Question

if kB of a weak base is 5.9 x 10 -6 what is the pH of a 0.35 M solution of this base?

Answer 1

Assume weak base is B. Consider it's dissociation in water.

B ( aq) + H₂O (l) BH ⁺ (aq) + OH ^- (aq)

Equilibrium constant for above reaction is K b = [ BH ⁺ ] [ OH ^- ] / [ B ] = 5.9 10 ^-06

Let's use ICE table.

Concentration ( M )	B ( aq) BH + (aq) + OH - (aq)
I	0.35
C	- X	+ X	+ X
E	0.35- X	X	X

Substituting these values in K b = [ BH ⁺ ] [ OH ^- ] / [ B ] , we get

( X ) ( X ) / 0.35 - X = 5.9 10 ^-06

X ² / 0.35 - X = 5.9 10 ^-06  

X ² = ( 0.35 - X )  5.9 10 ^-06  

X ² = 2.065 10 ^-06  -  5.9 10 ^-06  X

X ² +  5.9 10 ^-06  X - 2.065 10 ^-06 = 0

Comparing above equation with a X ² + b X+ c = 0 , we get a = 1 , b =  5.9 10 ^-06 , c = - 2.065 10 ^-06

Now , solve equation for X.

X = - b +/- b ² -4 ac / 2 a  

X = - b +/-   (  5.9 10 ^-06 ) ² - 4 (1) ( - 2.065 10 ^-06 ) / 2 ( 1)

X = - b +/-   8.26 10 ^-06 /2

X = - b +/- 2.874 10 ^-03 / 2

X = -  5.9 10 ^-06 + 2.874 10 ^-03 / 2 = 0.001434

or X = -  5.9 10 ^-06 - 2.874 10 ^-03 / 2 = - 0.001440

Acceptable value of X is 0.001434 .

[ BH ⁺ ] = [ OH ^- ] = 0.001434 M

We have relation, pOH = -log  [ OH ^- ]

pOH = - log 0.001434

pOH = 2.84

We have relation, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 2.84

pH = 11.16