calculate[OH-],[H+] of a .20 M solution of triethylamine Kb=4.0*10^-4
To calculate the concentration of hydroxide ions ([OH-]) and hydronium ions ([H+]) in a solution of triethylamine, we first need to determine the concentration of hydroxide ions.
Triethylamine (C6H15N) is a weak base, so it will partially dissociate in water to form hydroxide ions (OH-) and the corresponding conjugate acid. The dissociation reaction of triethylamine in water can be represented as follows:
C6H15N + H2O ⇌ C6H15NH+ + OH-
The equilibrium constant for this reaction is called the base dissociation constant or Kb. In this case, Kb = 4.0 × 10^-4.
The equilibrium expression for Kb is:
Kb = [C6H15NH+][OH-]/[C6H15N]
However, since we are starting with a 0.20 M solution of triethylamine, we can assume that the concentration of triethylamine ([C6H15N]) does not change significantly during the dissociation reaction. Therefore, we can simplify the equilibrium expression to:
Kb = [C6H15NH+][OH-]/(0.20)
Now we have an equation with two unknowns ([C6H15NH+] and [OH-]), but we can solve for one of them by using the Kb value.
Rearranging the equation to solve for [OH-]:
[OH-] = (Kb * 0.20) / [C6H15NH+]
Substituting the known values:
[OH-] = (4.0 × 10^-4 * 0.20) / [C6H15NH+]
Now, we need to determine the concentration of the conjugate acid ([C6H15NH+]). Since triethylamine partially dissociates, we can assume that the concentration of the conjugate acid is equal to the concentration of the hydroxide ions:
[C6H15NH+] = [OH-]
Substituting this equality into the equation:
[OH-] = (4.0 × 10^-4 * 0.20) / [OH-]
Simplifying the equation:
[OH-]^2 = 8.0 × 10^-5
Taking the square root of both sides:
[OH-] = √(8.0 × 10^-5)
[OH-] = 2.83 × 10^-3 M
Therefore, the concentration of hydroxide ions ([OH-]) in the 0.20 M solution of triethylamine is 2.83 × 10^-3 M.
To calculate the concentration of hydronium ions ([H+]), we can use the fact that water dissociates into equal concentrations of hydronium and hydroxide ions in pure water:
[H+] = [OH-]
Therefore, the concentration of hydronium ions ([H+]) in the solution is also 2.83 × 10^-3 M.
.......CH3NH2 + HOH ==> CH3NH3^+ + OH^-
initial..0.20.............0..........0
change....-x...............x.........x
equil...0.20-x.............x..........x
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Substitute into Kb expression from the ICE chart above and solve for x = OH^-, then convert to pOH and to H.