Consider the titration of 35.0 mL of 0.260 M HF with 0.215 MNaOH. Calculate the
ID: 958359 • 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
Answer – We are given, 35.0 mL of 0.260 M HF, [NaOH] =0.215 M
Part A) Milliliters of base are required to reach the equivalence point-
We know at equivalence point moles of acid and base are equal
Moles of HF = 0.260 M *0.0350 L
= 0.0091 moles
So, moles of NaOH = 0.0091 moles
So, volume of NaOH = 0.0091 moles / 0.215 M
= 0.0423 L
= 42.3 mL
So, 42.3 milliliters of base are required to reach the equivalence point.
Part B) the pH after the addition of 8.75 mL of base –
Moles of NaOH = 0.215 M * 0.00875 L
= 0.00188 moles
HF + NaOH -----> F- + H2O
0.0091 0.00188 0.00188
New moles -
Moles of HF = 0.0091 moles - 0.00188 moles = 0.00722 moles
Moles of F- = 0.00188 moles
Total volume = 35.0+8.75 mL = 43.75 L
So, [HF] = 0.00722 moles / 0.04375 L = 0.165 M
[F-] = 0.00188 mol / 0.04375 L = 0.043 M
We know the pKa value for the HF and it is 3.14
Now we need to use Henderson Hasselbalch equation
pH = pKa + log [F-] / [HF]
= 3.14 + log 0.043 / 0.165
= 2.55
Part C) the pH at halfway to the equivalence point-
We know at halfway to the equivalence point there is
pH = pKa
so, pH = 3.14
Part D) pH at the equivalence point –
At equivalence point moles of acid and base are equal
So, moles of HF = moles of NaOH = 0.0091 moles
We already calculated volume of of base are required to reach the equivalence point
Volume of NaOH = 42.3 mL
So, after the reaction there is only remaining
Moles of F- = 0.0091 moles
Total volume = 35+42.3 = 77.3 mL
[F-] = 0.0091 moles / 0.0773 L
= 0.118 M
Now we need to calculate Kb
We know,
Kb = 1E-14/Ka
= 1E-14 / 7.2E-4
= 1.39*10-11
Now we need to put ICE chart -
F- + H2O <----> HF + OH-
I 0.118 0 0
C -x +x +x
E 0.118-x +x +x
We know ,
Kb = [HF][OH-] /[F-]
1.39*10-11 = x*x /0.118-x
We can neglect x in the 0.118-x, since Kb value is too small
So, x2 = 1.39*10-11 *0.118 M
x = 1.28*10-6 M
, so, [OH-] = 1.28*10-6 M
pOH = -log [OH-]
= - log 1.28*10-6 M
= 5.89
So, pH = 14 – pOH
= 14-5.89
= 8.11
Part E) pH after the addition of 70.0 mL of base-
Moles of NaOH = 0.215 M * 0.070 L
= 0.0151 moles
So, moles of F- = 0.0151 moles
Total volume = 35+70 = 105 mL
[F-] = 0.0151 moles / 0.105 L
= 0.143 M
Now we need to put ICE chart -
F- + H2O <----> HF + OH-
I 0.143 0 0
C -x +x +x
E 0.143-x +x +x
We know ,
Kb = [HF][OH-] /[F-]
1.39*10-11 = x*x /0.143-x
We can neglect x in the 0.143-x, since Kb value is too small
So, x2 = 1.39*10-11 *0.143 M
x = 1.41*10-6 M
, so, [OH-] = 1.41*10-6 M
pOH = -log [OH-]
= - log 1.41*10-6 M
= 5.85
So, pH = 14 – pOH
= 14-5.85
= 8.15