Asked by anonymous
1.) What must be the molarity of an aqueous solution of trimethylamine (CH3)3N
if it has a Ph of 11.04? Kb = 6.3*10^-5
if someone can show the mechanics of doing this, will be appreciated much.
if it has a Ph of 11.04? Kb = 6.3*10^-5
if someone can show the mechanics of doing this, will be appreciated much.
Answers
Answered by
DrBob222
Think ammonia, NH3.
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3). You've probably solved 100 problem where you know pH of a NH3 solution and you calculate (NH3).
(CH3)3N does EXACTLY the same thing.
(CH3)3N + HOH ==> (CH3)3NH^+ + OH^-
Write Kb expression, look up Kb, calculate OH from pH and solve for (CH3)3N.
Post your work if you get stuck.
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3). You've probably solved 100 problem where you know pH of a NH3 solution and you calculate (NH3).
(CH3)3N does EXACTLY the same thing.
(CH3)3N + HOH ==> (CH3)3NH^+ + OH^-
Write Kb expression, look up Kb, calculate OH from pH and solve for (CH3)3N.
Post your work if you get stuck.
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