0.10 M methylamine (CH3NH2)
kb= 4.38e-4
calc the percent ionization
so i did the ICE table to find that x=6.26e-3
and then i divided by .1 and then multiplied by 100. but i got the wrong answer whats did i do wrong?
So far you haven't done anything wrong. But without showing your work I can't tell where you have gone wrong. The procedure you have used is correct But I get 6.62 x 10^-3 for (OH^-) and not 6.26 x 10^-3. That may be the error.
CH3NH2(aq) + H2O <=> CH3NH3+(aq) + OH-(aq)
Kb = [CH3NH3+][OH-] / [CH3NH2] = 4.38x10^-4
Let [OH-] = x, and [CH3NH2] = 1.00M
x^2 / (1.00-x) = 4.38x10^-4
If x is much smaller than 1.00, the above equation can be simplified to
x^2/1.00 = 4.38x10^-4
x = sqrt(4.38x10^-4)
solve for x
% ionization = (x/1.00)(100)
NOTE: if the above answer is not precise enough, the equation:
x^2 / (1.00-x) = 4.38x10^-4 must be simplified to a quadratic trinomial which is equal to 0. That must be solved. Of the two roots, only one would work.
A clarifying note here:
GK has used 1.00 M for the initial concentration but the problem gives 0.10 M.
To calculate the percent ionization of a weak base, you need to know the initial concentration of the base, the Kb value, and the final concentration of hydroxide ions (OH-) produced.
In this case, you have a 0.10 M solution of methylamine (CH3NH2) and a Kb value of 4.38e-4. To find the concentration of hydroxide ions, you can assume that the base dissociates completely into methylammonium ions (CH3NH3+) and hydroxide ions (OH-).
To start, let's set up the balanced chemical equation for the dissociation of methylamine:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Initially, the concentration of hydroxide ions is zero, and the concentration of methylamine is 0.10 M. Therefore, we can assume that the change in hydroxide ion concentration is x.
Now, we can set up the ICE table:
- CH3NH2 + H2O ⇌ CH3NH3+ + OH-
---------------------------
Initial 0.10 0 0
Change -x -x +x
---------------------------
Equilibrium 0.10 - x -x x
Since methylamine is a weak base, we can assume that the change in concentration is much smaller than the initial concentration (0.10 M). Therefore, we can neglect -x in comparison to 0.10 M and simplify the expression:
0.10 ≈ x
Now, we can substitute the value of x into the equation to find the actual concentration of hydroxide ions:
[OH-] = x = 0.10 M
Finally, we can calculate the percent ionization:
Percent Ionization = ([OH-] / Initial Concentration) x 100
= (0.10 M / 0.10 M) x 100
= 100%
So, the percent ionization of the methylamine is 100%.
It seems like you made a calculation mistake during the division. Make sure to double-check your calculations and ensure that you are dividing the values correctly as shown above.