Question

A 35.00−mL solution of 0.2500 M HF is titrated with a standardized 0.1825 M solution of NaOH at 25°C.

Be sure to answer all parts. A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1825 M solution of NaOH at 25° C. (a) What is the pH of the HF solution before titrant is added? 1.9 (b) How many milliliters of titrant are required to reach the equivalence point? mL (e) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?

Answer 1

The corresponding reaction is:

HClO + KOH ==> H2O + KClO

Since they react in a 1:1 mole ratio,

(a) before addition of any KOH :

We have 0.110 M HClO. For weak acids,

[H3O+] = (Ka x [HA]o)^0.5

And ka of HF = 6.6 x 10^-4
[H3O+] = ((6.6 x 10^-4)(0.25))^0.5 = 0.0128 M

pH = -log[H3O+]

= -log (0.0128)

= 1.89

b)

Using, M1V1 = M2V2

Where, M1, M2 = Molarities of HF and NaOH

V1, V2 = Volumes of HF and NaOH

Putting the respective values, we get

0.25 M x 35 mL = 0.1825 M x V2

50 mL = V2 = Volume of titrant ie NaOH to reach equivalence point

c) At 0.5 mL before equivalence point

at 0.5 mL before equivalence point, ie we have added 50 - 0.5 = 49.5 mL NaOH and at this point we are at the 99% mark of completion of the titration (49.5 / 50 x 100 = 99%) .

So the ratio of [NaF] to [HF] is 99 / 1.

Since we have a weak acid (HClO) and its conjugate base (ClO- in the form of NaClO) in the same solution, which constitutes a buffer system. So, using the Henderson-Hasselbalch equation for buffers,

pH = pKa + log ([conj. base] / [weak acid])

pH = pKa + log ([NaF] / [HF])

Where, pKa = -log Ka = -log (6.6x 10^-4) = 3.18

So, putting the value of pKa in above equation,

pH = 3.18 + log (99 /1) = 3.18 + 1.99 = 5.17

d) at equivalence point

At this point, you have reacted all of the HF and converted it to NaF

Initial mmoles HF = Molarity HClO x mL HClO

= (0.25)(35 ) = 8.75 mmoles HClO = mmol of NaF

[NaF] at equivalence point = mmoles NaF / mL of solution = 8.75 / (35 + 50)

= 0.102 M

Since NaF is made using weak acid and strong base, the resulting NaF solution will be basic (i.e.F- ion will form HF and OH- in aqueous solution)

For solution of weak bases,

[OH-] = (Kb x [B]o)^0.5

Where, Kb = Kw / Ka

= (1 x 10^-14) / (6.6x 10^-4)

= 1.51 x 10^-11

Putting this value in above equation :

[OH-] = ((1.51x 10^-11)(0.102 M))^0.5

= 1.24 x 10^-6

pOH = -log [OH-]

= -log (1.24 x 10^-6)

= 5.91

pH = 14.00 - pOH

= 14.00 - 5.91

= 8.09

e) after addition of 0.5 mL of NaOH after equivalence point :

Mol of OH- in extra 0.5 mL = 5x 10^-4 L X 0.1825 mol/L

= 9.12 x 10^-5mol OH- to the solution. So,

[OH-] = 9.12 x 10^-5 mol / 0.0855 L = 1.06 x 10^-3 M
pOH = - log [OH-]

pOH = - log (1.06 x 10^-3)

pOH = 2.97
pH = 14.00 - 2.97

pH = 11.03