Question

A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1408 M solution of NaOH at 25C.

a.What is the PH of the HF solution before titrant is added?

b.How many milliliters of titrant are required to reach the equivalence point?

c.What is the PH at 0.5ml before the equivalence point?

d.What is the PH at the equivalence point? e.What is the PH at 0.5 ml after 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)

7.2*10^-4 = x*x/(0.25-x)

This is quadratic equation

x =0.01306

For pH

pH = -log(H+)

pH =-log(0.01306)

pH in a = 1.88

b)

V required for equivalence

mmol of acid = mmol of base

35*0.25 = 0.1408*Vbase

Vbase = 35*0.25 /0.1408 = 62.1448 mL

c) V 0.5 mL before

V = 62.1448-0.5 = 61.6448 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 = 35*0.25 = 8.75 mmol of acid

mmol of base = MV = 61.6448 *0.1408 = 8.679 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 8.75 -8.679= 0.071 mmol

mmol of conjguate left = 0 + 8.679 = 8.679

Get pKa

pKa = -log(Ka)

pKa = -log(7.2*10^-4) = 3.1426

Apply equation

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

pH = 3.1426 + log (8.679/0.071 ) = 5.2298

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)/(7.2*10^-4) = 1.388*10^-11

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

1.388*10^-11 = [x^2]/[M-x]

recalculate M:

mol of conjugate/ vtotal = 0.35*0.25 / (35+62) = 0.00090 M

1.388*10^-11 = [x^2]/[0.00090  -x]

x = 1.47*10^-8

[OH-] =1.47*10^-8

Get pOH

pOH = -log(OH-)

pOH = -log 1.47*10^-8) = 7.832

pH = 14-pOH = 14-7.832= 6.168

e) Addition of base

There will be finally an Excess of Base!

mol of acid < mol of base

Calculate pOH directly

[OH-] = M*V / Vt

mmol of acid = MV = 35*0.25 = 8.75

mmol of base = MV =( 62.1448+0.5) * 0.1408 = 8.820

mmol of base left = 8.820 - 8.75 = 0.07 mmol

pOH = -log(0.07 / (62.14+0.5+35)) = 3.14452

pH = 14-3.14452 = 10.85548