Question

You are titrating 25.00 mL of a 0.100 M solution of a weak acid with a 0.250 M solution of potassium hydroxide. Assume the pKa of the weak acid is 5.6.

A. What is the pH after adding 3.0 mL of 0.250 M potassium hydroxide?

B. What is the pH at the midpoint of the titration?

C. What is the pH at the equivalence point of the titration?

Answer 1

This is a titration of monoprotic acid so

A) 3 ml KOH

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)

Vt = V1+V2

[A-] = mol A- / Vt = M*V1 / (Vt) = 3*0.25/ (50+3) = 0.01415 M

[HA] = mol HA / Vt = M*V2 /(Vt) = 25*0.1 / (50+3) = 0.047169M

since pKa = 5.6 then

Apply equation; Henderson hasselbaclh

pH = pKa + log(conjugate/acid) = 5.6 + log (0.01415/0.047169) = 5.0770

pH = 5.0770

B)

in the midpoint

pH = pKa + log(conjugate/acid)

since it is the midpoint then

[Conjugate] = [Acid]

this is a very special point since

pH = pKa + log(conjugate/acid)

turn sto

pH = pKa + log(1)

pH =pKa = 5.6

pH = 5.6

C)

In the equivalence point

We will need Kb

Ka*Kb = Kw

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

Ka = 10^-pKa = 10^-5.6

Kb = Kw/Ka = (10^-14)/(10^-5.6) = 3.981*10^-9

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

Volume needed

M1V1 = M2V2

V2 = M1V1/M2 = 25*0.1/0.25 = 10

M final = 25*0.1/(25+10) = 0.07142857

3.981*10^-9 = [x^2]/[0.07142857-x]

x = 1.68*10^-5

[OH-] =1.86x10^-4

Get pOH

pOH = -log(OH-)

pOH = -log ( 1.68*10^-5) = 4.7746

pH = 14-pOH = 14-4.7746= 9.2254

ph = 9.2254