Asked by Anonymous
what is the pH of .115 M methylammonium bromide?
Answers
Answered by
DrBob222
This is a hydrolysis question.
CH3NH3^+ + H2O ==> H3O^+ + CH3NH2
Set up an ICE chart and substitute into the below equilibrium expression.
Ka = (Kw/Kb) = (H3O^+)(CH3NH2)/(CH3NH3^+).
Solve for (H3O^+) and convert to pH.
CH3NH3^+ + H2O ==> H3O^+ + CH3NH2
Set up an ICE chart and substitute into the below equilibrium expression.
Ka = (Kw/Kb) = (H3O^+)(CH3NH2)/(CH3NH3^+).
Solve for (H3O^+) and convert to pH.
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