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

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