Question

a. Consider the titration of 25.0 mL of 0.065 M HCN with 0.065 M NaOH.

What is the pH before any NaOH is added?

What is the pH at the halfway point of the titration?

What is the pH when 95% of the NaOH required to reach the equivalence point has been added?

What is the pH at the equivalence point?

pH initial----------------------------------

pH at halfway--------------------------

pH when 95% is added--------------------------

pH at the equivalence point---------------------

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)

6.2*10^-10 = x*x/(0.065-x)

This is quadratic equation

x =6.34*10^-6

For pH

pH = -log(H+)

pH =-log(6.34*10^-6)

pH in a = 5.20

b)

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 = 25*0.065 = 1.625 mmol of acid

mmol of conjugate= 1.625

Get pKa

pKa = -log(Ka)

pKa = -log(6.2*10^-10) = 9.21

Apply equation

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

pH = 9.21+ log (1.625/1.625) = 9.21

c)

95% before

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 = 25*0.065 = 1.625 mmol of acid

mmol of conjugate= 0

0.95% --> 1.54375

so

mmol of acid left = 1.625-1.54375 = 0.08125

mmol of conjguate formed = 1.54375

Apply equation

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

pH = 9.21+ log (1.54375/0.08125)= 10.488

d)

finally

equivalence

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)/(6.2*10^-10) = 1.612*10^-5

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

1.612*10^-5 = [x^2]/[M-x]

recalculate M:

mmol of conjugate = 1.625 mmol

Total V = V1+V2 = 25+25 = 50

[M] = 1.62/50 = 0.0325M

1.612*10^-5 = [x^2]/[0.0325-x]

x = 7.13*10^-4

[OH-] =7.13*10^-4

Get pOH

pOH = -log(OH-)

pOH = -log (7.13*10^-4) = 3.1469

pH = 14-pOH = 14-3.1469= 10.8531