Question

Ka for hypochlorous acid, HCIO, is 3.0 x 10^-8. Calculate the pH after 10.0, 20.0, 30.0, and 40.0 mL of 0.100 M NaOH have been added to 40.0 mL of 0.100 M HCIO.

Answer 1

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

M = moles / volume

pH = -log[H+]

a) HA -> H+ + A-

Ka = [H+][A-]/[HA]

a) no NaOH

Ka = [H+][A-]/[HA]

Assume [H+] = [A-] = x

[HA] = M – x

Substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

3*10^-8 = x*x/(0.1-x)

This is quadratic equation

x =5.47*10^-5

For pH

pH = -log(H+)

pH =-log(5.47*10^-5)

pH in a = 4.26

b) 10 ml NaOH

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

pH = pKa + log (A-/HA)

initially

mmol of acid = MV = 40*0.1 = 4 mmol of acid

mmol of base = MV = 10*0.1 = 1 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 4-1 = 3 mmol

mmol of conjguate left = 0 + 1 = 1

Get pKa

pKa = -log(Ka)

pKa = -log(3*10^-8) = 7.52

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 7.52+ log (1/3) = 7.0428

c) for 20 ml

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

NOTE that in this case [A-] < [HA]; Expect pH lower than pKa

pH = pKa + log (A-/HA)

initially

mmol of acid = MV =4 mmol of acid

mmol of base = MV = 20*0.1 = 2 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 4-2= 2mmol

mmol of conjguate left = 0 + 2= 2

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 7.52+ log (2/2) = 7.52

d) for 30 ml

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

pH = pKa + log (A-/HA)

initially

mmol of acid = MV =4 mmol of acid

mmol of base = MV = 30*0.1 = 3 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 4-3= 1 mmol

mmol of conjguate left = 0 + 3= 3

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 7.52+ log (3/1) = 7.997

d) Addition of Same quantitie of Acid/Base

This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium

We will need Kb

Ka*Kb = Kw

And Kw = 10^-14 always at 25°C for water so:

Kb = Kw/Ka = (10^-14)/(3*10^-8) = 3.33333*10^-7

Now, proceed to calculate the equilibrium

H2O + A- <-> OH- + HA

Then K equilibrium is given as:

Kb = [HA][OH-]/[A-]

Assume [HA] = [OH-] = x

[A-] = M – x

Substitute in Kb

3.33333*10^-7= [x^2]/[M-x]

recalculate M:

mmol of conjugate = 4 mmol

Total V = V1+V2 = 40+40= 80

[M] = 4mmol/80= 0.05 M

3.33333*10^-7 = [x^2]/[0.05 -x]

x = 1.288*10^-4

[OH-] =1.288*10^-4

Get pOH

pOH = -log(OH-)

pOH = -log (1.288*10^-4) = 3.890

pH = 14-pOH = 14-3.890= 10.11