ± √(1.7531914e-4) / (2)
x ≈ -4.38e-4 ± 4.1856e-3 / (2)
x ≈ -4.38e-4 ± 2.0928e-3
x ≈ 1.7094e-3 or x ≈ -6.1106e-3
Since the concentration cannot be negative, we discard the negative solution.
8. Now, substitute the value of x into the equation for percent ionization:
Percent ionization = (x / [CH3NH2]) * 100
Percent ionization = (1.7094e-3 / 0.10) * 100
Percent ionization ≈ 1.71%
Therefore, the correct percent ionization of methylamine in the given solution is approximately 1.71%.