Calculate the poh of a 0.200 m (c2h5)3 aqueous solution given kb= 5.2 x 10^-4

The dissociation of (C2H5 )3 N takes place as follows:

(C2H5 )3 N --> (C2H5)2N- + C2H5 +

0.2 – x x x

Whereas, dissociation of ethanol is follows:

C2H5 OH --> C2H5 + + OH-

Thus, the concentration of C2H5 + is equivalent to OH-

Kb = [(C2H5)2N- ] [C2H5 + ] / [(C2H5 )3 N ]

5.2 * 10-4 = x^2 / 0.2

x = [OH- ] = 5.01 * 10^-2 M

The concentration of hydroxyl ion is 5.01 * 10^-2 M

P (OH) = - log [OH- ]

= - log [5.01 * 10^-2 ]

= 1.31

The Poh is 1.31

the answer is true.