In: Chemistry
The hydronium ion concentration of an aqueous solution of 0.497 M triethanolamine (a weak base with the formula C6H15O3N) is ... [H3O+] = _ M
The pOH of an aqueous solution of 0.462 M caffeine (a weak base with the formula C8H10N4O2) is .
-------- 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