4 Calculate the pH of a 0300 L Buffer containing o moles Nac

4- Calculate the pH of a 0.300 L Buffer containing o moles Nacio. (Ka (HClO) = 2.8 x 108) a) Before the addition of any strong acid or strong base. b) After the addition of 375.0 mL of 0.20 M HO c) After the addition of 375.0 mL of 0.20 M Sr(OH)2

Solution

4. pKa of HClO = -log(Ka)

                         = -log(2.8 x 10^-8) = 7.55

a) initial pH

Using Hendersen-Hasselbalck equation,

pH = pKa + log(base/acid)

     = 7.55 + log(0.09/0.15) = 7.33

b) after HCl = 0.20 M x 0.375 L = 0.075 mol was added

new moles HClO = 0.15 + 0.075 = 0.225 mol

new moles NaClO = 0.09 - 0.075 = 0.015 mol

pH = 7.55 + log(0.015/0.225) = 7.38

c) after Sr(OH)2 = 0.20 M x 0.375 L = 0.075 mol was added

[OH-] = 2 x 0.075 = 0.150 mol added

new moles HClO = 0.15 - 0.15 = 0.0 mol

new moles NaClO = 0.09 + 0.15 = 0.24 mol

[NaClO] = 0.24 mol/0.675 L = 0.355 M

NaClO --> Na+ + ClO-

ClO- + H2O <==> HClO + OH-

let x amount hydrolyzed

Kb = Kw/Ka = [HClO][OH-]/[ClO-]

1 x 10^-14/2.8 x 10^-8 = x^2/0.355

x = [OH-] = 3.56 x 10^-4 M

pOH = -log[OH-] = 3.45

pH = 14 - pOH = 10.55

pH = 7.55 + log(0.015/0.225) = 7.38