Asked by Mariana
Calculate the pH at the equivalence point for the titration of 0.120 M methylamine with 0.120 M HCl (kb of methyalamine is 5.0 x 10^-4)
Answers
Answered by
DrBob222
If we call methylamine, MeNH2, then
the pH at the equivalence point is determined by the hydrolysis of the salt, MeNH3Cl formed at the equivalence point. This is Ka for MeNH3Cl
.........MeNH3^+ + H2O -->MeNH2+ + H3O^+
initial..0.06M..............0........0
change.....-x................x.......x
equil.....0.06-x.............x.......x
Ka = (Kw/Kb) = (H3O^+)(MeNH2)/(MeNH3^+)
Kw = 1E-14
Kb = Kb for MeNH2
Substitute into the Ka expression and solve for x, then convert to pH.
(Note: the concn of the salt at the equivalence point is just 1/2 the concn initially. )
the pH at the equivalence point is determined by the hydrolysis of the salt, MeNH3Cl formed at the equivalence point. This is Ka for MeNH3Cl
.........MeNH3^+ + H2O -->MeNH2+ + H3O^+
initial..0.06M..............0........0
change.....-x................x.......x
equil.....0.06-x.............x.......x
Ka = (Kw/Kb) = (H3O^+)(MeNH2)/(MeNH3^+)
Kw = 1E-14
Kb = Kb for MeNH2
Substitute into the Ka expression and solve for x, then convert to pH.
(Note: the concn of the salt at the equivalence point is just 1/2 the concn initially. )
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