Consider the titration of 35.0 mL of 0.260 M HF with 0.215 MNaOH. Calculate the
ID: 958051 • Letter: C
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.
Explanation / Answer
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