Question

Calculate the pH of the solutions below. Include all appropriate reactions as part of your answer.

a. 0.10 M NaOCl (pKa= 7.53)

b. 0.10 M SrF2 (pKa= 28)

c. 0.10 M (CH3)3N (pKb= 4.30)

Answer 1

NaOCl is the conjugate base of the weak acid HOCl.

Ka = 10^-7.53 = 2.95 x 10^-8

We need to find the K_b value, we simply use the expression K_w = K_aK_b so K_b = K_w / K_a = (1 x 10^-14)/(2.95 x 10^-8) = 3.39 x 10^-7.

let´s build and ICE table for the equilibrium expression as follows:

	NaOCl	OH-	HOCl
I	0.10 M	0	0
R	-x	+x	+x
E	0.10 - x	x	x

Then we set up the equilbrium expression:

remember that equilibrium expression goes like:

,

the equilibrium expression we want is:

let´s assume that 0.1 -x is very close to the value of 0.1, this is because the dissociation constant is very small and the value of x will be very small too so

3.39 x 10^-7.= x² / 0.1

x² = 3.39 x10^-8

x = 0.000184, this will be the concentration for OH ion

POH = -log (0.000184) = 3.73

remember that PH + POH = 14

PH = 14 - 3.73 = 10.27

*Just 1 question at a time please =)