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The hydronium ion concentration of an aqueous solution of 0.497 M triethanolamine (a weak base with...

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Chemistry

The hydronium ion concentration of an aqueous solution of 0.497M triethanolamine (a weak base with the formula C6H15O3N) is ...[H3O+] = _ M

The pOH of an aqueous solution of 0.462 M caffeine (a weak basewith the formula C8H10N4O2) is .

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3.9 Ratings (698 Votes)

-------- C6H15O3N(aq) + H2O(l) ---------------> C6H15O3NH^+ (aq) + OH^- (aq)

I---------- 0.497 -------------------------------------------------- 0 ------------------------ 0

C-------- -x ----------------------------------------------------- +x ---------------------- +x

E-------- 0.497-x ---------------------------------------------- +x --------------------- +x

             Kb   =   [C6H15O3NH^+][OH^-]/[C6H15O3N]

           5.8*10^-7   = x*x/(0.497-x)

          5.8*10^-7(0.497-x) = x^2

           x   = 0.000536

[OH^-]   = x   = 0.000536M

[H3O^+]   = Kw/[OH^-]

               = 1*10^-14/(0.000536   = 1.86*10^-11 M

-------- C8H10N4O2(aq) + H2O(l) ---------------> C8H10N4O2H^+ (aq) + OH^- (aq)

I---------- 0.462 -------------------------------------------------- 0 ------------------------ 0

C-------- -x ----------------------------------------------------- +x ---------------------- +x

E-------- 0.462-x ---------------------------------------------- +x --------------------- +x

             Kb   =   [C8H10N4O2H^+][OH^-]/[C8H10N4O2]

         5.3*10^-14   = x*x/(0.462-x)

         5.3*10^-14(0.462-x) = x^2

        x   = 1.56*10^-7

[OH^-] = x   = 1.56*10^-7 M

POH   = -log[OH^-]

          = -log(1.56*10^-7)

          = 6.8068 >>>>answer
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