what is the percent ionization at equilibrium in a 0.31M solution of dimethylamine (Kb=4.2x10^-10) at 25 Celsius
......CH3NH2 + H2O ==> CH3NH3^+ + OH^-
I.....0.31M..............0.........0
C.......-x...............x.........x
E.....0.31-x.............x.........x
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Substitute the E line into the Kb expression and solve for x = (OH^-).
Then % ion = [(OH^-)/(CH3NH2)]*100
To find the percent ionization at equilibrium in a solution of dimethylamine, we need to use the given Kb value.
1. Write out the balanced chemical equation for the ionization of dimethylamine (C2H5)2NH:
(C2H5)2NH ⇌ (C2H5)2NH2+ + OH-
2. Set up an ICE (Initial, Change, Equilibrium) table:
Initial: (C2H5)2NH = 0.31 M, (C2H5)2NH2+ = 0 M, OH- = 0 M
Change: -x, +x, +x
Equilibrium: 0.31 - x, x, x
3. Write the expression for the base ionization constant Kb:
Kb = [(C2H5)2NH2+][OH-] / [(C2H5)2NH]
4. Substitute the equilibrium concentrations into the expression:
Kb = (x)(x) / (0.31 - x)
5. Simplify the equation by assuming that x is very small compared to 0.31:
Kb = x^2 / 0.31
6. Rearrange the equation to solve for x:
x^2 = Kb * 0.31
x = √(Kb * 0.31)
7. Calculate the value of x using the given Kb value and concentration:
x = √(4.2x10^-10 * 0.31) = 1.377x10^-6
8. Calculate the percent ionization at equilibrium:
% ionization = (x / initial concentration) * 100
% ionization = (1.377x10^-6 / 0.31) * 100 = 4.44x10^-4 %
Therefore, the percent ionization at equilibrium in a 0.31 M solution of dimethylamine at 25°C is approximately 4.44x10^-4%.