Question

Calculate the pH of a weak base solution ([B]₀ > 100 • K_b

Calculate the pH of a 0.106 M aqueous solution of triethanolamine (C6H15O3N, Kb = 5.8×10-7) and the equilibrium concentrations of the weak base and its conjugate acid.

Answer 1

answer :-
C₅H₁₅O₃N + H₂O C₅H₁₅O₃NH⁺ + OH^-

initial concentration   0.106 M 0 0

equillibrium concerntration   0.106 -x x x

Kb = [ C₅H₁₅O₃NH⁺ ] [OH^- ] / [ C₅H₁₅O₃N ] = x² /   0.106 -x

5.8 * 10^-7 (  0.106 -x )= x²

0.6148 - 5.8 * 10^-7 x = x²

x² +  5.8 * 10^-7x  - 0.6148 * 10^-7 = 0 ( using quadratic equation we solve )

x = - b +_ (b² - 4ac ) /2a

= - 5.8 * 10^-7  +_ ((5.8 * 10^-7)² - 4( 1 ) 0.6148 * 10^-7 ) /2*1

= - 5.8 * 10^-7 +_ 33.64 *  10^-14 - 2.46 * 10^-7 /2 ( neglecting   33.64 *  10^-14 in comparition to 0.6148 * 10^-7 )

=  - 5.8 * 10^-7 +_ 24.6 * 10^-8 /2

= - 0.0058 * 10^-4   +4.9598 * 10^-4 /2 = 4.954 * 10^-4 /2 = 2.477 * 10^-4

as [OH-] = x =  2.477 * 10^-4

pOH = -log [OH] = -log [ 2.477 * 10^-4 ]= 3.60

as pH+pOH= 14 or

pH = 14 -pOH= 14 - 3.60 = 10.4