Question

Consider the titration of 35.0 mL of 0.260 M HF with 0.215 MNaOH. Calculate the pH at each of the following points.

Part A.)

How many milliliters of base are required to reach the equivalence point?

Express your answer using three significant figures.

Part B.)

Calculate the pH after the addition of 8.75 mL of base

Express your answer using two decimal places.

Part C.)

Calculate the pH at halfway to the equivalence point.

Express your answer using two decimal places.

Part D.)

Calculate the pH at the equivalence point.

Express your answer using two decimal places.

Part E.)

Calculate the pH after the addition of 70.0 mL of base.

Express your answer using two decimal places.

Answer 1

Part A) :

Reaction of HF and NaOH

HF+ NaOH ----------> NaF + H2O

Moles of HF:

0.035 L* 0.260 M= 9.1*10^-3 mol HF

Moles of NaOH = 9.1*10^-3 mol because 1 mole HF reacts with 1 mol NaOH.

Volume = moles / molarity

= 9.1*10^-3 mol / 0.215 mol/L

= 0.0423 L or 42.3 mL

Part B.)

Calculate the pH after the addition of 8.75 mL of base

Express your answer using two decimal places.

HF(aq) --> H+(aq) + F-(aq) Ka = 7.08 * 10^-4

you add 8.75 ml 0.215 M NaOH

8.75 ml * 0.215 moles NaOH/1000ml = 0.00188 moles OH-

[HF] initial = Moles of HF:

0.035 L* 0.260 M= 9.1*10^-3 mol HF

by adding the base you deplete the HF but you add F- so you are now building a buffer

mole HF left = 9.1*10^-3 mol - 0.00188 moles = 0.0072 moles HF
moles F- generated = 0.0018

final volume = 35 ml + 8.75 ml = 43.75 ml = 0.04375 L

[HF] final = 0.0072 moles HF/0.04375 L = 0.165 M
[F-] final = 0.00188 moles F-/0.04375 L = 0.043 M

pH = pKa + log[F-]/[HF] = 3.15 + log(0.043/0.165)

= 3.15 -0.58

= 2.57

Part C.)

Calculate the pH at halfway to the equivalence point.

Express your answer using two decimal places.

at the half equivalent point [acid]= [salt]

The Handerson-Hasselbalch equation :

pH = pKa + log [salt]/ [acid]

= 3.15 + log 1

= 3.15

Part D.)

Calculate the pH at the equivalence point.

Express your answer using two decimal places.

HF+ NaOH = NaF + H2O

Molarity of NaF solution = 9.1*10^-3 mol/ 0.0773 L (total volume)

=0.112 M
Calculate pH as follows:
The aim is to find the [OH-] of the solution

The NaF dissociates in water :
NaF ↔ F- + Na+

The F- reacts with water:
F- + H2O ↔ HF + OH-

Ka for HF= 7.08 * 10^-4
Kw = Ka * Kb - we want Kb
Kb = Kw* Ka
Kb = 1.0^-14 / (7.08 * 10^-4)
Kb = 1.41*10^-11

Kb = [F-] *[[OH] / [NaF]
We know that [F-] = [OH-] so product is [OH-]²
[NaF] = 0.112 M

Kb = [OH]²/0.112

(1.41*10^-11 ) * 0.112 = [OH[²
[OH]² = 1.58*10^-12
[OH-] = 1.26*10^-6

pOH = -log[OH-]

pOH = -log ( 1.26*10^-6)
pOH = 5.90

pH + pOH=14

pH = 14-5.90

= 8.10