pH = 12.54 so pOH = 14.00 - 12.54 = 1.46 = - log (OH^-) so
(OH^-) = 0.0347 M
Note that I have added the missing + sign to the equation on the right.
..........CH3NH2(aq) + H2O(l) → CH3NH3^+(aq) + OH−(aq)
I.............2.65M................................0.......................0
C..........-0.0347..........................0.0347..............0.0347
E.............2.65-0.0347.................0.0347...............0.0347
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
Substitute the E line into the Kb expression and solve for Kb.
The following chemical equation represents the reaction that occurs when methylamine dissolves in water to form a basic solution.
CH3NH2(aq)+H2O(l)→CH3NH3+(aq)+OH−(aq)
The pH of 2.65M CH3NH2(aq) is 12.54. Determine the value of Kb for methylamine.
2 answers
[ HB] eq = [HB ] initial = 2.65M
pOH= 14-pH = 14 - 12.54 = 1.46
OH- = 10^-pOH = 10^-1.46 = 0.035
Kb = [B-][OH-] / [HB]
Assume [B]=[OH-] (weak base)
Kb = [OH-]^2 / [HB]
Kb= [0.035]^2 / [2.65]
Kb= 4.6 * 10^-4
pOH= 14-pH = 14 - 12.54 = 1.46
OH- = 10^-pOH = 10^-1.46 = 0.035
Kb = [B-][OH-] / [HB]
Assume [B]=[OH-] (weak base)
Kb = [OH-]^2 / [HB]
Kb= [0.035]^2 / [2.65]
Kb= 4.6 * 10^-4
1 answer