the pH of a 0.050 M triethylamine, is 11.70
For triehtylamine, [tex](C_{2}H_{5})_{3}N[/tex], the reaction will be
[tex](C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-} [/tex]
 and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value  = 5.3 x 10^-4 
And [tex]kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]} [/tex]
 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
                    =0.00516 M 
Thus, pOH = 2.30 
 pH = 14 - pOH = 11.7